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A157162 1/Product_{n>=1} (1 - a(n)*x^n) = 1 + Sum_{k>=1} F(k+1)*x^k = 1/(1-x-x^2), where F(n) = A000045(n) (Fibonacci numbers).

%I A157162 #20 Mar 02 2020 09:39:22
%S A157162 1,1,1,1,2,2,4,5,8,10,18,24,40,52,88,125,210,286,492,702,1144,1638,
%T A157162 2786,3986,6704,9640,16096,23964,39650,57794,97108,144245,236880,
%U A157162 353010,589298,880828,1459960,2179068,3604880,5471094,9030450,13561742,22542396,34277634
%N A157162 1/Product_{n>=1} (1 - a(n)*x^n) = 1 + Sum_{k>=1} F(k+1)*x^k = 1/(1-x-x^2), where F(n) = A000045(n) (Fibonacci numbers).
%C A157162 A formal infinite product representation for the o.g.f. series of the Fibonacci numbers (A000045).
%C A157162 In the context of Witt rings the o.g.f. is called associated unital series for the (infinite dimensional) Witt vector (a(1),a(2),...). Sometimes also called inverse Somos transform, here for the Fibonacci numbers.
%C A157162 1-x-x^2 = product(1 - a(n)*x^n, n=1..infinity).
%H A157162 Alois P. Heinz, <a href="/A157162/b157162.txt">Table of n, a(n) for n = 1..2000</a>
%H A157162 Giedrius Alkauskas, <a href="http://arxiv.org/abs/0801.0805">One curious proof of Fermat's little theorem</a>, arXiv:0801.0805 [math.NT], 2008.
%H A157162 Giedrius Alkauskas, <a href="https://www.jstor.org/stable/40391097">A curious proof of Fermat's little theorem</a>, Amer. Math. Monthly 116(4) (2009), 362-364.
%H A157162 Giedrius Alkauskas, <a href="https://arxiv.org/abs/1609.09842">Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems</a>, arXiv:1609.09842 [math.NT], 2016.
%H A157162 H. Gingold, H. W. Gould, and Michael E. Mays, <a href="https://www.researchgate.net/publication/268023169_Power_product_expansions">Power Product Expansions</a>, Utilitas Mathematica 34 (1988), 143-161.
%H A157162 H. Gingold and A. Knopfmacher, <a href="http://dx.doi.org/10.4153/CJM-1995-062-9">Analytic properties of power product expansions</a>, Canad. J. Math. 47 (1995), 1219-1239.
%H A157162 Wolfdieter Lang, <a href="/A157162/a157162.txt">Recurrences for the general problem</a>.
%F A157162 Recurrence I. With P(n,m) the set of partitions of n with m parts:
%F A157162 a(n)= F(n+1) - sum(sum(product(a(j)^e(j),j=1..m), p from P(n,m)), m=2..n), n>=2, with sum(j*e(j),j=1..n)=n, sum(e(j),j=1..n)=m for the partition p of n with m parts. F(n) = A000045(n) (Fibonacci numbers). Input a(1)=F(2)=1. See the array A008284(n,m) for the cardinalities of the sets P(n,m).
%F A157162 Recurrence II (simplified version). With the Lucas numbers L(n)=A000035(n), n>=1, as input (found by V. Jovovic, Mar 10 2009):
%F A157162 a(n) = (- sum(d*a(d)^(n/d), d|n with 1<=d<n) + L(n))/n, n>=2, a(1)=1.
%F A157162 Recurrence II. With the number array M0(n,vec(e)) given for any partition in A048996.
%F A157162 a(n) = - sum((d/n)*(a(d))^(n/d),d|n with 1<=d<n) + sum(((-1)^(m-1))*(1/m)*sum(M0(p)*F(2)^e(1)*...*F(n+1)^e(n),p=(1^e(1),...,n^e(n)) from P(n,m)), m=1..n) for n>=2; a(1)=F(2)=1. See recurrence 1 for the set P(n,m). The M0 numbers are m!/product(e(j)!,j=1..n).
%e A157162 Recurrence I: a(2) = F(3) - a(1)^2 = 1; a(4) = F(5) - (a(1)*a(3) + a(2)^2 +a(1)^2*a(2) + a(1)^4) = 5 - 4 = 1.
%e A157162 Recurrence II (simplified): a(4) = (-(a(1)^4 + 2*a(2)^2) + L(4))/4 = (-3 + 7)/4 = 1.
%e A157162 Recurrence II: a(4)= (-(a(1)^4 + 2*a(2)^2)/4 + 1*1*F(5) - (1/2)*(2*F(2)*F(4)+ 1*F(3)^2) +(1/3)*3*F(2)^2*F(3)-(1/4)*1*F(2)^4 = -3/4 +7/4 = 1.
%t A157162 a[n_] := a[n] = If[n == 1, 1, (-Sum[d a[d]^(n/d), {d, Most@ Divisors@ n}] + LucasL[n])/n];
%t A157162 Array[a, 50] (* _Jean-François Alcover_, Mar 02 2020 *)
%Y A157162 Cf. A147542 (with the product instead of the reciprocal one).
%Y A157162 Cf. A220418.
%K A157162 nonn,easy
%O A157162 1,5
%A A157162 _Wolfdieter Lang_, Aug 10 2009