A147542 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).

1, 2, 1, 4, 2, 1, 4, 18, 8, 8, 18, 17, 40, 50, 88, 396, 210, 296, 492, 690, 1144, 1776, 2786, 3545, 6704, 10610, 16096, 25524, 39650, 63544, 97108, 269154, 236880, 389400, 589298, 956000, 1459960, 2393538, 3604880, 5739132, 9030450, 14777200

Offset: 1

Author

Neil Fernandez, Nov 06 2008

Keywords

• nonn

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]

Crossrefs

Cf. A000045, A137852, A006973, A157159.

Programs

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 *)

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=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)

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