A351704 Sums of the ascending diagonals of the triangle A162609.

1, 1, 2, 3, 7, 10, 20, 26, 45, 55, 86, 101, 147, 168, 232, 260, 345, 381, 490, 535, 671, 726, 892, 958, 1157, 1235, 1470, 1561, 1835, 1940, 2256, 2376, 2737, 2873, 3282, 3435, 3895, 4066, 4580, 4770, 5341, 5551, 6182, 6413, 7107, 7360, 8120, 8396, 9225, 9525, 10426, 10751, 11727, 12078, 13132, 13510, 14645

Offset

0,3

Comments

Each term is the sum of an ascending diagonal of the triangle A162609.

Links

Andrew Howroyd, Table of n, a(n) for n = 0..1000

Eddie Gutierrez, The Wheel Octagon Triangle - Ascending Diagonals (Part VIIIb)

Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Formula

a(n) = (n^3 + 2*n + 12)/12, for even n.

a(n) = (2*n^3 - 3*n^2 + 10*n + 15)/24, for odd n.

G.f.: (1 - 2*x^2 + x^3 + 4*x^4)/((1 - x)^4*(1 + x)^3). - Andrew Howroyd, Nov 15 2025

E.g.f.: ((24 + 9*x + 6*x^2 + 2*x^3)*cosh(x) + (15 + 6*x + 3*x^2 + 2*x^3)*sinh(x))/24. - Stefano Spezia, Nov 15 2025

Example

a(4) = (64 + 8 + 12)/12 = 7
a(5) = (250 - 75 + 50 + 15)/24 = 10.

Prog

(C) // Calculates and prints out the triangle and terms of ascending diagonals (on first line). To get more terms increment j.
#include <stdio.h>
int main()
{
   int n, j=8, k, C, F1, F2, s;
   F1=1; F2=1;
   printf("%d ", F1);
   printf("%d ", F2);
   for (s=0; s<=j; s++)
   {
      F1=F1 + 2*s*s + 2*s + 1;
      F2=F2 + 2*s*s + 3*s + 2;
      printf("%d ", F1);
      printf("%d ", F2);
   }
   printf("\n");
   return 0;
}
(PARI) a(n) = if (n%2, 2*n^3 - 3*n^2 + 10*n + 15, 2*(n^3 + 2*n + 12))/24; \\ Andrew Howroyd, Nov 15 2025

Crossrefs

Cf. A162609.

Sequence in context: A355726 A356364 A173132 * A320675 A336414 A095010

Adjacent sequences: A351701 A351702 A351703 * A351705 A351706 A351707

Keyword

nonn,easy

Author

Eddie Gutierrez, May 05 2022

Status

approved