A217536 Square array read by antidiagonals, where the top row is the nonnegative integers and the other numbers are the sum of the neighbors in the preceding row.

0, 1, 1, 2, 3, 4, 3, 6, 10, 14, 4, 9, 18, 32, 46, 5, 12, 27, 55, 101, 147, 6, 15, 36, 81, 168, 315, 462, 7, 18, 45, 108, 244, 513, 975, 1437, 8, 21, 54, 135, 324, 736, 1564, 3001, 4438, 9, 24, 63, 162, 405, 973, 2222, 4761, 9199, 13637, 10, 27, 72, 189, 486, 1215, 2924, 6710, 14472, 28109, 41746

Offset: 0

Author

WG Zeist, Oct 06 2012

Keywords

• nonn
• tabl

Comments

Each number in the top row of the array is determined by the pre-defined sequence (in this case, the nonnegative integers). Each number in lower rows is the sum of the numbers vertically or diagonally above it (so, the number at the left end of each row is the sum of two numbers, and all other numbers the sum of three).

Replacing the top row with A000012 (the all 1's sequence) and constructing the rest of the array the same way produces A062105. Similarly, replacing the top row with A000007 (a(n) = 0^n) produces A020474. - WG Zeist, Aug 24 2024

For any array constructed with this method, regardless of the sequence chosen for the top row, the sequence in the first column of the array can be computed from the sequence in the top row as follows: let a(0), a(1), a(2), ... be the terms in the top row, and b(0), b(1), b(2), ... the terms in the first column. Then b(n) = Sum_{k=0..n} A064189(n,k) * a(k). The inverse operation, to compute the top row from the first column, is given by a(n) = Sum_{k=0..n} A104562(n,k) * b(k). - WG Zeist, Aug 26 2024

Examples

			The array starts:
  0  1  2  3
  1  3  6  9
  4  10 18 27
  14 32 55 81
		

Links

• Alois P. Heinz, Antidiagonals n = 0..150, flattened

Crossrefs

Main diagonal gives A036290. First column gives A330796.

Cf. A020474, A062105, A064189, A104562.

Programs

Maple

A:= proc(n, k) option remember; `if`(k<0, 0,
     `if`(n=0, k, add(A(n-1, k+i), i=-1..1)))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, Aug 24 2024

Formula

T(m+1,n) = sum(T(m,k), |k-n| <= 1) (and T(0,n)=n), m, n >= 0. - M. F. Hasler, Oct 09 2012

Extensions

Offset 0 from Alois P. Heinz, Aug 24 2024