A304357 Antidiagonal sums of the first quadrant of array A(k,m) = F_k(m), F_k(m) being the k-th Fibonacci polynomial evaluated at m.

0, 1, 1, 3, 5, 13, 32, 94, 297, 1036, 3911, 15918, 69350, 321779, 1582745, 8220349, 44925187, 257563819, 1544896976, 9671289892, 63051738167, 427254561854, 3003872526303, 21876513464296, 164790822258172, 1282198404741305, 10292007232817249, 85126350266370355

Offset: 0

Author

Alois P. Heinz, May 11 2018

Keywords

• nonn

Comments

Equivalently, antidiagonal sums of the third quadrant of array A(k,m).

It seems that: a(n+1) is the sum of the n-th antidiagonal of triangle A101494; a(n)-(n mod 2) is the sum of the n-th antidiagonal of array A172236; and a(n+1)+(n mod 2) is the sum of row n of triangle A157103. - Mathew Englander, Feb 28 2021

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..600
• Wikipedia, Fibonacci polynomials
• Wikipedia, Quadrant (plane geometry)

Crossrefs

Cf. A000045, A084844, A304359, A101494, A172236, A157103.

Programs

Maple

F:= (n, k)-> (<<0|1>, <1|k>>^n)[1, 2]:
a:= n-> add(F(j, n-j), j=0..n):
seq(a(n), n=0..30);
# second Maple program:
F:= proc(n, k) option remember;
      `if`(n<2, n, k*F(n-1, k)+F(n-2, k))
    end:
a:= n-> add(F(j, n-j), j=0..n):
seq(a(n), n=0..30);
# third Maple program:
a:= n-> add(combinat[fibonacci](j, n-j), j=0..n):
seq(a(n), n=0..30);

Mathematica

a[n_] := Sum[Fibonacci[j, n - j], {j, 0, n}];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 02 2018, from 3rd Maple program *)

Formula

a(n) = Sum_{j=0..n} F_j(n-j).

a(n+1) = Sum_{j = 0..n} Sum_{i = j..floor((n+j)/2)} binomial(i,j)*(n+j-2*i)^j (empirically). - Mathew Englander, Feb 28 2021