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).

1, 1, 1, 1, 2, 2, 4, 5, 8, 10, 18, 24, 40, 52, 88, 125, 210, 286, 492, 702, 1144, 1638, 2786, 3986, 6704, 9640, 16096, 23964, 39650, 57794, 97108, 144245, 236880, 353010, 589298, 880828, 1459960, 2179068, 3604880, 5471094, 9030450, 13561742, 22542396, 34277634

Offset: 1

Author

Wolfdieter Lang, Aug 10 2009

Keywords

• nonn
• easy

Comments

A formal infinite product representation for the o.g.f. series of the Fibonacci numbers (A000045).

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.

1-x-x^2 = product(1 - a(n)*x^n, n=1..infinity).

Examples

			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.
Recurrence II (simplified): a(4) = (-(a(1)^4 + 2*a(2)^2) + L(4))/4 = (-3 + 7)/4 = 1.
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.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..2000
• Giedrius Alkauskas, One curious proof of Fermat's little theorem, arXiv:0801.0805 [math.NT], 2008.
• Giedrius Alkauskas, A curious proof of Fermat's little theorem, Amer. Math. Monthly 116(4) (2009), 362-364.
• Giedrius Alkauskas, Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems, arXiv:1609.09842 [math.NT], 2016.
• H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143-161.
• H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canad. J. Math. 47 (1995), 1219-1239.
• Wolfdieter Lang, Recurrences for the general problem.

Crossrefs

Cf. A147542 (with the product instead of the reciprocal one).

Cf. A220418.

Programs

Mathematica

a[n_] := a[n] = If[n == 1, 1, (-Sum[d a[d]^(n/d), {d, Most@ Divisors@ n}] + LucasL[n])/n];
Array[a, 50] (* Jean-François Alcover, Mar 02 2020 *)

Formula

Recurrence I. With P(n,m) the set of partitions of n with m parts:

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).

Recurrence II (simplified version). With the Lucas numbers L(n)=A000035(n), n>=1, as input (found by V. Jovovic, Mar 10 2009):

a(n) = (- sum(d*a(d)^(n/d), d|n with 1<=d=2, a(1)=1.

Recurrence II. With the number array M0(n,vec(e)) given for any partition in A048996.

a(n) = - sum((d/n)*(a(d))^(n/d),d|n with 1<=d=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).