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.

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A210472 Number A(n,k) of paths starting at {n}^k to a border position where one component equals 0 using steps that decrement one component by 1; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 6, 1, 0, 1, 4, 33, 20, 1, 0, 1, 5, 196, 543, 70, 1, 0, 1, 6, 1305, 22096, 10497, 252, 1, 0, 1, 7, 9786, 1304045, 3323092, 220503, 924, 1, 0, 1, 8, 82201, 106478916, 1971644785, 574346824, 4870401, 3432, 1, 0
Offset: 0

Views

Author

Alois P. Heinz, Jan 22 2013

Keywords

Examples

			A(0,3) = 1: [(0,0,0)].
A(1,1) = 1: [(1), (0)].
A(1,2) = 2: [(1,1), (0,1)], [(1,1), (1,0)].
A(1,3) = 3: [(1,1,1), (0,1,1)], [(1,1,1), (1,0,1)], [(1,1,1), (1,1,0)].
A(2,1) = 1: [(2), (1), (0)].
A(2,2) = 6: [(2,2), (1,2), (0,2)], [(2,2), (1,2), (1,1), (0,1)], [(2,2), (1,2), (1,1), (1,0)], [(2,2), (2,1), (1,1), (0,1)], [(2,2), (2,1), (1,1), (1,0)], [(2,2), (2,1), (2,0)].
Square array A(n,k) begins:
  0, 1,   1,      1,         1,             1, ...
  0, 1,   2,      3,         4,             5, ...
  0, 1,   6,     33,       196,          1305, ...
  0, 1,  20,    543,     22096,       1304045, ...
  0, 1,  70,  10497,   3323092,    1971644785, ...
  0, 1, 252, 220503, 574346824, 3617739047205, ...
		

Crossrefs

Columns k=0-4 give: A000004, A000012, A000984, A209245, A209288.
Rows n=0-3 give: A057427, A001477, A093964, A210486.
Main diagonal gives A276490.
Cf. A089759 (unrestricted paths), A225094, A262809, A263159.

Programs

  • Maple
    b:= proc() option remember; `if`(nargs=0, 0, `if`(args[1]=0, 1,
          add(b(sort(subsop(i=args[i]-1, [args]))[]), i=1..nargs)))
        end:
    A:= (n, k)-> b(n$k):
    seq(seq(A(n, d-n), n=0..d), d=0..10);
  • Mathematica
    b[] = 0; b[args__] := b[args] = If[First[{args}] == 0, 1, Sum[b @@ Sort[ReplacePart[{args}, i -> {args}[[i]] - 1]], {i, 1, Length[{args}]}]]; a[n_, k_] := b @@ Array[n&, k]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 12 2013, translated from Maple *)

A209245 Main diagonal of the triple recurrence x(i,j,k) = x(i-1,j,k) + x(i,j-1,k) + x(i,j,k-1) with x(i,j,k) = 1 if 0 in {i,j,k}.

Original entry on oeis.org

1, 3, 33, 543, 10497, 220503, 4870401, 111243135, 2602452993, 61985744967, 1497148260033, 36566829737727, 901314269530113, 22385640256615743, 559574590912019457, 14065064484334380543, 355222860485671141377, 9008982166319523972903, 229325469394627488082497
Offset: 0

Views

Author

Jon Perry, Jan 13 2013

Keywords

Comments

Level sums are defined as the sum of x(i,j,k) with i,j,k >= 0 and i+j+k = n. This gives 3*A164039(n-1) for n>0.
Slice x(1,j,k) with j,k >= 0 of the cube begins:
1, 1, 1, 1, 1, 1, 1, 1, ... A000012
1, 3, 5, 7, 9, 11, 13, 15, ... A005408
1, 5, 11, 19, 29, 41, 55, 71, ... A028387
1, 7, 19, 39, 69, 111, 167, 239, ... A108766(k+1)
1, 9, 29, 69, 139, 251, 419, 659, ...
1, 11, 41, 111, 251, 503, 923, 1583, ...
1, 13, 55, 167, 419, 923, 1847, 3431, ...
1, 15, 71, 239, 659, 1583, 3431, 6863, ...
The main diagonal of the slice is A134760.

Crossrefs

Column k=3 of A210472. - Alois P. Heinz, Jan 23 2013

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, 2*n+1,
          ((888-3020*n+3668*n^2-1912*n^3+364*n^4) *a(n-1)
           +3*(3*n-4)*(7*n-5)*(2*n-3)*(3*n-5) *a(n-2)) /
           ((2*n-1)*(7*n-12)*(n-1)^2))
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Jan 17 2013
  • Mathematica
    b[] = 0; b[args__] := b[args] = If[{args}[[1]] == 0, 1, Sum[b @@ Sort[ ReplacePart[{args}, i -> {args}[[i]] - 1]], {i, 1, Length[{args}]}]];
    a[n_] := b @@ Table[n, 3];
    Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jun 03 2018, from Alois P. Heinz's Maple code for A210472 *)

Formula

a(n) = x(n,n,n) with x(i,j,k) = 1 if 0 in {i,j,k} and x(i,j,k) = x(i-1,j,k) + x(i,j-1,k) + x(i,j,k-1) else.
a(n) ~ 3^(3*n+1/2) / (8*Pi*n). - Vaclav Kotesovec, Sep 07 2014
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