cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A300946 Rectangular array A(n, k) = (-1)^k*hypergeom([-k, k + n/2 - 1/2], [1], 4) with row n >= 0 and k >= 0, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 19, 1, 5, 33, 239, 1, 7, 51, 387, 3011, 1, 9, 73, 587, 4737, 38435, 1, 11, 99, 847, 7123, 59523, 496365, 1, 13, 129, 1175, 10321, 89055, 761121, 6470385, 1, 15, 163, 1579, 14499, 129367, 1135005, 9854211, 84975315
Offset: 0

Views

Author

Peter Luschny, Mar 16 2018

Keywords

Examples

			Array starts:
[0] 1,  1,  19,  239,  3011,  38435,  496365,  6470385, ... [A299864]
[1] 1,  3,  33,  387,  4737,  59523,  761121,  9854211, ... [A299507]
[2] 1,  5,  51,  587,  7123,  89055, 1135005, 14660805, ... [A245926]
[3] 1,  7,  73,  847, 10321, 129367, 1651609, 21360031, ... [A084768]
[4] 1,  9,  99, 1175, 14499, 183195, 2351805, 30539241, ... [A245927]
[5] 1, 11, 129, 1579, 19841, 253707, 3284737, 42924203, ...
[6] 1, 13, 163, 2067, 26547, 344535, 4508877, 59402397, ...
		

Crossrefs

Programs

  • Mathematica
    Arow[n_, len_] := Table[(-1)^k Hypergeometric2F1[-k, k + n/2 - 1/2, 1, 4], {k, 0, len}]; Table[Print[Arow[n, 7]], {n, 0, 6}];

A299507 a(n) = (-1)^n*hypergeom([-n, n], [1], 4).

Original entry on oeis.org

1, 3, 33, 387, 4737, 59523, 761121, 9854211, 128772609, 1694927619, 22437369633, 298419470979, 3984500221569, 53376363001731, 717044895641121, 9656091923587587, 130310873022310401, 1761872309456567811, 23861153881099854369, 323634591584064809859
Offset: 0

Views

Author

Peter Luschny, Mar 16 2018

Keywords

Crossrefs

Programs

  • Maple
    seq(simplify( (-1)^n*hypergeom([-n, n], [1], 4)), n = 0..20); # Peter Bala, Apr 18 2024
  • Mathematica
    a[n_] := (-1)^n Hypergeometric2F1[-n, n, 1, 4]; Table[a[n], {n, 0, 19}]

Formula

From Vaclav Kotesovec, Jul 05 2018: (Start)
Recurrence: n*(2*n-3)*a(n) = 2*(14*n^2 - 28*n + 11)*a(n-1) - (n-2)*(2*n-1)*a(n-2).
a(n) ~ 2^(-3/2) * 3^(1/4) * (7 + 4*sqrt(3))^n / sqrt(Pi*n). (End)
From Peter Bala, Apr 18 2024: (Start)
a(n) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k-1, k-1)*3^k = R(n, 3) for n >= 1, where R(n, x) denotes the n-th row polynomial of A253283.
a(n) = 3*n* hypergeom([1 - n, n + 1], [2], -3) for n >= 1.
a(n) = (1/2)*(LegendreP(n, 7) - LegendreP(n-1, 7)) for n >= 1.
a(n) = [x^n] ( (1 - x)/(1 - 4*x) )^n.
It follows that the Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
G.f.: (sqrt(x^2 - 14*x + 1) - x + 1)/(2*sqrt(x^2 - 14*x + 1)) = 1 + 3*x + 33*x^2 + 387*x^3 + .... (End)
Showing 1-2 of 2 results.