A261721 - OEIS

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, 1, 14, 55, 140, 280, 476, 714, 960, 1155, 1210, 1001, 1

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

1,3

COMMENTS

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).

E.g.f. as array: exp(y)*(exp(x)*(24 + 24*(3 + x)*y + 36*(1 + x)*y^2 + 4*(1 + 3*x)*y^3 + x*y^4) - 4*(6 + 18*y + 9*y^2 + y^3))/24. - Stefano Spezia, Aug 15 2024

EXAMPLE

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

...

(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: */