A067418 Triangle A067330 with rows read backwards.
1, 2, 1, 5, 3, 2, 10, 7, 5, 3, 20, 15, 12, 8, 5, 38, 30, 25, 19, 13, 8, 71, 58, 50, 40, 31, 21, 13, 130, 109, 96, 80, 65, 50, 34, 21, 235, 201, 180, 154, 130, 105, 81, 55, 34, 420, 365, 331, 289, 250, 210, 170, 131, 89, 55, 744, 655, 600, 532, 469, 404, 340, 275, 212, 144, 89, 1308, 1164, 1075, 965
Offset: 0
Examples
{1}; {2,1}; {5,3,2}; {10,7,5,3}; ...; p(2,n)=5+3*x+2*x^2.
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..10000
Programs
-
Mathematica
Reverse /@ Table[Sum[Fibonacci[k + 1] Fibonacci[n - k + 1], {k, 0, m}], {n, 0, 11}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 11 2016 *)
Formula
a(n, m)=A067330(n, n-m), n>=m>=0, else 0.
a(n, m)= (((3*(n-m)+5)*F(n-m+1)+(n-m+1)*F(n-m))*F(m+1)+((n-m)*F(n-m+1)+2*(n-m+1)*F(n-m))*F(m))/5.
G.f. for column m=0, 1, ...: (x^m)*(F(m+1)+F(m)*x)/(1-x-x^2)^2, with F(m) := A000045(m) (Fibonacci).
a(n, m) = ((-1)^m*F(n-2*m+1)-m*L(n+2)+n*L(n+2)+5*F(n)+4*F(n-1))/5, with F(-n) = (-1)^(n+1)*F(n), hence a(n, m) = (2*(n-m+1)*L(n+2)-A067990(n, m))/5, n>=m>=0. - Ehren Metcalfe, Apr 11 2016
Comments