A110035 Row sums of an unsigned characteristic triangle for the Fibonacci numbers.
1, 2, 5, 12, 31, 80, 209, 546, 1429, 3740, 9791, 25632, 67105, 175682, 459941, 1204140, 3152479, 8253296, 21607409, 56568930, 148099381, 387729212, 1015088255, 2657535552, 6957518401, 18215019650, 47687540549, 124847601996
Offset: 0
Examples
G.f. = 1 + 2*x + 5*x^2 + 12*x^3 + 31*x^4 + 80*x^5 + 209*x^6 + ... - _Michael Somos_, Mar 03 2023
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).
Programs
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Mathematica
LinearRecurrence[{3,0,-3,1},{1,2,5,12},50] (* Harvey P. Dale, May 01 2022 *) a[ n_] := With[{F = Fibonacci}, (1 + F[n+1]*F[n+2] + F[n+n])/2]; (* Michael Somos, Mar 03 2023 *) -
PARI
{a(n) = my(F = fibonacci); (1 + F(n+1)*F(n+2) + F(n+n))/2}; /* Michael Somos, Mar 03 2023 */
Formula
G.f.: (1-x-x^2)/((1-x^2)(1-3x+x^2));
a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4);
a(n) = F(2n) + 1 + Sum_{k=0..n-1} F(k)*F(k+1).
From R. J. Mathar, Jul 22 2010: (Start)
a(n) = Sum_{i=0..n} A061646(i).
a(n) = (5 + (-1)^n + 4*A002878(n))/10. (End)
a(n) = A110034(-n) = 1 - A110034(1+n) = A236438(n) + (n mod 2) = (1 + F(n+1)*F(n+2) + F(2*n))/2 for all n in Z. - Michael Somos, Mar 03 2023
Comments