A261721 Fourth-dimensional figurate numbers.

1, 1, 5, 1, 6, 15, 1, 7, 20, 35, 1, 8, 25, 50, 70, 1, 9, 30, 65, 105, 126, 1, 10, 35, 80, 140, 196, 210, 1, 11, 40, 95, 175, 266, 336, 330, 1, 12, 45, 110, 210, 336, 462, 540, 495, 1, 13, 50, 125, 245, 406, 588, 750, 825, 715

Offset

1,3

Comments

The array as shown in A257200:

1, 5, 15, 35, 70, 126, 210, 330, ... A000332

1, 6, 20, 50, 105, 196, 336, 540, ... A002415

1, 7, 25, 65, 140, 266, 462, 750, ... A001296

1, 8, 30, 80, 175, 336, 588, 960, ... A002417

1, 9, 35, 95, 210, 406, 714, 1170, ... A002418

1, 10, 40, 110, 245, 476, 840, 1380, ... A002419

...

Generating polygons for the sequences are: Triangle, Square, Pentagon, Hexagon, Heptagon, Octagon, ... .

n-th row sequence is the binomial transform of the fourth row of Pascal's triangle (1,n) followed by zeros; and the fourth partial sum of (1, n, n, n, ...).

n-th row sequence is the binomial transform of:

((n-1) * (0, 1, 3, 3, 1, 0, 0, 0) + (1, 4, 6, 4, 1, 0, 0, 0)).

Given the n-th row of the array (1, b, c, d, ...), the next row of the array is (1, b, c, d, ...) + (0, 1, 5, 15, 35, ...)

References

Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 195 (Table 80)

Links

Alois P. Heinz, Antidiagonals n = 1..141, flattened

Index to sequences related to pyramidal numbers

Formula

G.f. for row n: (1 + (n-1)*x)/(1 - x)^5.

A(n,k) = C(k+3,3) + n * C(k+3,4) = A080852(n,k).

Example

(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of the fourth row of Pascal's triangle (1,3) followed by zeros: (1, 6, 12, 10, 3, 0, 0, 0, ...); and the fourth partial sum of (1, 3, 3, 3, 0, 0, 0).
(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of: ((2 * (0, 1, 3, 3, 1, 0, 0, 0, ...) + (1, 4, 6, 4, 1, 0, 0, 0, ...)); that is, the binomial transform of (1, 6, 12, 10, 3, 0, 0, 0, ...).
Row 2 of the array is (1, 5, 15, 35, 70, ...) + (0, 1, 5, 15, 35, ...), = (1, 6, 20, 50, 105, ...).

Maple

A:= (n, k)-> binomial(k+3, 3) + n*binomial(k+3, 4):
seq(seq(A(d-k, k), k=0..d-1), d=1..13);  # Alois P. Heinz, Aug 31 2015

Mathematica

row[1] = LinearRecurrence[{5, -10, 10, -5, 1}, {1, 5, 15, 35, 70}, m = 10];
row1 = Join[{0}, row[1] // Most]; row[n_] := row[n] = row[n-1] + row1;
Table[row[n-k+1][[k]], {n, 1, m}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 26 2016 *)

Prog

(PARI) A(n, k) = binomial(k+3, 3) + n*binomial(k+3, 4)
table(n, k) = for(x=1, n, for(y=0, k-1, print1(A(x, y), ", ")); print(""))
/* Print initial 6 rows and 8 columns as follows: */
table(6, 8) \\ Felix Fröhlich, Jul 28 2016

Crossrefs

Cf. A257200, A261720 (pyramidal numbers), A000332, A002415, A001296, A002417, A002418, A002419.

Similar to A080852 but without row n=0.

Main diagonal gives A256859.

Sequence in context: A007397 A362489 A204203 * A275490 A052345 A197733

Adjacent sequences: A261718 A261719 A261720 * A261722 A261723 A261724

Keyword

nonn,tabl,easy

Author

Gary W. Adamson, Aug 30 2015

Status

approved