OFFSET
1,2
COMMENTS
A formal infinite product representation for the Fibonacci numbers (A000045(n+1)).
For references see A147541. [R. J. Mathar, Mar 12 2009]
LINKS
Jean-François Alcover, Table of n, a(n) for n = 1..200
Wolfdieter Lang Two recurrences for the general problem.
R. J. Mathar, Re: polynomial-to-product transform, Maple code (2008). [From R. J. Mathar, Mar 12 2009]
FORMULA
From Wolfdieter Lang, Mar 06 2009: (Start)
Recurrence I: With FP(n,m) the set of partitions of n with m distinct parts (which could be called fermionic partitions (fp)):
a(n)= F(n+1) - sum(sum(product(a(k[j]),j=1..m),fp from FP(n,m)),m=2..maxm(n)), with maxm(n):=A003056(n) and the distinct parts k[j], j=1,...,m, of the partition fp of n, n>=3. Inputs a(1)=F(2)=1, a(2)=F(3)=2. See the array A008289(n,m) for the cardinality of the set FP(n,m).
Recurrence II: With the definition of FP(n,m) from the above recurrence I, P(n,m) the general set of partitions of n with m parts, and the multinomial numbers M_0 (given for every partition under A048996):
a(n) = sum((d/n)*(-a(d)^(n/d)),d|n with 1<d<n) + sum(((-1)^(m-1))*(1/m)*sum(M_0(p)*F(2)^e(1)*...*F(n+1)^e(n),p=(1^e(1),...,n^e(n)) from P(n,m)),m=1..n-1), n>=2; a(1)=F(2)=1. The exponents e(j)>=0 satisfy sum(j*e(j),j=1..n)=n and sum(e(j),j=1..m). The M_0 numbers are m!/product(e(j)!,j=1..n).
Example of recurrence I: a(4) = F(5) - a(1)*a(3) = 5 - 1*1 = 4.
Example of recurrence II: a(4)= 2*(-1)^2 + (1*F(5)-(1/2)*(2*F(2)*F(4) + 1*F(3)^2) + (1/3)*3*F(2)^2*F(3)) = 4. (End)
MATHEMATICA
m = 200;
sol = Thread[CoefficientList[Sum[Log[1 + a[n] x^n], {n, 1, m}] - Log[1/(1 - x - x^2)] + O[x]^(m + 1), x] == 0] // Solve // First;
Array[a, m] /. sol (* Jean-François Alcover, Oct 22 2019 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Neil Fernandez, Nov 06 2008
EXTENSIONS
More terms and revised description from Wolfdieter Lang Mar 06 2009
Edited by N. J. A. Sloane, Mar 11 2009 at the suggestion of Vladeta Jovovic
More terms from R. J. Mathar, Mar 12 2009
STATUS
approved