for the m th moment of the probability distribution of k = 0 , . . . , n . Define also so that the m th moment becomes 〈| x | $^{m}$〉 = ϱ$_{m}$/ϱ$_{0}$ . For m = 0 we have so that Next, for m = 1 where − 1 < erf ( x ) < 1 is the error function. and thus In general we have for m ≥ 0 that where ϱ$_{m}$ is given by $^{φ}$$^{−}$$^{(}$$^{m}$$^{+1)}$$^{/}$$^{4}$ 2 4 Here $_{1}$F$_{1}$ ( a, b, c ) and U( a, b, c ) denote the confluent hypergeometric functions of the first and second kind respectively. If we want to compute cumulant ratios we first need moment ratios which of course is easy now. For example, in the case of a = 0 we have These moment ratios are the same as obtained in the 5-dimensional Ising model, see e.g. $^{11}$. 26 σ$_{2}$ σ 2 1 ∼ ϱ$_{0}$ ϱ$_{2}$ ϱ 2 1 = √$_{2 = 1}$$_{.}$$_{4142}$$_{. . .}$ (98) σ$_{4}$ σ 2 ∼ ϱ$_{0}$ ϱ$_{4}$ ϱ 2 = Γ( 1 / 4 ) 4 8 π 2 = 2 . 1884 . . . (99) σ$_{6}$ σ 3 2 ∼ ϱ 2 0 ϱ$_{6}$ ϱ 3 2 = 3 Γ( 1 / 4 ) 4 8 π 2 = 6 . 5653 . . . (100) σ$_{m}$ = 〈∣ ∣ k − n 2 ∣ ∣$^{m}$ 〉 ∼ n 3 m/ 4 ϱ$_{m}$ ϱ$_{0}$ (96) σ$_{1}$ = 〈∣ ∣ k − n 2 ∣ 〉 ∼ n 3 / 4 ϱ$_{1}$ ϱ$_{0}$ (95) ϱ$_{1}$ = √$_{π}$ $_{2}$√$_{φ}$ exp ( φ a $^{2}$) ( 1 + erf ( a √ φ )) (94) 1 = n/ 2 ∑ k $^{=}$− n/ 2 P$_{p,q}$ ( n, n/ 2 + k ) ∼ n 3 / $^{4}$P$_{p,q}$ ( n, n/ 2) ϱ$_{0}$ (93) ϱ$_{0}$ =      $_{π}$√$_{a}$ 2 exp ( φ a $^{2}$/ 2) ( I$_{1}$$_{/}$$_{4}$ ( φ a $^{2}$/ 2 ) + I − 1 / $_{4}$( φ a $^{2}$/ 2 ) ) , for a > 0 √ − a √$_{2}$ exp ( φ a $^{2}$/ 2) K$_{1}$$_{/}$$_{4}$ ( φ a $^{2}$/ 2 ) , for a < 0 Γ(1 / 4) 2 φ 1 / $_{4}$, for a = 0 (92) ϱ$_{m}$ = $^{+}$∞ ∫ −∞ | x | m exp ( − φ ( x 4 − 2 a x $^{2}$) ) d x, m ≥ 0 (91)     2 ( Γ( m +1 4 ) $_{1}$F$_{1}$ ( m +1 4 , 1 $_{2}$, φ a 2 ) + 2 $_{a}$√$_{φ}$$_{Γ(}$ m +3 4 ) $_{1}$F$_{1}$ ( m +3 4 , 3 $_{2}$, φ a 2 ) ) , a > 0 φ − ( m +1) / 4 2 ( m +1) / 2 Γ( m +1 2 ) U( m +1 4 , 1 $_{2}$, φ a 2 ) , a < 0 φ − ( m +1) / 4 2 Γ( m +1 4 ) , a = 0 (97)