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A261721
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Fourth-dimensional figurate numbers.
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3
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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
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OFFSET
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1,3
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COMMENTS
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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, ...)
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REFERENCES
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Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 195 (Table 80).
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LINKS
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FORMULA
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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
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EXAMPLE
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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, ...).
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MAPLE
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A:= (n, k)-> binomial(k+3, 3) + n*binomial(k+3, 4):
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MATHEMATICA
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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;
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PROG
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(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: */
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CROSSREFS
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Similar to A080852 but without row n=0.
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KEYWORD
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AUTHOR
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STATUS
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approved
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