cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A253217 Number of n X n nonnegative integer arrays with upper left 0 and lower right its king-move distance away minus 2 and every value within 2 of its king move distance from the upper left and every value increasing by 0 or 1 with every step right, diagonally se or down.

Original entry on oeis.org

0, 0, 1, 19, 268, 3568, 47698, 649712, 9023385, 127419681, 1823918697, 26398702645, 385582981615, 5674890516295, 84060883775765, 1252066289632643, 18738613233957420, 281620474177057788, 4248088188086420832
Offset: 1

Views

Author

R. H. Hardin, Dec 29 2014

Keywords

Comments

Diagonal of A253223.

Examples

			Some solutions for n=4:
  0  1  1  1    0  1  1  1    0  0  0  1    0  0  0  1    0  0  0  1
  0  1  1  1    0  1  1  1    0  0  1  1    0  0  0  1    0  0  1  1
  0  1  1  1    1  1  1  1    0  1  1  1    0  0  1  1    0  0  1  1
  1  1  1  1    1  1  1  1    1  1  1  1    1  1  1  1    1  1  1  1
		

Formula

Recurrence: 32*(1 + n)*(1 + 2*n)^2*(161046 + 465785*n + 551943*n^2 + 343020*n^3 + 117954*n^4 + 21285*n^5 + 1575*n^6)*a(n) - 8*(4443102 + 33718283*n + 105734340*n^2 + 180574335*n^3 + 186866686*n^4 + 122556360*n^5 + 51280818*n^6 + 13267683*n^7 + 1933470*n^8 + 121275*n^9)*a(n+1) + 2*(12137328 + 91378536*n + 283626704*n^2 + 478464380*n^3 + 488415476*n^4 + 315713355*n^5 + 130145646*n^6 + 33170868*n^7 + 4763070*n^8 + 294525*n^9)*a(n+2) + (10688508 + 80866406*n + 252913504*n^2 + 431097970*n^3 + 445804136*n^4 + 292620525*n^5 + 122735586*n^6 + 31877118*n^7 + 4668570*n^8 + 294525*n^9)*a(n+3) - (4877748 + 36871922*n + 114948300*n^2 + 194784258*n^3 + 199650088*n^4 + 129484209*n^5 + 53503836*n^6 + 13655808*n^7 + 1961820*n^8 + 121275*n^9)*a(n+4) + 2*(3 + n)^2*(7 + 2*n)*(2428 + 16118*n + 41382*n^2 + 52554*n^3 + 35154*n^4 + 11835*n^5 + 1575*n^6)*a(n+5) = 0. - conjectured by Manuel Kauers and Christoph Koutschan, Mar 02 2023; proved by Robert Dougherty-Bliss and Manuel Kauers
Conjecture: a(n) ~ 2^(4*n - 2) / (81 * Pi * n), based on the above recurrence - Vaclav Kotesovec, Mar 02 2023