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Best Anti-Spam Solution: ASSP
This is a discussion on the "ASSP" entry in the Best Anti-Spam Solution SpiceList
7 Replies
Nov 29, 2008 at 6:25 UTC
Anyone willing to share your assp.cfg file with me? I promise you it'll be completely secure - I just want to see the settings that are making a successful ASSP installation - successful.
It seems like changing the most innocuous-sounding settings, like 'tagging' or 'monitoring' are making the most profound difference in how ASSP performs.
Maybe I'm changing too many at one time - but when the president is coming to you bit**ing about being bombarded with spam, well, it just makes you just want to kinda skip ahead to the end, you know?
And I would LOVE to see a screenshot of your statistics page so I can use that to compare the effectiveness (reality-check) of what I'm seeing.
Thanks so much!
Mike Rodgers
Triangle United Way
Morrisville, NC
Nov 30, 2008 at 3:41 UTC
Plus Technologies is an IT service provider.
ASSP - One of the best programs I have come across - been using it for a number of years, although not using it to its full potential...
Dec 1, 2008 at 11:41 UTC
Those are very impressive stats! And as you can see from mine, it's not coming close to yours. But, it's better than it was week before last, by more than double!
I changed these two: useHeloBlacklist changed from '3' to '1' and DoBayesian changed from '0' to '3' and the block rate immediately went from the upper 30's up to high 80's or so at one point. And that just makes me wonder what other settings would improve it even more.
Anyway, thanks for posting the stats, it gives me something to shoot for.
Dec 1, 2008 at 2:35 UTC
We tried several on-site installations & were VERY dis-satisfied. We get over 40,000 messages a day & the volume was killing our bandwidth. So, we found that the absolute best option was a hosted service called MXLogic. We have eliminated 95% of our SPAM (over 20k a day) & all virus emails. It has seriously lessened the load on our servers & people have been so impressed with it. If it catches a false-positive, people can automatically release it from a daily email.
It also acts as a good back up if our servers or internet goes down (it queues it all up).
GREAT product with ZERO administration.
Dec 2, 2008 at 11:57 UTC
I have used 3 hosted Anti-Spam companies in the past year and am now satisified with GMSI (Global Messaging Solution). The biggest issue for me, beyond the usual spam problem, was "lost" emails - this company went way above and beyond to not help, but to take care of issues with other companies who were having issues sending us emails. We also get daily emails allowing us to release from quarantine. The management console is nice - allowing me to block/whitelist, etc...
Dec 2, 2008 at 5:23 UTC
I am using a double tripple spam solution and it works GREAT.
FREE Open Source Untangle at Firewall
Then NoSpamToday running on Exchange box with SP2 IMF...
I can manage the whole thing easily.. check it out.. and best of all it's LOCAL with NO Montly Payments.. We have 250 users
Dec 4, 2008 at 4:06 UTC
Barracuda Anti-Spam firewall. I use the 200 model, but if you get a higher model your users can manage their own Quarantine folders. But guess what? In 2 1/2 years I have found only a handful of good emails in quarantine (all were mailing lists that changed their sending addresses.
This stops nearly all spam prior to hitting my Exchange server, so no extra load on my systems. What isn't purely blocked or quarantined is delivered marked as SPAM. These can be easily re-classified in the message log on Barracuda.
They use two different anti-spam engines and all is updated hourly. Very nice.
This discussion has been inactive for over a year.
You may get a better answer to your question by starting a new discussion.

subset: commoncrawl


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Mohali (Punjab), Sep 16: Still trying to find its foothold in the cricketing landscape, the Champions League Twenty20 will enter its fifth season here tomorrow with the qualifying matches during which Pakistan's domestic champions Faisalabad Wolves are likely to hog the limelight.
A league which was born out of the hugely successful IPL, CLT20 is gradually establishing its identity but it would take some time for it to truly find a connect with the fans.
The event will kick off with Faisalabad Wolves, led by Pakistan's Test skipper Misbah-ul-Haq, taking on New Zealand's Otago Volts in the first qualifier tomorrow.
This would be followed by a star-studded Sunrisers Hyderabad, led by Delhi-dasher Shikhar Dhawan, squaring off against Sri Lanka's Kandurata Maroons in the second qualifier at the PCA stadium. Barring Volts, the other three teams in the Qualifiers come from the sub-continent, but none among the four feels that either has any big advantage in the event.
The top two teams will advance to the main draw, which has three Indian teams in fray -- IPL champions Mumbai Indians, runners-up Chennai Super Kings and third-place finishers Rajasthan Royals.
In tomorrow's opening match, the Wolves would mainly rely on their young players including promising medium-pacer Ehsan Adil, Ali Waqas, Khuram Shahzad, Asad Ali, besides banking on the experience of skipper Misbah and Saeed Ajmal, who has been a national asset in all three formats after his Test debut four years back.
The Wolves, who had some visa issues when they landed here, will draw strength from the fact that at the domestic level, they beat teams that had many stars. The Wolves are only the second team from Pakistan to compete in the event after Sialkot Stallions last year.
Tuesday's Matches
4PM IST: Faisalabad Wolves Vs Otago Volts
8PM IST: Sunrisers Hyderabad Vs Kandurata Maroons


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The wonders of alcoholic beverage.
The achilles heel for various society. The 'norm' for many others. And poison if you have CFS/PVFS/ME ...
I retrieve once I original got Post Viral Fatigue Syndrome/CFS final in 1995, I idea that having a helping or two wouldn't affect me any otherwise to my PRE-CFS years. Unfortunately I didn't agnise what I was doing to myself.
I didn't know how severely drug of abuse can affect the rigidity of this CFS ...
Nowadays I don't touch alcoholic beverage at all, and haven't through with so for a few old age now. I'd to some extent see at tiniest a few hours in the day alternatively of being bedridden for weeks on end because of one wet drink!
Because once you have CFS/M.E., you may possibly as healed be running arsenic descending your oesophagus...
Okay - so that examination mightiness be a slender 'dramatic', but you get what I close-fisted right?
You see once you have CFS/M.E., you are apparent to come along **alcohol intolerance**...
The least wet get drunk - even a minuscule bit,
can send away you into a reversion - as I learnt the rocky way many another geezerhood ago.
Having an intoxicating party now and over again was decent to tip the roughness of my PVFS/CFS done the rim and put me in bed indefinitely.
When a **normal** causal agency gets besotted they may get the impression a bit bad for a day or so, after which, they're hindermost to 'normal', bouncing back and raring to go. But once a character next to M.E./CFS gets half-seas-over - or even has meet one drink, they're promising to grain approaching 'death warmed up' for what seems close to an eternity!!!
Personally, I don't certainly suchlike potable much so I don't decline it. But even if I did,
I have complete that beverage no longer makes me touch the way it utilised to up to that time I had CFS...
Drinking beverage now feels suchlike I'm feeding myself poison, and my article reacts as expected - i.e. a flare, or worse, a reversion - and I experience the results for a long, long case in a while.
Unfortunately, most of us CFS sufferers are honorable not rugged satisfactory to escape the poisons of potable.
So for utmost of us, it's sadly a satchel of acceptive it, or feat much, so much worse.
As Dr Shepherd writes in his work of fiction 'Living With M.E.':
"Some group who in the past enjoyed and tolerated equal activity of alcohol short any untoward effects, now discovery that even teeny amounts generate them exceedingly off colour."
- p214, Dr Shepherd, 'Living With M.E.' -
So if you have M.E./CFS, it is probable that you may have developed an intolerance to potable.
And that's not forgetting that drug of abuse as well affects the personal effects of antidepressants (often unarbitrary to CFS sufferers to kickshaw their weariness and to aid sufferers take a nap)! So if you're attractive antidepressants, it's decidedly thing to accept in awareness.
Having CFS/ME can be a incredibly lonely and grave undertake and reduction can be a very actual and capital evidence for a number of CFS sufferers. So the closing state of affairs you need is to pilfer substances that sort you grain worsened.
And assume what?
Yep, you've gather it, inebriant is besides a ataractic. So it's not a tremendous situation to drink if you're depressed!
And according to Dr Shepherd, galore sufferers mournfully do crook to drinkable...
But drug of abuse is no statement. Apart from belike fashioning you grain depressed, you could as well change an drinkable dependency!
What's more, if you **are** street drug intolerant, then it would formulate it by a long way harder for you to rest from M.E./CFS (pretty much impractical even), time you hold on to swing alcoholic beverage in your article.
And here's both more 'food for thought'...
When you infer roughly how disappointingly a CFS sufferer can be false by **one** alcoholic drink, consequently how severely could drug of abuse be affecting you if you're drinking more???
So at hand you go - a unbroken document of reasons why you may perhaps contemplate doubly going on for ingestion thing remotely spirituous time you have CFS/M.E.
While you may no longest be inebriant bigoted past you have well from CFS/M.E., it may capably be a starring culprit for preventing you from **recovering** from your CFS/M.E time you stagnant have it.
I didn't cognise roughly speaking the viable personal estate of drinkable once I prototypic had Post Viral Fatigue Syndrome ...
... and I suffered the knock-on effect. My disorder got worse. A lot worse. And I was lone having a small indefinite quantity of drinks now and again! So I confidence this article armaments you with satisfactory information for you to brand name an schooled judgment about drinkable any way.
You never cognize - it may well engineer the unlikeness...
To discovery out more than just about diet and biological process for ME/CFS visit:
And to read reviews of my favourite supplements & items for ME/CFS, visit:
Copyright, Claire Williams, 2003-2005. All Rights Reserved.
** Publication Guidelines **
The article above may be gladly reproduced provided that:
(1) you with the sole purpose correspondence to a 100% opt-in list; (2) ALL golf links are
LIVE hyperlinks (3) the nonfictional prose is published in its entirety
including the title, government grant notice, & the author's
bio & assets box (which essential be situated straight below
the nonfictional prose).


text:
Demandware shares climb as revenue rises 27 percent
Burlington-based Demandware saw its shares climb more than 6 percent today after the e-commerce software technology maker reported first quarter total revenue of $20.5 million, a 27 percent increase from the same period a year ago.
The company also saw a 42 percent increase in live customers from last year to 156 at the end of March, and a 57 percent spike in live sites to 630.
During the quarter, the company signed several new customers, including Cole Haan, Moet Hennessy Asia Pacific, Rocky Brands, Scotch & Soda, SkyMall and Vilebrequin, officials said.
|The article you requested has been archived


text:
In 1868 during the late Qing Dynasty, rampant corruption on the Imperial Court inflicts much suffering in people's lives. For years, the Black Tiger's fearsome boss Lei Gong has been trying to get rid of the leader of the Northern Sea. One of his latest recruits is Fei, a fearless fighter who takes the Northern Sea leader's head after a fierce fight. Just as Lei Gong believes he has total control of the port, a new gang called the Orphans rises in power. Led by Fei's childhood friend Huo, the Orphans are out to eliminate all the criminal power from the port…
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This air-conditioned business hotel has 133 rooms. Guests are welcomed in the lobby, which offers a 24-hour reception and a 24-hour check-out service, as well as a hotel safe, cloakroom, a complimentary newspaper stand and lift access to the upper floors. Guests can dine in the breakfast room and business travellers can make use of the conference facilities and small meeting rooms. High-speed wired and wireless Internet access is complimentary. The pet-friendly business hotel also offers room service and dry cleaning/laundry services. There is a car park for those arriving by car.
This business hotel is situated in Fort Worth's North Fort Worth neighbourhood, about 8 km from The Golf Club at Fossil Creek. Other area attractions include Billy Bob's Texas and Fort Worth Stockyards.
All rooms are en suite and feature a shower, bath and hairdryer, as well as a double or king-size bed, a direct dial telephone, a satellite/cable flat-screen TV, a clock radio and complimentary wired and wireless high-speed Internet access. Additional amenities include tea and coffee making facilities, an ironing set and individually regulated air conditioning/heating.
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Previous: Take Him Out! (CYOAH)
It was decided to send troops to Iceland, the last Bastion of Freedom in Europe. However, leftists in the US Government in employ to the Soviet Union betray these plans to the Russians, and Stalin orders the island be taken, no matter the cost. It is now a race to figure out who reaches Iceland first, and if they can hold it from their opponent.
The Soviets arrive first, but only a few ships, roughly 4500 soldiers, on March 17, 1939. The Americans arrive two days later, with over 12,000 Marines and manage to land on the west coast, and push the Soviets out of Reykjavik, and soon the Battle of Iceland begins.
The Americans push farther, until the Soviets send reinforcements and stop the US Marines outside the small town of Akureyri, and the march to the West begins, until US soldiers arrive on April 2, and hold the town of Selfoss.
The two armies, fighting in rocky and treacherous conditions, are exhausted after four weeks. the question, however, is who will reach Iceland first, the US, or the Soviets?
Tie! They both get their at the same time
created by Tbguy1992 06:45, September 6, 2010 (UTC)


text:
Photos by Ron Edelstein/NYC. Twitter: uniquelyron Facebook: https://www.facebook.com/ron.edelstein
Item: LV-Lisbeth
Sultry Gypsy Look. Big, Curly Drag Queen Wig. Teased layers. Fluffy & wavy.. 4" high from the top to the crown. 22" in total from top to bottom.
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Price: $ 198.99 plus shipping & handling


text:
Butterick 6805; ca. 1954; Soft One Piece Casual: Raglan Sleeves: An easy-to-wear casual with smart raglan sleeves and a smooth rising neckline. Dress buttons to below the waist. Uncuffed sleeves are short (A), below-elbow length (B). The four-gore skirt is gathered on either side.
Links to reviews/blog postsEdit
Add a photo to the gallery by clicking the "modify" button below.
[insert your username, and make sure your preferences allow for people to contact you via e-mail. Do not list your email address here!]


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The Primrose winter storage bag keeps your awning in pristine condition when ever it is not being used, either during the winter months or for any long period of time, significantly extending the life of your awning.
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text:
Positive Integers n Such That σ(φ(n)) = σ(n)
- Oswin Lamb
- 2 years ago
- Views:
1 Journal of Integer Sequences, Vol. (2008), Article Positive Integers n Such That σ(φ(n)) = σ(n) Jean-Marie De Koninck Départment de Mathématiques Université Laval Québec, PQ GK 7P4 Canada Florian Luca Instituto de Matemáticas Universidad Nacional Autónoma de Méico C.P Morelia, Michoacán Méico Abstract In this paper, we investigate those positive integers n for which the equality σ(φ(n)) = σ(n) holds, where σ is the sum of the divisors function and φ is the Euler function. Introduction For a positive integer n we write σ(n) and φ(n) for the sum of divisors function and for the Euler function of n, respectively. In this note, we study those positive integers n such that σ(φ(n)) = σ(n) holds. This is sequence A03363 in Sloane s Online Encylopedia of Integer Sequences. Let A be the set of all such positive integers n and for a positive real number we put A() = A [,]. Our result is the following.
2 Theorem. The estimate holds for all real numbers >. ( ) #A() = O (log ) 2 The above upper bound might actually be the correct order of magnitude of #A(). Indeed, note that if m is such that σ(2φ(m)) = 2σ(m) () and if q > m is a Sophie Germain prime, that is a prime number q such that p = 2q + is also prime, then n = mp A. The numbers m = 238, 2806, 5734, 5937, 798, 8097,... all satisfy relation (), and form Sloane s sequence A More generally, if k and m are positive integers such that σ(2 k φ(m)) = 2 k σ(m) (2) and q <... < q k are primes with p i = 2q i + also primes for i =,...,k and q > m, then n = p...p k m A. Now recall that the Prime K-tuplets Conjecture of Dickson (see, for instance, [2, 4, 8]) asserts that, ecept in cases ruled out by obvious congruence conditions, K linear forms a i n+b i, i =,...,K, take prime values simultaneously for about c/(log ) K integers n, where c is a positive constant which depends only on the given linear forms. Under this conjecture (applied with K = 2 and the linear forms n and 2n+), we obtain that there should be /(log ) 2 Sophie Germain primes q, which suggests that #A() /(log ) 2. We will come back to the Sophie Germain primes later. Throughout, we use the Vinogradov symbols and and the Landau symbols O and o with their regular meanings. We use log for the natural logarithm and p, q and r with or without subscripts for prime numbers. 2 Preliminary Results In this section, we point out a subset B() of all the positive integers n of cardinality O(/(log ) 2 ). For the proof of Theorem we will work only with the positive integers n A()\B(). Further, 0 is a sufficiently large positive real number, where the meaning of sufficiently large may change from a line to the net. We put ( ) log y = ep. log log For a positive integer n we write P(n) for the largest prime factor of n. It is well known that Ψ(,y) = #{n P(n) y} = ep( ( + o())u log u) (u ), (3) where u = log / log y, provided that u y /2 (see [], Corollary.3 of [6], or Chapter III.5 of [9]). In our case, u = log log, so, in particular, the condition u y /2 is satisfied for > 0. We deduce that u log u = (log log )(log log log ). 2
3 Thus, if we set B () = {n P(n) y}, then We now let #B () = Ψ(,y) = ep ( ( + o())(log log )(log log log )) < for > (log ) 0 0. (4) z = (log ) 26 and for a positive integer n we write ρ(n) for its largest squarefull divisor. Recall that a positive integer m is squarefull if p 2 m whenever p is a prime factor of m. It is well known that if we write S(t) = {m t m is squarefull}, then #S(t) = ζ(3/2) ζ(3) t/2 + O(t /3 ), where ζ is the Riemann Zeta-Function (see, for eample, Theorem 4.4 in [7]). By partial summation, we easily get that m t/2. (5) m t m squarefull We now let B 2 () be the set of positive integers n such that one of the following conditions holds: (i) ρ(n) z, (ii) p n for some prime p such that ρ(p ± ) z, (iii) there eist primes r and p such that p n, p ± (mod r) and ρ(r ± ) z. We will find an upper bound for #B 2 (). Let B 2, () be the set of those n B 2 () for which (i) holds. We note that for every n B 2, () there eists a squarefull positive integer d z such that d n. For a fied d, the number of such n does not eceed /d. Hence, #B 2, () d z d squarefull d (log ) 3, (6) where we have used estimate (5) with t = z. Now let B 2,2 () be the set of those n B 2 () for which (ii) holds. We note that each n B 2,2 () has a prime divisor p such that p ± (mod d), where d is as above. Given d and p, the number of such n does not eceed /p. Summing up over all choices of p and d we get that #B 2,2 () d z d squarefull p ± (mod d) p (log log ) 2 d z d squarefull 3 p d d z d squarefull log log φ(d) (log log )2 (log ) 3, (7)
4 where in the above estimates we used aside from estimate (5), the fact that the estimate ( ) p log log p (a,b) + O (8) φ(b) p a (mod b) p holds uniformly in a, b and when b, where a and b are coprime and p (a,b) is the smallest prime number p a (mod b) (note that p (,b) b+ and p (,b) = p (b,b) b φ(b) for all b 2), together with the well known minimal order φ(n)/n / log log, valid for n in the interval [,]. Let B 2,3 () be the set of those n B 2 () for which (iii) holds. Then there eists r such that r ± (mod d) for some d as above, as well as p n such that r p or r p +. Given d, r and p, the number of such n does not eceed /p, and now summing up over all choices of d, r and p, we get that #B 2,3 () d z r ± (mod d) d squarefull r d z r ± (mod d) d squarefull r (log log ) 2 d z d squarefull (log log ) 3 d z d squarefull (log log ) 4 d z d squarefull p ± (mod r) p log log φ(r) r ± (mod d) r φ(d) d r p (log log )4 (log ) 3, (9) where in the above estimates we used again estimate (5), estimate (8) twice as well as the minimal order of the Euler function on the interval [,]. Hence, using estimates (6) (9), we get #B 2 () #B 2, () + #B 2,2 () + #B 2,3 () < (log ) 0 for > 0. (0) We now put and set w = 0 log log S(w,) = ω(m) w m m, 4
5 where ω(m) denotes the number of distinct prime factors of the positive integer m. Note that, using the fact that p = log log t + O() and Stirling s formula k! = (+o())kk e k 2πk, p t we have S(w,) = k w ω(m)=k m m = k w k! ( p α ) k p α = ( ( )) k k! p + O p 2 k w p p 2 k w ( ) k e log log + O() ( e log log + O() k w k w k w ( ) w e log log + O() w for > 0 because 0 log(0/e) >. (log log + O()))k k! ) k (log ) 0 log(0/e) < (log ) () We now let B 3 () be the set of positive integers n such that one of the following conditions holds: (i) ω(n) w, (ii) p n for some prime p for which ω(p ± ) w, (iii) there eist primes r and p such that p n, p ± (mod r) and ω(r ± ) w. Let B 3, (), B 3,2 () and B 3,3 () be the sets of n B 3 () for which (i), (ii) and (iii) hold, respectively. To bound the cardinality of B 3, (), note that, using (), we have #B 3, () = ω(n) w n ω(n) w n n = S(w,) < (log ) (2) for > 0. To bound the cardinality of B 3,2 (), note that each n B 3,2 () admits a prime divisor p such that ω(p ±) w. Fiing such a p, the number of such n does not eceed /p. Summing up over all such p we have, again in light of (), #B 3,2 () ω(p±) w p p ω(p+) w p+ + 2 p + + ω(p ) w p p (2S(w, + ) + S(w,)) < 3S(w,) + 2 (log ) (3) 5
6 for > 0. To bound the cardinality of B 3,3 (), note that for each n B 3,3 () there eist some prime r with ω(r ± ) w and some prime p n such that p ± (mod r). Given such r and p, the number of such n does not eceed /p. Summing up over all choices of r and p given above we get, again using (), for > 0. #B 3,3 () ω(r±) w r = (log log ) (log log ) p ± (mod r) p ω(r±) w r p ω(r ) w r r = (log log ) r + ω(r+) w r+ (log log )(S(w,) + 3S(w, + )) 4(log log )S(w,) + O(log log ) Hence, using estimates (2) to (4), we get We now let #B 3 () #B 3, () + #B 3,2 () + #B 3,3 () < ω(r±) w r 3 r + φ(r) (log log ) (log ) (4) (log ) 0 for > 0. (5) B 4 () = {n n (B () B 2 ()) and p 2 φ(n) for some p > z}. Let n B 4 () and let p 2 φ(n) for some prime p. Then it is not possible that p 2 n (because n B 2 ()), nor is it possible that p 2 q for some prime factor q of n (again because n B 2 ()). Thus, there must eist distinct primes q and r dividing n such that q (mod p) and r (mod p). Fiing such p, q and r, the number of acceptable values of such n does not eceed /(qr). Summing up over all the possible values of p, q and r we arrive at #B 4 () z p z p q (mod p) r (mod p) q<r, qr (log log )2 (log ) 3, qr z p 2 (log log ) 2 (p ) 2 (log log ) 2 6 q (mod p) q z p p 2 q 2
7 where we used estimates (8) and (5). Hence, We now let #B 4 () < (log ) 0 for all > 0. (6) τ = ep ( (log log ) 2) and let B 5 () stand for the set of n which are multiples of a prime p for which either p or p + has a divisor d > τ with P(d) < z. Fi such a pair d and p. Then the number of n divisible by p is at most /p. This shows that #B 5 () P(d)<z τ<d P(d)<z τ<d p ± (mod d) p p log log P(d)<z τ<d d. (7) It follows easily by partial summation from the estimates (3) for Ψ(,v), that if we write v = log τ/ log z, then S = d log ep(( + o())v log v). Since v = (log log )/26, we get that and therefore that v log v = (/26 + o())(log log )(log log log ) S for all > 0, which together with estimate (7) gives Thus, setting log ep(( + o())v log v) < (8) (log ) #B 5 () < (log ) 0 for > 0. (9) B() = we get, from estimates (4), (0), (5), (6) and (9) that 5 B i (), (20) i= #B() 5 i= #B i () (log ) 0 for all > 0. (2) 3 The Proof of Theorem We find it convenient to prove a stronger theorem. 7
8 Theorem 2. Let a and b be any fied positive integers. Setting A a,b = {n σ(aφ(n)) = bσ(n)}, then the estimate holds for all 3. #A a,b () a,b (log ) 2 Proof. Let be large and let B() be as in (20). We assume that n is a positive integer not in B(). We let A () be the set of n A a,b ()\B() for which (P(n) )/2 is not prime but such that there eists a prime number r > z and another prime number q P(n) for which r gcd(p(n) +,q + ). To count the number of such positive integers n, let r and q be fied primes such that r q +, and let P be a prime such that q P and r P +. The number of positive integers n such that P(n) = P does not eceed /P. Note that the congruences P (mod r) and P (mod q) are equivalent to P a q,r (mod qr), where a q,r is the smallest positive integer m satisfying m (mod r) and m (mod q). We distinguish two instances: Case : qr < P. Let A () be the set of such integers n A (). Then #A () z<r q (mod r) q log log z<r log log z<r P a q,r (mod qr) qr<p q (mod r) q r (log log ) 2 z<r (log log ) 2 z<r q (mod r) q rφ(r) where in the above inequalities we used estimates (8) and (5). Case 2: qr P. P φ(qr) q (log log )2, (22) r2 (log ) Let A () be the set of such integers n A (). Here we write n = Pm. Note that P > P(m) because y > z for large and n B () B 2 (). Furthermore, since r q +, we may write q = rl. Since q P, we may write P = sq + = s(rl )+ = srl+ s. Since r P +, we get that s (mod r), and therefore that s 2 (mod r). Hence, there eists a nonnegative integer λ such that s = λr + 2. If λ = 0, then s = 2 leading to 8
9 P = 2q +, which is impossible. Thus, λ > 0. Let us fi the value of λ as well as that of r. Then rl = q and (λr 2 + 2r)l (λr + ) = P (23) are two linear forms in the variable l which are simultaneously primes. Note that P /m and since P lr(λr + 2), we get that l /(mr(λr + 2)). In particular, mr(λr + 2). Recall that a typical consequence of Brun s sieve (see for eample Theorem 2.3 in [3]), is that if L (m) = Am + B and L 2 (m) = Cm + D are linear forms in m with integer coefficients such that AD BC 0 and if we write E for the product of all primes p dividing ABCD(AD BC), then the number of positive integers m y such that L (m) and L 2 (m) are simultaneously primes is Ky ( E ) 2 (log y) 2 φ(e) for some absolute constant K. Applying this result for our linear forms in l shown at (23) for which A = r, B =, C = λr 2 + 2r and D = (λr + ), we get that the number of acceptable values for l does not eceed K mr(λr + 2) (log(/(mr(λr + 2))) 2 (log log )2 mr(λr + 2), ( ) 2 (λr + 2)(λr + )r φ((λr + 2)(λr + )r) where for the rightmost inequality we used again the minimal order of the Euler function on the interval [,]. Here, K is some absolute constant. Summing up over all possible values of λ, r and m, we get #A () (log log ) 2 mr(λr + 2) z<r λ m ( ) ( < (log log ) 2 r 2 where we used again estimate (5). z<r (log log )2 (log ) 2 (log ) 3 = From estimates (22) and (24), we get #A () #A () + A () < λ ) ( λ m ) m (log log )2 (log ), (24) (log ) 0 for all > 0. (25) Now let A 2 () be the set of those n A a,b ()\(A () B()) and such that (P(n) )/2 is not prime. With n A 2 (), we get that n = Pm, where P > P(m) for > 0 because 9
10 y > z for large and n B () B 2 (). Then bσ(n) = bσ(m)(p + ). Let d be the largest divisor of P + such that P(d ) z. Then P + = d l, and bσ(n) = bσ(m)d l. Furthermore, φ(n) = φ(m)(p ), so that aφ(n) = aφ(m)(p ). Let d be the largest divisor of P which is z-smooth; that is, with P(d) z. Then P = dl. Note that l is squarefree (because n B 2 ()) and l and φ(m) are coprime (for if not, there would eist a prime r > z such that r l and r φ(m), so that r 2 φ(n), which is impossible because n B 4 ()). Since z > a for > 0, we get that aφ(m)d and l are coprime, and therefore that σ(φ(n)) = σ(aφ(m)d)σ(l). We thus get the equation σ(aφ(m)d)σ(l) = bσ(m)d l. Now note that l and σ(l) are coprime. Indeed, if not, since l is squarefree, there would eist a prime factor r of l (necessarily eceeding z) dividing q + for some prime factor q of l, that is of P. But this is impossible because n A (). Thus, l σ(aφ(m)d). Note now that Pm, so that m /P /y. Furthermore, ma{d,d } τ because n B 5 (). Let us now fi m, d and d. Then l σ(aφ(m)d), and therefore the number of choices for l does not eceed τ(σ(aφ(m)d)), where τ(k) is the number of divisors of the positive integer k. The above argument shows that if we write M for the set of such acceptable values for m, then #A 2 () τ2 y ma{τ(σ(aφ(m)d)) m M, d τ, d τ}. (26) To get an upper bound on τ(σ(aφ(m)d)), we write aφ(m)d as aφ(m)d = AB, where A is squarefull, B is squarefree, and A and B are coprime. Clearly, so that Since B is squarefree, it is clear that σ(b) Furthermore, σ(aφ(m)d) = σ(ab) = σ(a)σ(b), τ(σ(aφ(m)d)) τ(σ(a))τ(σ(b)). q aφ(m)d (q + ). ω(aφ(m)d) ω(aφ(n)) ω(a) + ω(n) + p n ω(p ) w 2 + w + O(), (27) so that we have ω(aφ(m)d) 2w 2 if > 0. The above inequalities follow from the fact that n B 3 (). Also, for each one of the at most 2w 2 prime factors q of aφ(m)d, the number q+ has at most w prime factors itself and its squarefull part does not eceed z, again because 0
11 n B 3 (). In conclusion, σ(b) C, where C is a number with at most 2w 3 prime factors whose squarefull part does not eceed z 2w2. Thus, where ω(c) 2w 3 and ρ(c) z 2w2. Clearly, and since also we finally get that τ(σ(b)) τ(c) 2 ω(c) τ(ρ(c)), τ(ρ(c)) ρ(c) z w2 = ep(w 2 log z) = ep(o(log log ) 3 ), 2 ω(c) 2 2w3 = ep(o(log log ) 3 ), τ(σ(b)) = ep(o(log log ) 3 ). (28) We now deal with σ(a). We first note that P(A) z for > 0. Indeed, if q > z divides A, then q 2 aφ(n). If > 0, then z > a, in which case the above divisibility relation forces q 2 φ(n) which is not possible because n B 3 (). By looking at the multiplicities of the prime factors appearing in A, we easily see that A ρ(a)ρ(d) ρ(p ) p m q aφ(m)d q z q ω(aφ(m)d) As we have seen at estimate (27), ω(aφ(m)d) 2w 2, in which case the above relation shows that A ρ(a)z ω(aφ(m)d)+ω(aφ(m)d)2 z 4w4 +2w 2 = ep(o(log log ) 5 ). But since σ(a) < A 2 and τ(σ(a)) σ(a) A 2, we get that which together with estimate (28) gives τ(σ(a)) = ep(o(log log ) 5 ), (29) τ(σ(aφ(m)d)) τ(σ(a))τ(σ(b)) = ep(o(log log ) 5 ). Returning to estimate (26), we get that ( #A 2 () τ2 y ep(o(log log )5 ) = ep log ) log log + O((log log )5 ) < for all > (log ) 0 0. (30) Thus, writing C() = A a,b ()\(A () A 2 () B()), we get that ( A a,b () = #C() + O (log ) 0. ). (3)
12 Moreover, if n C(), then n = Pm, with P > P(m) for > 0, and (P )/2 is a prime. Hence, P = 2q, where q is a Sophie Germain prime. Since also P /m, it follows by Brun s method that the number of such values for P is m(log /m) 2 (log log )2 = m(log y) 2 m(log ). 2 In the above inequalities we used the fact that /m P y. Summing up over all m, we get #C() (log log ) 2 (log log )2. (32) m(log ) 2 log m In particular, #A a,b () (log log )2. (33) log This is weaker than the bound claimed by our Theorem 2. However, it implies, by partial summation, that n + 2 t d(#a a,b(t)) n A a,b () = + #A a,b(t) t ( t= + O t=2 t(log log t) 2 dt 2 t 2 log t = O((log log ) 3 ). (34) To get some improvement, we return to C(). Let n C() and write it as n = Pm, where P > P(m). Write also P = 2q +. Then φ(n) = 2φ(m)q. Moreover, q > (y )/2 > z for > 0 so that q does not divide aφ(m) (because q > a and n B 3 ()). Hence, σ(aφ(n)) = σ(2aφ(m))(q + ). On the other hand, bσ(n) = bσ(m)(p + ) = 2bσ(m)(q + ). Thus, the equation σ(aφ(n)) = σ(bn) forces σ(2aφ(m)) = 2bσ(m), implying that m A 2a,2b (/P) A 2a,2b (/y). The argument which leads to estimate (32) now provides the better estimate In particular, we get #C() m A 2a,2b () #A a,b () (log log ) 2 m(log ) 2 (log log )5 (log ) 2. (log log )5 (log ) 2. (35) This is still somewhat weaker than what Theorem 2 claims. However, it implies that the sum of the reciprocals of the numbers in A a,b is convergent. In fact, by partial summation, ) 2
13 it follows, as in estimates (34), that n A a,b n y n y = #A a,b(t) t t d(#a a,b(t)) t= t=y ( + O y ) t(log log t) 5 t 2 (log t) dt 2 (log log y)5 + (log y) 2 y t(log t) 3/2dt (log y) /2. (36) We now take another look at C(). Let again n C() and write n = Pm. Fiing m and using the fact that P /m is a prime such that (P )/2 is also a prime, we get that the number of choices for P is m(log(/m)) 2. Hence, #C() m A 2a,2b (/y) m(log(/m)) 2. We now split the above sum at m = /2 and use estimate (36) to get #C() m A 2a,2b ( /2 ) (log /2 ) 2 m(log(/m)) 2 + m A 2a,2b m + (log y) 2 m A 2a,2b /2 m /y m A 2a,2b m /2 m(log(/m)) 2 (log ) + 2 (log y) = (log log )5/2 + 5/2 (log ) 2 (log ) 5/2 (log ) 2, (37) which together with estimate (3) leads to the desired conclusion of Theorem 2. m 4 Acknowledgments This work was done in April of 2006, while the second author was in residence at the Centre de recherches mathématiques of the Université de Montréal for the thematic year Analysis and Number Theory. This author thanks the organizers for the opportunity of participating in this program. He was also supported in part by grants SEP-CONACyT 46755, PAPIIT IN04505 and a Guggenheim Fellowship. The first author was supported in part by a grant from NSERC. 3
14 References [] E. R. Canfield, P. Erdős and C. Pomerance, On a problem of Oppenheim concerning Factorisatio Numerorum, J. Number Theory 7 (983), 28. [2] L. E. Dickson, A new etension of Dirichlet s theorem on prime numbers, Messenger of Math. 33 (904), [3] H. Halberstam and H.-E. Richert, Sieve Methods, Academic Press, London, 974. [4] G. H. Hardy and J. E. Littlewood, Some problems on partitio numerorum III. On the epression of a number as a sum of primes, Acta Math. 44 (923), 70. [5] A. Hildebrand, On the number of positive integers and free of prime factors > y, J. Number Theory 22 (986), [6] A. Hildebrand and G. Tenenbaum, Integers without large prime factors, J. de Théorie des Nombres de Bordeau 5 (983), [7] A. Ivić, The Riemann Zeta-Function, Theory and Applications, Dover Publications, [8] A. Schinzel and W. Sierpiński, Sur certaines hypothèses concernant les nombres premiers, Acta Arith. 4 (958), Erratum, Acta Arith. 5 (959), 259. [9] G. Tenenbaum, Introduction to analytic and probabilistic number theory, Cambridge Univ. Press, Mathematics Subject Classification: Primary A25; Secondary N37. Keywords: Euler function, sum of divisors. (Concerned with sequences A03363 and A37733.) Received June ; revised version received February Published in Journal of Integer Sequences, February Return to Journal of Integer Sequences home page. 4
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text:
Nevada Drops 44-27 Contest at Texas A&M
Wolfpack drops their second game of the season
Wolf Pack set to graduate 49 student-athletes in spring and summer Commencement
Every Nevada sport represented on graduation list.
Blue tops Silver in the spring game for the Wolf Pack football team.
Defense has the upper hand with specialized spring game scoring format.
Halfway through spring, Pack is making progress
Saturday's practice will be held at Douglas High School in Minden
Nevada Struggles in Second Half, Loses to Boise State, 34-17
Wimberly leads team with 11 catches for 121 yards.
In 35 career games with the Wolf Pack, boasts 241 tackles and 13.5 tackles for loss.
Senior (2015): Played in 11 games with nine starts in the regular season … Posted 72 tackles, 6.0 TFL, one sack and one fumble recovery … Ranked third on the team in tackles and 15th in the Mountain West among linebackers … Recorded his first collegiate touchdown with a 30-yard fumble recovery in Nevada's win at Fresno State … Earned Nevada's defensive player of the week honor with one sack and one fumble recovery against the Bulldogs … Matched a career-high with 16 tackles in Nevada's victory over Buffalo … Tallied 12 tackles and 1.5 TFL against rivals UNLV … Letter winner.
Junior (2014): Played in 12 games at linebacker with 10 starts ... Ranked third on the team and 20th in the Mountain West in tackles with 80 ... Credited with 80 tackles, 1.5 tackles for loss, half a sack, two fumble recoveries and one pass breakup on the season ... Tied for the game-high in tackles with nine against Louisiana in the R+L Carriers New Orleans Bowl ... Posted double-digit tackle totals in consecutive games with 11 at San Jose State and a season-high 14 against Boise State ... Recovered a fumble in victories against BYU and San Diego State ... Posted five or more tackles in 8-of-11 games.
Sophomore (2013): Joined Nevada after playing one season at L.A. Pierce College in Southern California ... Played in all 12 games with four starts ... Recorded four double-digit tackling performances with 11 each against Air Force, UNLV and BYU ... Set a career high with a 15-tackle performance in a win against San Jose State ... Registered six tackles for loss on the year.
Before Nevada
Played one season for coach Efrain Martinez at L.A. Pierce College after originally signing with Washington.
High School
A graduate of Dorsey High School in Los Angeles, Calif. ...Earned three varsity letters ... Named all-Coliseum League in 2010 and 2011 ... Also earned all-Los Angeles county honors as a senior.
Born Matthew Dean Lyons on May 15, 1993 ... Son of Kathryn Wainwright ... Has one sister Mariah, and four brothers, Mikey, Mitchell, Max and Tsalta ... Enjoys spending time with his family and friends when he is not on the field ... Majoring in general studies.


text: 4. As of noon trading, Anglogold Ashanti ( AU) is down $0.99 (-5.5%) to $16.87 on average volume Thus far, 1.6 million shares of Anglogold Ashanti exchanged hands as compared to its average daily volume of 3.0 million shares. The stock has ranged in price between $16.80-$17.22 after having opened the day at $17.17 as compared to the previous trading day's close of $17.86. AngloGold Ashanti Limited engages in the exploration, production, and marketing of gold. It also produces by-products, such as silver, uranium, and sulfuric acid. Anglogold Ashanti has a market cap of $6.9 billion and is part of the metals & mining industry. The company has a P/E ratio of 11.1, below the S&P 500 P/E ratio of 17.7. Shares are down 43.1% year to date as of the close of trading on Thursday. Currently there is 1 analyst that rates Anglogold Ashanti a buy, no analysts rate it a sell, and 2 rate it a hold. TheStreet Ratings rates Anglogold Ashanti as a hold. The company's strengths can be seen in multiple areas, such as its expanding profit margins and largely solid financial position with reasonable debt levels by most measures. However, as a counter to these strengths, we also find weaknesses including a generally disappointing performance in the stock itself, weak operating cash flow and deteriorating net income. Get the full Anglogold Ashanti Ratings Report now. STOCKS TO BUY: TheStreet Quant Ratings has identified a handful of stocks that can potentially TRIPLE in the next 12-months. Learn more.


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Winter Triangle
The Winter Triangle, or the Great Southern Triangle, is an asterism formed by three bright stars in three prominent winter constellations. These stars are Betelgeuse in Orion, Procyon in Canis Minor and Sirius in Canis Major constellation. The Winter Triangle is prominent in the night sky in the northern hemisphere during the winter months, from … [Read more…]
Highlights of the August Sky
Betelgeuse, Alpha Orionis, is the second brightest star in Orion constellation and the ninth brightest star in the sky. It is a supergiant star, distinctly red in colour, located at an approximate distance of 643 light years from Earth. It is an evolved star, one expected to explode as a supernova in a relatively near … [Read more…]
Northern Cross
Highlights of the Night Sky in August
Here are some of the highlights of the night sky in August: 9 PM southern sky – Mars and Saturn moving toward the western horizon after sunset, moving closer together as the month progresses 9 PM western sky – August 31 – Mars, Saturn and the crescent Moon in a triple conjunction 5:45 AM eastern … [Read more…]
Summer Triangle
The Summer Triangle is one of the most familiar patterns in the northern summer night sky. The three bright stars that mark the vertices of the Summer Triangle are Altair, Deneb, and Vega, the brightest stars in the constellations Aquila, Cygnus, and Lyra. The star pattern makes it easy to locate each of the three … [Read more…]
Deneb, Alpha Cygni, is the brightest star in Cygnus constellation and the most distant of all first magnitude stars. Deneb is a luminous blue supergiant. It is one of the stars that form the Summer Triangle, along with Altair in Aquila constellation and Vega in Lyra. Deneb is also the brightest star of the Northern … [Read more…]
Altair, or Alpha Aquilae, is the brightest star in the constellation Aquila, the Eagle. It lies only 16.7 light years from Earth. The star has a visual magnitude of 0.77. Along with the stars Vega in Lyra constellation and Deneb in Cygnus, Altair forms the Summer Triangle, a prominent asterism in the northern hemisphere. It … [Read more…]
The Southern Cross
The Southern Cross is one of the best known asterisms in the night sky, and the most familiar star pattern in the southern hemisphere. Composed of five stars forming the shape of a cross, it is the most distinctive feature of the constellation Crux, the smallest constellation in the sky. The Southern Cross is notable … [Read more…]


text:
Survivor Blood vs. Water predictions.
Aras: (winner of survivor: Exile Island) I predict that Ara's will do pretty good for a while, he will play both sides a while, and then he makes a crucial mistake and that could be his downfall. I think it will take a while for people to realize he's playing both sides and he will be blind-sided.
vytas: I think vtyas looks like a useful person to have on a tribe. Being brothers with Ara's one would assume that he has fishing skills, and that he could be a very "safe" player in the game. Who knows how far that would take him.
Candice: Why she's back, I still don't know. Most players who play their third time are generally "great" players but I don't consider Candice a great player, but maybe being out there for a third time she will learn from her past mistakes on survivor and try to not rub anyone the wrong way and it would make her chances for survival easier.
John Cody: I think John will lack the social skills but will be really strong in challenges. I predict that win it comes to the game, and everyone decides to vote him out, he'll think up this clever plan to get out a huge player, maybe like Rupert but it backfires and his torch will be deflamed.
Colton: In a really devious way, I sort of like him. I think he's going to try and control the game and get people to stupidly give him the immunity idol, or tell them they have it because he is somehow trustworthy to them, and then he cut's their head off. Either that or he's nicer this season, because of all the crap the press gave him about being a racist and stuff, but if he's nice I don't think he'll make it as far as he would if he was mean. And I don't think he was a racist, he was just mean.
Caleb: I think he'll try to be like colton and try and manipulate people by alligning himself with someone like Hayden or Vtyas. He might make the merge but probably not more than that.
Gervase: I think his only downfall in the game might be that he hasn't played the game in a long time, but with that being said, I still think he has it in him. He's nice, he's not afraid to make big moves and he's strong in challenges(except eating gross stuff) I think he'll make it far.
Marissa Peterson: I think she's athletic and really likes winning. I think she'll vote for a stronger team rather than a weaker team, but I don't think she'll make the merge, but maybe so.
Kat: I think Kat won't play a good social game but if she makes it far it will probably be because someone carried her there, and if she makes it to the merge(I hope not) then Hayden and his alliance will most likely carry her to the finals.
Hayden: I've watched him in big brother and he's a force to be reckon with. He has a good social game and is good at challenges, if he somehow alligns himself with aras and Gervase after the merge they could be unstoppable. I predict he'll make it as far as malcom did to the final four before they realize that he could win the game and Jeff snuffs his torch.
Laura: I don't really know why she's back, but I think she'll just be the person that everyone hates, and she'll just be another vote to keep someone. I don't think she'll make the merge.
Ciera: I think that Ciera is like Natalie in redemption island, she'll want to have her way and allign herself with the guys and demand that she be taken to the finals. She'll make the merge but will go not longer after.
Monica: She'll probably be a physical threat and want her out, unless their tribe sucks and they need her. She won't win because her husbands rich, that's her downfall in the game.
Brad: He will try to not tell people his secret NFL life but they find out and vote him out before the merge.
Rupert: Sadly, as much as we love rupert I see him going out before the merge, simply because he played too many times and although Boston Rob could do it, Rupert is too well liked and they will want to vote him out. I hope my prediction is wrong here.
Laura B: I think she'll be the older one in the tribe and she might not find everything in common with the younger people, and she may be targeted on multiple occasions, but I think she'll play a loyal and honest game and someone would respect that.
Tina: She's iffy for me. I don't know if she'll do really good or really bad, because both her seasons she was one or the other. All-Stars she was targeted because shew was a winner, so will she be targeted as a winner this season since only two winners are in it. If she's not targeted she could play a great social game.
Katie: She will make it pass the merge. She seems like a good person to have around. She's cute and seems like she'd be a hard worker, and if she makes it to the finals she might win because of her laying low and somehow getting to the end without ruffling feathers.
Tyson: My favorite this season. I hope he wins because if he looses the episodes would be dull, so they'll keep him around until the merge and then he'll dominate and hopefully this is his season to win, and he doesn't make a stupid move. (we can only hope) Hopefully he brings back his humor because everyone loves Tyson, he's just Tyson.
Rachel: I feel like Rachel could win. She seems strategic and cunning and she's not afraid to cut someone's throat, maybe she would even cut tyson's throat. She'll make it far, a likable third place.
Final Talleys
19th- Monica
17th- Brad
16th- Laura M.
15th- Laura B.
13th- Marissa
11th- Caleb
9th- vtyas
1st- Tyson (just because I want him to win and usually i'm right about winners. Positive energy)


text:
I can't believe that the time is almost upon us after years of waiting. In less that three days, the biggest and most-anticipated movie of 2008 will finally be released and I couldn't be more excited. "The Dark Knight" has been thrown left and right between fanboys and critics alike as to whether it's the greatest comic book movie ever made or an epic crime drama but there one thing that both parties can agree on, the movie KICKS ASS!
As a celebration of the film's release on Friday morning, the worldwide premiere was held last night and all the big stars of the movie were in attendance. It's very sad however that the biggest star of the movie could not attend because I'm absolutely sure that the spotlight would most certainly have been on him. However, the entire cast and crew looked great for the event and Maggie Gyllenhaal looked quite delicious as well. So enjoy the wicked hottie photos from the event below and remember, try not to see the movie more than 3-4 times this weekend if you can help it.
The hottie photos below are in the following order: Maggie Gyllenhaal, Lauren Conrad and Blake Lively.


text:
"This is the paradox of what happened after Mary said yes: everything turned out wrong—and, then again, everything turned out right." (Heather King on The Annunciation, Magnificat, March 2012)
From the fall of 1976 to the spring of 1977, I was in my second year of college, at Temple University in Philadelphia. I decided at some point during that year to transfer to the University of Scranton, a Jesuit institution a couple of hours to the north. During the summer prior to that transition I participated in a Newman Center retreat at the Jersey shore.
We were gathered in a room filled with tables, chairs, and booths, listening to a speaker talk to us about something, I don't remember what, when a thought popped into my head, as clear as though someone were standing right next to me, leaning into me, and speaking it directly into my ear: "Go over to that priest and ask him to pray over you and bless you for the ministry I've called you to and on which you are about to embark."
"What?!" I silently but vehemently questioned the invisible speaker.
The voice repeated its instructions.
"I can't do that. He'll laugh at me, or scold me, or something."
Again I heard, "Go and ask to be blessed."
My heart raced and I began to feel like crying.
I made my way over to the booth where the priest was sitting and slid in next to him. He looked at me questioningly but with a smile. I leaned toward him and whispered what I had heard, or what I thought I had heard.
I was amazed when he simply put his hand on my shoulder, bowed his head, and quietly prayed aloud for me.
"Everything turned out wrong...." I could write a long, long list of all the things that any reasonable person—whether he or she believes in annunciations or not—would easily say "turned out wrong" in the thirty-five years since that priest unquestioningly prayed that blessing in response to my questionable request. When I am alone I can easily descend into rehearsing that paralyzing list in my mind and tagging it as evidence, if not proof, that there was no annunciation and no blessing after all—or at least that I ruined any benefit intended or provided.
The most effective weapon I have found for defending myself against such deterioration has been meditating on the Mysteries of the Rosary, especially the ones that are called Joyful.
The Angel Gabriel's Annunciation to Mary that her Creator wanted her to be the mother of Jesus is known as the First Joyful Mystery. "How can this be?" and "Let it be done unto me according to your word" are the responses attributed to Mary in this conversation. Her question comforts me as much as her yes. It makes sense to me that Mary, especially at the foot of the cross, could have questioned her faith in the Angel Gabriel's words to her on that ancient pivotal day. There were many, I'm sure, who told her that her son's life was a failure. Maybe some even accused her of having filled his head with crazy ideas.
"...and, then again, everything turned out right." The final set of Rosary Mysteries is called the Glorious Mysteries—the hardest ones for us to believe. The first three are the Resurrection, the Ascension of Jesus to His Father, and Pentecost. Mary reportedly was present at all these events, and if she had even a crumb of doubt in her, surely it would have been gone at her first sight of His resurrected body.
The list of what has "turned out right" in my life is much harder to conjure than the list of what has (or seems to have) gone wrong. But as Easter approaches, that's the list I'm feeling called to write—or at least to write as much of it as I am able to know. Because, I think, it is in that list that I will see the proof that Jesus is indeed who the Angel Gabriel told Mary he would be; it is in that list that I will recognize, in myself and my life, His resurrected body.


text:
Harmonix wanted to bring Rock Band to the PC and decided the way to do it was a crowdfunding campaign but the whole affair tanked pretty spectacularly.
The project over on Fig was aiming for a fairly low figure of $1.5 million to create a fully functional PC version of the popular game where you hammer plastic peripherals and pretend you can actually play guitar. The intention was to allow the use of whatever older accessories you have lying around and to sell new ones online. You could even get a signed guitar for $1000. The video was so incredibly cheery!
Well the campaign just ended with only $793,000 collected, far short of the goal, after 35 days. It's a disappointment but yet another lesson in how fickle crowdfunding can be. Maybe there just wasn't enough publicity, maybe choosing Fig over Kickstarter was a mistake (it certainly gets less attention) or perhaps people just aren't that interested in a PC version of a title that's always been associated with consoles.
For whatever reason, this is yet another nice thing you can't have.


text:
I believe the success of a person always derived from good planning and innovative thinking. From the view point of fatalism, your life is controlled by your congenital fate.
Feel yourself are in darkness or have a decision to take?
Want to know how different are you from everyone else?
Have an empty feeling, want have change and better life?
No one can give you suggestion or no suggestion seems helpful?
Want to know more information about other people or the person you love?
You don't have to give up! You know you are not incompetent, but need more advices. You know problems are always solved easily when you are looking those from other directions.
I will provide you a detailed forecast report (at least 500 words) for your life or other's life if you like.
You may even learn something about yourself or other people in analysis. Anyway, I don't expect you have to take every word from me, but I think these are good cues for you to contemplate and plan your future.
If you would like to have forecast report, please send your payment to firstname.lastname@example.com using PayPal. The fee is USD10.00/person.
Report will send to your email.
Off line consultation:
Green Dragon Temple:
8/12 Sukhumvit Road Sriracha Chonburi 20110 Thailand
Consultation fee: THB 880
Time: Friday, Saturday and Sunday from 8.08am to 11.48am
If there is fate for us to meet, we will meet.


text:
The Dublin University Magazine: A Literary and Political Journal, Volume 92
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Popular passages
Page 154 - There is grandeur in this view of life, with its several powers, having been originally breathed by the Creator into a few forms or into one; and that, whilst this planet has gone cycling on according to the fixed law of gravity, from so simple a beginning endless forms most beautiful and most wonderful have been, and are being evolved.
Page 180 - Then shall we know, if we follow on to know the Lord: his going forth is prepared as the morning; and he shall come unto us as the rain, as the latter and former rain unto the earth.
Page 562 - Dreamer of dreams, born out of my due time, Why should I strive to set the crooked straight? Let it suffice me that my murmuring rhyme Beats with light wing against the ivory gate, Telling a tale not too importunate To those who in the sleepy region stay, Lulled by the singer of an empty day.
Page 524 - Lo, children are an heritage of the LORD and the fruit of the womb is his reward. As arrows are in the hand of a mighty man; so are children of the youth. Happy is the man that hath his quiver full of them: they shall not be ashamed, but they shall speak with the enemies in the gate.
Page 642 - Both riches and honour come of thee, and thou reignest over all ; and in thine hand is power and might ; and in thine hand it is to make great, and to give strength unto all.
Page 638 - For I have learned To look on Nature not as in the hour Of thoughtless youth; but hearing oftentimes The still, sad music of humanity, Nor harsh, nor grating, though of ample power To chasten and subdue. And I have felt A presence that disturbs me with the joy Of elevated thoughts; a sense sublime Of something far more deeply interfused Whose dwelling is the light of setting suns And the round ocean, and the living air, And the blue sky, and in the mind of man...
Page 523 - Enlarge the place of thy tent, and let them stretch forth the curtains of thine habitations: spare not, lengthen thy cords, and strengthen thy stakes ; for thou shalt break forth on the right hand and on the left ; and thy seed shall inherit the Gentiles, and make the desolate cities to be inhabited.
Page 180 - Come, and let us return unto the Lord: for he hath torn, and he will heal us; he hath smitten, and he will bind us up. After two days will he revive us: in the third day he will raise us up, and we shall live in his sight.
Page 463 - ... to demonstrate, that the most effectual plan for advancing a people to greatness, is to maintain that order of things which nature has pointed out...
Page 154 - I believe that animals have descended from at most only four or five progenitors, and plants from an equal or lesser number.