A227583 Number of lattice paths from {n}^4 to {0}^4 using steps that decrement one component such that for each point (p_1,p_2,...,p_4) we have p_1<=p_2<=...<=p_4.

1, 1, 42, 3532, 456033, 77767945, 16104165970, 3848596333400, 1026843977181745, 298985252352030713, 93462550593036735356, 30988255904733937513266, 10797084830552485796491313, 3924879717822914648655364113, 1479988943331198988162565625314

Offset: 0

Author

Alois P. Heinz, Jul 16 2013

Keywords

• nonn

Examples

			a(0) = 1: [(0,0,0,0)].
a(1) = 1: [(1,1,1,1),(0,1,1,1),(0,0,1,1),(0,0,0,1),(0,0,0,0)].
a(2) = 42: [(2,2,2,2),(0,2,2,2),(0,0,2,2),(0,0,0,2),(0,0,0,0)], ..., [(2,2,2,2),(1,2,2,2),(1,1,2,2),(1,1,1,2),(1,1,1,1),(0,1,1,1),(0,0,1,1),(0,0,0,1),(0,0,0,0)].
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..200

Crossrefs

Column k=4 of A227578.

Programs

Maple

b:= proc(l) option remember; `if`(l[-1]=0, 1, add(add(b(subsop(
      i=j, l)), j=`if`(i=1, 0, l[i-1])..l[i]-1), i=1..nops(l)))
    end:
a:= n-> `if`(n=0, 1, b([n$4])):
seq(a(n), n=0..16);
# second Maple program:
a:= proc(n) option remember; `if`(n<4, [1, 1, 42, 3532][n+1], (
      (n+2)*(176594184*n^7-1252171596*n^6+3635870170*n^5
      -5627048591*n^4 [phone]*n^3 -2520013645*n^2+664396290*n
      -69572160) *(n+1)^2* a(n-1) -3*(n-2)*(n+1)*(350669592*n^8
      -2376848760*n^7+6486604406*n^6-9505222590*n^5
      [phone]*n^4-4746390780*n^3+1536942539*n^2-253906050*n
      [phone]) *a(n-2) +(n-3)*(1575354456*n^7-8934392532*n^6
      [phone]*n^5-19551471603*n^4 [phone]*n^3
      -4838385873*n^2+1019903146*n-96686592)*(n-2)^2 *a(n-3)
      -5000*(n-4)*(139932*n^4-226356*n^3+87227*n^2-21684*n-1919)
      *(n-3)^2 *(n-2)^3 *a(n-4)) / (2*(n+3)*(139932*n^4-786084*n^3
      +1605887*n^2-1434934*n+473280)*(n+2)^2*(n+1)^3))
    end:
seq(a(n), n=0..20);

Mathematica

b[l_] := b[l] = If[l[[-1]] == 0, 1, Sum[Sum[b[ReplacePart[l, i -> j]], {j, If[i == 1, 0, l[[i - 1]]], l[[i]] - 1}], {i, 1, Length[l]}]];
a[n_] := b[Array[n&, 4]];
a /@ Range[0, 20] (* Jean-François Alcover, Dec 20 2020, after Alois P. Heinz in A227578 *)

Formula

a(n) ~ 3*5^(4*n+11)/(2^13*Pi^(3/2)*(3*n)^(15/2)). - Vaclav Kotesovec, Jul 19 2013