A363754 a(n) = Sum_{k=0..n} F(2k-1)*F(2k)*F(2k+1)/2, where F(n) is the Fibonacci number A000045(n).

0, 1, 16, 276, 4917, 88132, 1581196, 28372701, 509125596, 9135883240, 163936760185, 2941725767256, 52787126964456, 947226559367881, 16997290941068152, 305004010378316172, 5473074895864584141, 98210344115173624636, 1762313119177232976916, 31623425801074947486405

Offset: 0

Author

Hans J. H. Tuenter, Jun 19 2023

Keywords

• nonn
• easy

Comments

This is one of the triple Fibonacci sums that were considered by Subba Rao (1953).

Taking any of the given closed-form expressions for a(n) with Fibonacci numbers, one can extend a(n) to negative indices by using the property F(-n)=(-1)^(n+1). This gives a(-n)=a(n-1).

Links

• K. Subba Rao, Some properties of Fibonacci numbers, The American Mathematical Monthly, 60(10):680-684, December 1953.
• Index entries for linear recurrences with constant coefficients, signature (22,-77,77,-22,1).

Crossrefs

Cf. A000045, A256178, A363753.

Programs

Mathematica

LinearRecurrence[{22, -77, 77, -22, 1}, {0, 1, 16, 276, 4917}]

Formula

a(n) = (F(2n+1)^3 + F(2n+1) - 2)/8.

a(n) = (F(6*n+3)+8*F(2*n+1)-10)/40.

a(n) = 22*a(n-1) - 77*a(n-2) + 77*a(n-3) - 22*a(n-4) + a(n-5).

G.f.: x*(1 - 6*x + x^2)/((1 - x)*(1 - 3*x + x^2)*(1 - 18*x + x^2)).