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%I #33 Apr 15 2020 18:05:26
%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
%N Fourth-dimensional figurate numbers.
%C The array as shown in A257200:
%C 1, 5, 15, 35, 70, 126, 210, 330, ... A000332
%C 1, 6, 20, 50, 105, 196, 336, 540, ... A002415
%C 1, 7, 25, 65, 140, 266, 462, 750, ... A001296
%C 1, 8, 30, 80, 175, 336, 588, 960, ... A002417
%C 1, 9, 35, 95, 210, 406, 714, 1170, ... A002418
%C 1, 10, 40, 110, 245, 476, 840, 1380, ... A002419
%C ...
%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).
%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
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