This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A215037 #41 Mar 03 2024 18:48:00
%S A215037 1,4,19,79,339,1431,6072,25707,108922,461362,1954426,8278978,35070483,
%T A215037 148560678,629313573,2665814361,11292572005,47836100785,202636977730,
%U A215037 858384007525,3636173014596,15403076054964,65248477252164
%N A215037 a(n) = Sum_{k=0..n} fibonomial(k+3,3), n >= 0.
%C A215037 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
%C A215037 (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.
%C A215037 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.
%H A215037 Wolfdieter Lang, <a href="/A215037/a215037.pdf">Partial summation formula applied to sums over cubes of Fibonacci numbers.</a>
%H A215037 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (4, 3, -9, 2, 1).
%F A215037 Let F(n) be the Fibonacci number A000045(n).
%F A215037 a(n) = Sum_{k=0..n} F(k+3)*F(k+2)*F(k+1)/2.
%F A215037 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.
%F A215037 G.f.: 1/((1+x-x^2)*(1-4*x-x^2)*(1-x)) (from the g.f. of the Fibonomials A001655).
%F A215037 From _Hans J. H. Tuenter_, Jun 26 2023: (Start)
%F A215037 a(n) = (F(n+3)^2*F(n+2) + (-1)^n*F(n+1)-1)/4.
%F A215037 a(n) = (F(n+3)^3 + F(n+2)^3 + (-1)^n*F(n+1) - 2)/8.
%F A215037 a(n) = (F(3*n+8) + 4*(-1)^n*F(n+1) - 5)/20.
%F A215037 a(n) = 4*a(n-1) + 3*a(n-2) - 9*a(n-3) + 2*a(n-4) + a(n-5).
%F A215037 a(-n) = A363753(n-3).
%F A215037 (End)
%e A215037 a(3) = 2*1*1/2 + 3*2*1/2 + 5*3*2/2 + 8*5*3/2 = 1 + 3 + 15 + 60 = 79.
%t A215037 LinearRecurrence[{4, 3, -9, 2, 1}, {1, 4, 19, 79, 339}, 23] (* _Hans J. H. Tuenter_, Jun 26 2023 *)
%Y A215037 Cf. A000045, A008346, A001655, A215038, A363753.
%K A215037 nonn,easy
%O A215037 0,2
%A A215037 _Wolfdieter Lang_, Aug 09 2012