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.

A229465 Number of lattice paths from {2}^n to {0}^n using steps that decrement one component or all components by the same positive integer.

Original entry on oeis.org

1, 2, 22, 248, 6506, 292442, 19450082, 1781791202, 214899390722, 33007840951682, 6290830043769602, 1456812593474515202, 402910665233497344002, 131173228963457333452802, 49656810289226589524275202, 21628258853895326260083456002, 10739534026001485870629015552002
Offset: 0

Views

Author

Alois P. Heinz, Sep 24 2013

Keywords

Crossrefs

Row n=2 of A229345.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<5, [1, 2, 22, 248, 6506][n+1],
         ((64481193996*n^5 -656050382562*n^4 +1835465682464*n^3
          -3691825299357*n^2 +10428520019257*n -9978603085078)*a(n-1)
          -(64481193996*n^6 -251022627918*n^5 -4253631972584*n^4
          +29686486719123*n^3 -71916661134305*n^2 +77149141951487*n
          -30090569866279)*a(n-2) +(n-2)*(437268351642*n^5
          -5777340617365*n^4 +26203609431616*n^3 -50411340883791*n^2
          +38226810988733*n -9795152028455)*a(n-3) -(n-2)*(n-3)*
          (170273280324*n^4 -2136687453608*n^3 +8692120865702*n^2
          -11643795721897*n +4287224601259)*a(n-4) -(n-6)*(n-2)*(n-3)*
          (n-4)*(202513877322*n^2-310611483677*n+98391999767)*a(n-5))/
          (32240596998*n^3-328025191281*n^2+768115007074*n-189524735891))
        end:
    seq(a(n), n=0..20);
  • Mathematica
    b[l_] := b[l] = With[{m = Length[l]}, If[m == 0 || l[[m]] == 0, 1, If[m > 1, Sum[b[l - Array[j&, m]], {j, 1, l[[1]]}], 0] + Sum[Sum[b[Sort[ ReplacePart[l, i -> l[[i]] - j]]], {j, 1, l[[i]]}], {i, 1, m}]]];
    a[k_] := b[Array[2&, k]];
    a /@ Range[0, 20] (* Jean-François Alcover, Dec 22 2020, after Alois P. Heinz in A229345 *)

Formula

a(n) ~ sqrt(Pi) * 2^(n+1) * n^(2*n+1/2) / exp(2*n-1). - Vaclav Kotesovec, Jul 16 2014