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%I #18 Aug 02 2023 07:19:37
%S 1,2,1,4,2,1,4,18,8,8,18,17,40,50,88,396,210,296,492,690,1144,1776,
%T 2786,3545,6704,10610,16096,25524,39650,63544,97108,269154,236880,
%U 389400,589298,956000,1459960,2393538,3604880,5739132,9030450,14777200
%N Product(1 + a(n)*x^n, n=1..infinity) = sum(F(k+1)*x^k, k=1..infinity) = 1/(1-x-x^2), where F(n) = A000045(n) (Fibonacci numbers).
%C A formal infinite product representation for the Fibonacci numbers (A000045(n+1)).
%C For references see A147541. [_R. J. Mathar_, Mar 12 2009]
%H Jean-François Alcover, <a href="/A147542/b147542.txt">Table of n, a(n) for n = 1..200</a>
%H Wolfdieter Lang <a href="/A147542/a147542.txt">Two recurrences for the general problem.</a>
%H R. J. Mathar, <a href="http://list.seqfan.eu/oldermail/seqfan/2008-November/000113.html">Re: polynomial-to-product transform</a>, Maple code (2008). [From _R. J. Mathar_, Mar 12 2009]
%F From _Wolfdieter Lang_, Mar 06 2009: (Start)
%F Recurrence I: With FP(n,m) the set of partitions of n with m distinct parts (which could be called fermionic partitions (fp)):
%F 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).
%F 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):
%F 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).
%F Example of recurrence I: a(4) = F(5) - a(1)*a(3) = 5 - 1*1 = 4.
%F 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)
%t m = 200;
%t 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;
%t Array[a, m] /. sol (* _Jean-François Alcover_, Oct 22 2019 *)
%Y Cf. A000045, A137852, A006973, A157159.
%K nonn
%O 1,2
%A _Neil Fernandez_, Nov 06 2008
%E More terms and revised description from _Wolfdieter Lang_ Mar 06 2009
%E Edited by _N. J. A. Sloane_, Mar 11 2009 at the suggestion of _Vladeta Jovovic_
%E More terms from _R. J. Mathar_, Mar 12 2009
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