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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).
21
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
1,5
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).
LINKS
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.
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.
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<n) + L(n))/n, n>=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<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).
EXAMPLE
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.
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 *)
CROSSREFS
Cf. A147542 (with the product instead of the reciprocal one).
Cf. A220418.
Sequence in context: A240451 A206138 A241545 * A109434 A371840 A089299
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 10 2009
STATUS
approved