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Two Simple Ways of Generating the Partitions of (n+1) from the Partitions of n

I propose two simple ways of generating the partitions of (n+1) from the partitions of n. A recurrence relation for P(n+1), the number of partitions of (n+1), in terms of P(n) and Q(n), where Q(n) denotes the number of partitions of n having strictly different last two parts is obtained. Also a generating function for Q(n) is given. The other method for generating the partitions of (n+1) from the partitions of n is discussed at the end.

). So, by expanding the infinite products as power series, carrying out multiplication of these series, and then comparing the coefficients of q n one can generate as well as count the partitions of n.
In this paper I propose two new methods to generate the partitions of (n+1) from the partitions of n.
The first method that I have proposed in this short paper is based on the observation that among the partitions of n there are two kinds of partitions, say partitions of the "First Kind", having last two (or more) parts identical, and partitions of the "Second Kind", having only one part or having last two parts strictly different. We see that partitions of the "First Kind" generate one and only "one" partition of (n+1) while partitions of the "Second Kind" generate "two" partitions of (n+1).
As n is made large the count of partitions of the "Second Kind" becomes large. This fact leads to the astronomical growth of the number of partitions of n, P(n), with the increase in n.

The First Method:
This method is based on the following simple observations: Among the partitions of n, (a) Partitions of n of the "First Kind" produce exactly one partition of (n+1), obtained by simply adding "1" separately. (b) Partitions of n of the "Second Kind" produce exactly two partitions of (n+1), one obtained as above by adding "1" separately, while the other obtained by augmenting 1 in the last part (or in the only part), i.e. if the last part is k α then one has to make it "( k α +1)" by augmentation.

Lemma 1:
For every partition of (n+1) the method implies a unique predecessor in the partitions of n.
Proof: Let α be a k-partition (partition having k parts) of (n+1), then α is uniquely obtained, as per the method, from the (k-1)-partition of n, β say, by adding "1" separately where α > 1 then again α is uniquely obtained from the k-partition of n, γ say, by augmenting "1" in the last part where Thus, every partition of (n+1) is uniquely derivable from a partition of n. The word "every" ensures that all the partitions (without omission) of (n+1) are achievable by this method, and the word "uniquely" implies that every partition of (n+1) gets generated only once (without repetition). Hence the lemma. (2) From each partition of n in the first group obtain a partition of (n+1) by separately adding a unit i.e. "1" in each partition.
(3) From each partition of n in the second group obtain exactly two partitions of (n+1), one by adding "1" separately (as in (2)) and the other by augmenting "1" in the last part i.e. by changing k α to ( k α +1).
Theorem: Algorithm 1 generates all the partitions of (n+1) from the partitions of n, without omission or repetition.
Proof: Follows from the lemma 1.
Example: Let n = 5. Referring to the list of the partitions of 5 given above, the groups, obtained as per step (1)  Then we have, as an immediate consequence of the above algorithm, the following Theorem: P(n+1) = P(n) + Q(n).
Proof: (i) P(n) on the r.h.s. counts the partitions of (n+1) obtained by adding "1" separately, in all the partitions of n.
(ii) Q(n) on the r.h.s. counts the partitions of (n+1) obtained from the partitions of n having the only part or having the last two parts strictly different, by augmenting "1" in the only part i.e. n, or, by augmenting "1" in the last part where the last two parts strictly differ.

A generating function for Q(n):
Theorem: The following is the generating function for Q(n), the partitions of n having the only part i.e. n, or having strictly different last two parts: It is easy to check that in the sum of products on the r.h.s. of the above equation a product for (any) k counts the partitions with the only or the last part equal to k while all other parts strictly bigger than k. So, by collecting all terms whose indices add to n, for all k, we can obtain all the partitions of n that are of the "Second Kind".
It is straightforward to see that, after rewriting the sum of the products in the above equation, the equation can be rewritten as Hence the result.

The Second Method:
This method is similar to the first method. The difference is in the definition for classification of the partitions of n.
Consider a partition of n: where ℵ are the number of units present in the partition.
Hereafter, a partition of n will be called the partition of the "First Kind" if ℵ= 0, or ℵ ≥ k α , and a partition of n will be called the partition of the "Second Kind" when 1 ≤ ℵ < k α .
In this method, (a) Partitions of n of the "First Kind" produce exactly one partition of (n+1), obtained by simply adding "1" separately.
(b) Partitions of n of the "Second Kind" produce exactly two partitions of (n+1), one obtained as above by adding "1" separately, while the other obtained by augmenting 1 in the count of the number of units i.e.
replacing all units, ℵin number, by a single number (ℵ+1). Note that in this procedure of generating the partitions of (n +1) from the partitions of n we are not augmenting "1" to the last part but collecting the units present in the partition (of n) +1 into a single number. So, the partitions of n with ℵ = 0 produce only one partition, produced by adding "1" separately. Therefore, in this method the partition of (n+1) with only one part i.e. {n+1} is not generated and has to be added in the list separately.
Lemma 2: For every partition of (n+1) (except {n+1}) the method (second) implies a unique predecessor in the partitions of n.
Proof: Let α be a k-partition (partition having k parts) of (n+1), and suppose k > 1: Thus, every partition of (n+1) (except the partition with one part, i.e. {n+1}) is uniquely derivable from a partition of n. The word "every" ensures that all the partitions (without omission) of (n+1) are achievable by this method, and the word "uniquely" implies that every partition of (n+1) gets generated only once (without repetition). Hence the lemma.