A372015 Product of Fibonacci and self-convolution of Fibonacci numbers: a(n) = A000045(n+1)*A001629(n+1).

0, 1, 4, 15, 50, 160, 494, 1491, 4420, 12925, 37380, 107136, 304764, 861445, 2421700, 6775755, 18879734, 52413856, 145038890, 400183575, 1101277060, 3023462521, 8282790024, 22646131200, 61805595000, 168399404425, 458128878724, 1244567262471, 3376576740410, 9149594423200

Offset: 0

Author

Vladimir Kruchinin, Apr 15 2024

Keywords

• nonn
• easy

Comments

Conjecture: a(n) is the total number of pairs of adjacent parts that are the same color in all n-color compositions of n+1. - John Tyler Rascoe, Jul 30 2024

Links

• John Tyler Rascoe, Table of n, a(n) for n = 0..500
• Index entries for linear recurrences with constant coefficients, signature (5,-5,-5,5,-1).

Crossrefs

Cf. A000045, A001629, A088305.

Programs

Maple

a := proc(n) option remember; if n < 3 then return n^2 fi;
-((2 - 2*n^2 + n)*a(n - 1) + (1 - 2*n^2 + 3*n)*a(n - 2) + n^2*a(n - 3))/(n - 1)^2 end: seq(a(n), n = 0..29);  # Peter Luschny, Apr 16 2024

Mathematica

CoefficientList[Series[x(1-x)/((1+x)*(1-3*x+x^2)^2),{x,0,29}],x] (* Stefano Spezia, Apr 16 2024 *)

PARI

A_x(N)= {my(x='x+O('x^N)); concat([0],Vec(x*(1-x)/((1+x)*(1-3*x+x^2)^2)))}
A_x(40) \\ John Tyler Rascoe, Jul 29 2024

Formula

a(n) = F(n+1)*((n+2)*F(n) + (n)*F(n+2))/5 where F(n) = A000045(n) is the Fibonacci numbers.

G.f.: x*(1-x)/((1+x)*(1-3*x+x^2)^2).