A215037 - OEIS

A215037

a(n) = Sum_{k=0..n} fibonomial(k+3,3), n >= 0.

1, 4, 19, 79, 339, 1431, 6072, 25707, 108922, 461362, 1954426, 8278978, 35070483, 148560678, 629313573, 2665814361, 11292572005, 47836100785, 202636977730, 858384007525, 3636173014596, 15403076054964, 65248477252164

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

This sum is obtained from the m=2 member of the m-family of sums s(m;n) := Sum_{k=0..n} F(k+m)*F(k+1)*F(k), n>=0, given by

(F(n+m)*F(n+2)*F(n+1) - (-1)^n*F(m)*A008346(n))/2 with A008346(n) = (F(n) + (-1)^n), where F = A000045.

The formula for s(m;n), m>=0, n>=0, follows by induction on m, using the sums for m=0 and m=1. s(0,n) = F(n+1)*(F(n+1)^2 - (-1)^n)/2 = F(n+2)*F(n+1)*F(n)/2 (see A001655(n-1)), and s(1,n) = (F(n+2)*F(n+1)^2 - (-1)^n*A008346(n))/2 (see A215038). For the formulas for s(0,n) and s(1,n) see also the link on partial summation, eqs. (6) and (7). There Sum_{k=0..n} fibonomial(k+2,k) is obtained more directly in eq. (5) with the help of the partial summation formula.

LINKS

Table of n, a(n) for n=0..22.

Wolfdieter Lang, Partial summation formula applied to sums over cubes of Fibonacci numbers.

Index entries for linear recurrences with constant coefficients, signature (4, 3, -9, 2, 1).

FORMULA

Let F(n) be the Fibonacci number A000045(n).

a(n) = Sum_{k=0..n} F(k+3)*F(k+2)*F(k+1)/2.

a(n) = (F(n+3)^2*F(n+2) + (-1)^n*A008346(n+1))/4, n>=0, with A008346(n) = F(n) + (-1)^n. See a comment above.

G.f.: 1/((1+x-x^2)*(1-4*x-x^2)*(1-x)) (from the g.f. of the Fibonomials A001655).

From Hans J. H. Tuenter, Jun 26 2023: (Start)

a(n) = (F(n+3)^2*F(n+2) + (-1)^n*F(n+1)-1)/4.

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