cp's OEIS Frontend

Conjecture: polynomials with (positive) Fibonomial coefficients are reducible iff n odd > 1. - Ralf Stephan, Oct 29 2004

Row n+1 (n >= 1) of the signed triangle lists the coefficients of the recursion relation for the n-th power of Fibonacci numbers A000045: Sum_{m=0..n+1} T(n+1,m)*(Fibonacci(k-m))^n = 0, k >= n+1; inputs: (Fibonacci(k))^n, k=0..n.

The inverse of the row polynomial p(n,x) := Sum_{m=0..n} T(n,m)*x^m is the g.f. for the column m=n-1 of the Fibonomial triangle A010048.

The row polynomials p(n,x) factorize according to p(n,x) = G(n-1)*p(n-2,-x), with inputs p(0,x)= 1, p(1,x)= 1-x and G(n):= 1 - A000032(n)*x + (-1)^n*x^2. (Derived from Riordan's result and Knuth's exercise).

The row polynomials are the characteristic polynomials of product of the binomial matrix binomial(i,j) and the exchange matrix J_n (matrix with 1's on the antidiagonal, 0 elsewhere). - Paul Barry, Oct 05 2004

From Feryal Alayont, Apr 27 2023: (Start)

a(n) is the number of edge covers of a caterpillar graph with spine P_(3n-2), one pendant attached at vertex n counting from the left end of the spine and another pendant at vertex 2n-1. The caterpillar graph for n=3 is as follows:

* *

| |

*--*--*--*--*--*--*

v

Each pendant edge must be included in an edge cover. Every vertex except v is then incident with at least one edge. Therefore, of the remaining four edges in the spine, only two in the middle have restrictions. At least one of those two has to be in the edge cover to ensure v is incident with one edge in the edge cover, leaving us with 3*2*2 total edge covers. (End)

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.

Let L(n) = A000032(n) denote the n-th Lucas number.

For a fixed positive integer k, the power series expansion of exp( Sum_{n >= 1} L(k*n)x^n/n ) has integer coefficients given by the formula F(k*n)/F(k), where F(n) = A000045(n) [Johnson, 2.22].

The power series expansion of exp( Sum_{n >= 1} L(k*n)*L(2*k*n) *x^n/n ) has integer coefficients given by ( F(k*(n + 1))*F(k*(n + 2))*F(k*(n + 3)) )/( F(k)*F(2*k)*F(3*k) )

The present sequence is the particular case k = 2. See A001655 for the case k = 1.

Alternating sum of the product of three consecutive Fibonacci numbers, divided by two.

Can also be seen as the alternating sum of the Fibonomial coefficients (n+1,3), A001655.

This sequence is part of a suite of sums over triple products of Fibonacci numbers. Subba Rao (1953) gives closed-form expressions for several Fibonacci sums of this type.