A261721 - OEIS

%I #43 Aug 16 2024 08:36:04

%S 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,

%T 196,210,1,11,40,95,175,266,336,330,1,12,45,110,210,336,462,540,495,1,

%U 13,50,125,245,406,588,750,825,715,1,14,55,140,280,476,714,960,1155,1210,1001,1

%N Fourth-dimensional figurate numbers.

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

%C 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, ...).

%C n-th row sequence is the binomial transform of:

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

%C 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, ...)

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

%H Alois P. Heinz, <a href="/A261721/b261721.txt">Antidiagonals n = 1..141, flattened</a>

%H <a href="/index/Ps#pyramidal_numbers">Index to sequences related to pyramidal numbers</a>

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

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

%F 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

%e The array as shown in A257200:

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

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

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

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

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

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

%e   ...

%e (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).

%e (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, ...).

%e Row 2 of the array is (1, 5, 15, 35, 70, ...) + (0, 1, 5, 15, 35, ...), = (1, 6, 20, 50, 105, ...).

%p A:= (n, k)-> binomial(k+3, 3) + n*binomial(k+3, 4):

%p seq(seq(A(d-k, k), k=0..d-1), d=1..13);  # _Alois P. Heinz_, Aug 31 2015

%t row[1] = LinearRecurrence[{5, -10, 10, -5, 1}, {1, 5, 15, 35, 70}, m = 10];

%t row1 = Join[{0}, row[1] // Most]; row[n_] := row[n] = row[n-1] + row1;

%t Table[row[n-k+1][[k]], {n, 1, m}, {k, 1, n}] // Flatten (* _Jean-François Alcover_, May 26 2016 *)

%o (PARI) A(n, k) = binomial(k+3, 3) + n*binomial(k+3, 4)

%o table(n, k) = for(x=1, n, for(y=0, k-1, print1(A(x, y), ", ")); print(""))

%o /* Print initial 6 rows and 8 columns as follows: */

%o table(6, 8) \\ _Felix Fröhlich_, Jul 28 2016

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

%Y Similar to A080852 but without row n=0.

%Y Main diagonal gives A256859.

%K nonn,tabl,easy

%O 1,3

%A _Gary W. Adamson_, Aug 30 2015