A024458 a(n) = s(1)*s(n) + s(2)*s(n-1) + ... + s(k)*s(n+1-k), where k = floor((n+1)/2), s = (Fibonacci numbers).

1, 1, 3, 5, 12, 19, 40, 65, 130, 210, 404, 654, 1227, 1985, 3653, 5911, 10720, 17345, 31090, 50305, 89316, 144516, 254568, 411900, 720757, 1166209, 2029095, 3283145, 5684340, 9197455, 15855964, 25655489, 44061862, 71293590, 122032508

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

From Wolfdieter Lang, Jan 02 2012: (Start)

chat(n):=a(n+1), n>=0, is the half-convolution of the sequence A000045(n+1), n>=0, with itself. For the definition of half-convolution see a comment on A201204, where also the rule to find the o.g.f. is given. Here the o.g.f. is obtained from (U(x)^2 + U2(x^2))/2 with U(x)=1/(1-x-x^2), the o.g.f. of A000045(n+1), n>=0, and U2(x):=(1-x)/((1+x)*(1-3*x+x^2)) the o.g.f. of A007598(n+1), n>=0. This coincides with the o.g.f. given below in the formula section after x has been divided.

For the bisection of this half-convolution see A027991(n+1) and A001870(n), n>=0.

(End)

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,3,-2,0,-2,-3,1,1).

Crossrefs

Cf. A000032, A000045, A058071.

Programs

Magma

[(&+[Fibonacci(j+1)*Fibonacci(n-j): j in [0..Floor((n-1)/2)]]): n in [1..50]]; // G. C. Greubel, Apr 06 2022

Mathematica

Table[((13-5(-1)^n +10n)Fibonacci[n] + (1-(-1)^n +2n)LucasL[n] +8Sin[Pi*n/2])/40, {n, 30}] (* Vladimir Reshetnikov, Oct 03 2016 *)
LinearRecurrence[{1,3,-2,0,-2,-3,1,1},{1,1,3,5,12,19,40,65},40] (* Harvey P. Dale, Mar 02 2023 *)

SageMath

def A024458(n): return sum(fibonacci(j+1)*fibonacci(n-j) for j in (0..((n-1)//2)) )
[A024458(n) for n in (1..50)] # G. C. Greubel, Apr 06 2022

Formula

G.f.: x*(1-x^2+x^3)/((1+x^2)*(1+x-x^2)*(1-x-x^2)^2).

a(n) = ((13 - 5*(-1)^n + 10*n)*A000045(n) + (1 - (-1)^n + 2*n)*A000032(n) + 8*sin(Pi*n/2))/40. - Vladimir Reshetnikov, Oct 03 2016

From G. C. Greubel, Apr 06 2022: (Start)

a(2*n) = (1/5)*(n*Lucas(2*n+1) + Fibonacci(2*n)), n >= 1.

a(2*n+1) = (1/5)*((-1)^n + (n+1)*Lucas(2*n+2) + Fibonacci(2*n+1)), n >= 0.

a(n) = Sum_{j=0..floor((n-1)/2)} fibonacci(j+1)*Fibonacci(n-j). (End)

Extensions

More terms from James Sellers, May 03 2000