cp's OEIS Frontend

A110035 Row sums of an unsigned characteristic triangle for the Fibonacci numbers.

%I A110035 #25 Jan 01 2024 11:37:49
%S A110035 1,2,5,12,31,80,209,546,1429,3740,9791,25632,67105,175682,459941,
%T A110035 1204140,3152479,8253296,21607409,56568930,148099381,387729212,
%U A110035 1015088255,2657535552,6957518401,18215019650,47687540549,124847601996
%N A110035 Row sums of an unsigned characteristic triangle for the Fibonacci numbers.
%C A110035 Rows sums of abs(A110033).
%H A110035 Harvey P. Dale, <a href="/A110035/b110035.txt">Table of n, a(n) for n = 0..1000</a>
%H A110035 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-3,1).
%F A110035 G.f.: (1-x-x^2)/((1-x^2)(1-3x+x^2));
%F A110035 a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4);
%F A110035 a(n) = F(2n) + 1 + Sum_{k=0..n-1} F(k)*F(k+1).
%F A110035 From _R. J. Mathar_, Jul 22 2010: (Start)
%F A110035 a(n) = Sum_{i=0..n} A061646(i).
%F A110035 a(n) = (5 + (-1)^n + 4*A002878(n))/10. (End)
%F A110035 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
%e A110035 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
%t A110035 LinearRecurrence[{3,0,-3,1},{1,2,5,12},50] (* _Harvey P. Dale_, May 01 2022 *)
%t A110035 a[ n_] := With[{F = Fibonacci}, (1 + F[n+1]*F[n+2] + F[n+n])/2]; (* _Michael Somos_, Mar 03 2023 *)
%o A110035 (PARI) {a(n) = my(F = fibonacci); (1 + F(n+1)*F(n+2) + F(n+n))/2}; /* _Michael Somos_, Mar 03 2023 */
%Y A110035 Cf. A110034, A236438.
%K A110035 easy,nonn
%O A110035 0,2
%A A110035 _Paul Barry_, Jul 08 2005