A261720 Array of pyramidal (3-dimensional figurate numbers) read by antidiagonals.

1, 1, 4, 1, 5, 10, 1, 6, 14, 20, 1, 7, 18, 30, 35, 1, 8, 22, 40, 55, 56, 1, 9, 26, 50, 75, 91, 84, 1, 10, 30, 60, 95, 126, 140, 120, 1, 11, 34, 70, 115, 161, 196, 204, 165, 1, 12, 38, 80, 135, 196, 252, 288, 285, 220, 1, 13, 42, 90, 155, 231, 308, 372, 405, 385, 286

Offset: 1

Author

Gary W. Adamson, Aug 29 2015

Keywords

• nonn
• tabl
• easy

Comments

First few sequences in the array:

1, 4, 10, 20, 35, 56, 84, 120, 165, 220, ... A000292

1, 5, 14, 30, 55, 91, 140, 204, 285, 385, ... A000330

1, 6, 18, 40, 75, 126, 196, 288, 405, 550, ... A002411

1, 7, 22, 50, 95, 161, 252, 372, 525, 715, ... A002412

1, 8, 26, 60, 115, 196, 308, 456, 645, 880, ... A002413

1, 9, 30, 70, 135, 231, 364, 540, 765, 1045, ... A002414

1, 10, 34, 80, 155, 266, 420, 624, 885, 1210, ... A007584

...

The corresponding bases to rows are: Triangle, Square, Pentagon, Hexagon, Heptagon, Octagon, ...

Examples

			Row 2: (1, 5, 14, 30, 55, ...) = (1, 4, 10, 20, 35, ...) + (0, 1, 4, 10, 20, 35, ...).
(1, 7, 22, 50, ...) is the binomial transform of (1, 6, 9, 4, 0, 0, 0, ...) 3rd row in Pascal's triangle (1,4) followed by zeros. (1, 7, 22, 50, ...) is the third partial sum of (1, 4, 4, 4, ...).
		

References

• Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 194.

Crossrefs

Cf. A000292, A000330, A002411, A002412, A002413, A002414, A007584, A220084.

Similar to A080851 but without row n=0.

Programs

Mathematica

T[n_,k_]:=k(k+1)((k-1)n+3)/6; Flatten[Table[T[n-k+1,k],{n,11},{k,n}]] (* Stefano Spezia, Aug 15 2024 *)

Formula

T(n,k) = A080851(n,k).

Given: first sequence in the array is A000292: (1, 4, 10, 20, 35, ...) Subsequent rows are generated by adding (0, 1, 4, 10, 20, 35, ...) to the current row.

n-th row is the binomial transform of row 3 in Pascal's triangle (1,n) followed by zeros. Alternatively, begin with (1, 4, 10, 20, ...) being the binomial transform of (1, 3, 3, 1, 0, 0, 0, ...). Add (0, 1, 2, 1, 0, 0, 0, ...) to the latter to obtain the inverse binomial transform of the next row: (1, 5, 14, 30, 55,..); then repeat the operation.

The row starting (1, N, ...) is the 3rd partial sum of (1, (N-3), (N-3), (N-3), ...).

From Stefano Spezia, Aug 15 2024: (Start)

T(n,k) = k*(k + 1)*((k - 1)*n + 3)/6.

G.f. as array: x*y*(1 + x*(y - 1))/((1 - x)^2*(1 - y)^4).

E.g.f. as array: exp(y)*y*(exp(x)*(6 + 3*(1 + x)*y + x*y^2) - 3*(2 + y))/6. (End)