OFFSET
1,3
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, ...
REFERENCES
Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 194.
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)
EXAMPLE
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, ...).
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 *)
CROSSREFS
KEYWORD
AUTHOR
Gary W. Adamson, Aug 29 2015
STATUS
approved