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.

Previous Showing 11-15 of 15 results.

A368650 a(n) = (6*n + 1)!*(9*n + 1)!/((2*n)!*(3*n)!*((5*n + 1)!)^2).

Original entry on oeis.org

1, 2940, 27511848, 324265486545, 4234842288963000, 58626067532977225512, 842744763083824037236800, 12437726604034570811549435040, 187171833825593326056635733697560, 2859197188199406875783449346275416000, 44198453917285616202092687086145825181264, 689863061309915307698539343386922516078167200
Offset: 0

Views

Author

Karol A. Penson, Jan 02 2024

Keywords

Comments

a(n) can be rigorously proven to be an integer for n>=0.

Crossrefs

Cf. A304126, A368545, A082368, A113424.

Programs

  • Maple
    seq((6*n + 1)!*(9*n + 1)!/((2*n)!*(3*n)!*((5*n + 1)!)^2), n=0..12);

Formula

G.f.: hypergeometric10F9([2/9, 1/3, 4/9, 5/9, 2/3, 7/9, 5/6, 8/9, 10/9, 7/6], [2/5, 2/5, 3/5, 3/5, 4/5, 4/5, 1, 6/5, 6/5], (167365651248*z)/9765625).
O.g.f.: hypergeometric10F10([2/9, 1/3, 4/9, 5/9, 2/3, 7/9, 5/6, 8/9, 10/9, 7/6], [2/5, 2/5, 3/5, 3/5, 4/5, 4/5, 1, 1, 6/5, 6/5], (167365651248*z)/9765625).
a(n) = Integral_{x=0..167365651248/9765625} x^n*W(x) dx, n>=0, where W(x) = (78125*MeijerG([[], [-3/5, -3/5, -2/5, -2/5, -1/5, -1/5, 0, 0, 1/5, 1/5]], [[1/6, 1/9, -1/9, -1/6, -2/9, -1/3, -4/9, -5/9, -2/3, -7/9], []], (9765625*x)/167365651248))/(2066242608*Pi). MeijerG is the Meijer G - function. W(x) can be represented as a sum of 10 hypergeometric functions of type 10F9. W(x) can be proven to be a positive function in the interval [0, 167365651248/9765625]. W(x) is singular at x=0 and monotonically decreases to zero at x = 167365651248/9765625. This integral representation as the n-th power moment of the positive function W(x) in the interval [0, 167365651248/9765625] is unique, as W(x) is the solution of the Hausdorff moment problem.

A303790 G.f. satisfies: 120*(1-216*x)*A(x) + (1-3*(1-216*x)^2)*A'(x) - (1-216*x)*(2-216*x)*x*A''(x) = 0, a(0)=1.

Original entry on oeis.org

1, 60, 7380, 1090320, 176978340, 30471320880, 5461962826320, 1007754602437440, 189974650649174820, 36407481107391279600, 7068262344580438681680, 1386636913539840633652800, 274365765112318301005693200, 54676607910763730416065374400
Offset: 0

Views

Author

Bradley Klee, Apr 30 2018

Keywords

Comments

The surface "u = 2H = p^2 + q^2 - (4/27)*q^6" determines a Picard-Fuchs equation, "5*u*T(u) + 9*(3*u^2-1)*T'(u) + 9*(u^2-1)*u*T''(u) = 0", (cf. link to "Proof Certificate"). The Picard-Fuchs differential equation transforms to the defining relation by "u->1-216*x". G.f. A(x) generates coefficients of the complex period-energy function, while the real period-energy function can be written in terms of hypergeometric A113424. These results agree with Kreshchuk and Gulden, as "d/du(5*u*T(u) + 9*(3*u^2-1)*T'(u) + 9*(u^2-1)*u*T''(u)) = 5*T(u) + 59*u*T'(u) + 18*(3*u^2-1)*T''(u) + 9*u*(u^2-1)*T'''(u) = 0" (cf. Eq. 16).

Examples

			G.f. = 1 + 60*x + 7380*x^2 + 1090320*x^3 + 176978340*x^4 + 30471320880*x^5 + ... _Michael Somos_, Jun 22 2018

Links

Crossrefs

Real Period: A113424.

Programs

  • Mathematica
    a[0] = 1; a[1] = 60;
a[n0_] := a[n0] = ReplaceAll[Dot[Divide[
{5-27*n+27*n^2,(5-3*n)*(-1+3*n)},18*n^2],
{216*a[n0-1],(216^2)*a[n0-2]}],n->n0]
a /@ Range[0, 15]
(* Second program: *)
CoefficientList[Series[Hypergeometric2F1[1/6, 5/6, 1, 432*x - 46656*x^2],{x,0,20}], x]

Formula

G.f.: 2F1(1/6, 5/6; 1; 432*x - 46656*x^2).
D-finite with recurrence a(0) = 1; a(1) = 60; a(n) = (c1/c0)*216*a(n-1) + (c2/c0)*216^2*a(n-2); with c1 = 5-27*n+27*n^2; c2 = (5-3*n)*(-1+3*n); c0 = 18*n^2.
a(n) ~ 6^(3*n) / (Pi*n). - Vaclav Kotesovec, May 01 2018

A368513 a(n) = (6*n+1)!/(n!*(2*n)!*(3*n+1)!).

Original entry on oeis.org

1, 105, 25740, 7759752, 2574148500, 902522205585, 328074738591600, 122332313750680800, 46485667563689950596, 17924037162454524601500, 6991900927489809108938160, 2753354160571011216583946400, 1092796344333659321191117573200, 436609643814534385348768088729640, 175431288302508215774213129529432000
Offset: 0

Views

Author

Karol A. Penson, Dec 28 2023

Keywords

Crossrefs

Cf. A113424.

Programs

  • Maple
    seq((6*n+1)!/(n!*(2*n)!*(3*n+1)!),n=0..14)

Formula

G.f.: hypergeometric3F2([1/3, 5/6, 7/6], [1, 4/3], 432*z).
E.g.f.: hypergeometric3F3([1/3, 5/6, 7/6], [1, 1, 4/3], 432*z).
a(n) = Integral_{x=0..432} x^n*W(x) dx, n>=0, where W(x) = sqrt(3)/(12*Pi*x^(2/3)) - gamma(2/3)*gamma(5/6)*sqrt(3)*hypergeometric3F2([1/2, 5/6, 5/6], [2/3, 3/2], x/432)/(432*Pi^(5/2)*x^(1/6)) + x^(1/6)*hypergeometric3F2([5/6, 7/6, 7/6], [4/3, 11/6], x/432)/(12960*sqrt(Pi)*gamma(2/3)*gamma(5/6)). W(x) is positive on x = [0, 432], it diverges at x=0, and monotonically decreases for x>0. It appears that at x=432, W(x) tends to a constant value close to 0.0007368284. This integral representation as the n-th power moment of the positive function W(x) on the interval [0, 432] is unique, as W(x) is the solution of the Hausdorff moment problem.
The shape of W(x) in the above integral representation of a(n) resembles very much the shape of the corresponding W(x) in A113424.

A368692 a(n) = (12*n + 6)!*(6*n + 9)!/(108*(4*n + 2)!*(2*n + 3)!*((6*n + 5)!)^2).

Original entry on oeis.org

14, 563108, 54231252075, 6700034035890000, 928978310614152999200, 137569863175651804211692560, 21253098849879053645154605945160, 3381375421559384124434964404229384000, 549714622911935710495977183989400234273000
Offset: 0

Views

Author

Karol A. Penson, Jan 03 2024

Keywords

Comments

According to A. Adolphson and S. Sperber, "On the integrality of hypergeometric series whose coefficients are factorial ratios", ArXiv: 2001.03296, s.page 14, first equation after Eq.(7.4): for any two integers K, L, the ratios (3*K)!*(3*L)!/(K!*L!*((K+L)!)^2) are proven to be integers. 108*a(n) results from K = 4*n+2 and L = 2*n+3, n>=0. It is conjectured here that a(n) are integers.

Links

Crossrefs

Cf. A368650, A304126, A368545, A082368, A113424, A368875.

Programs

  • Maple
    seq((12*n + 6)!*(6*n + 9)!/(108*(4*n + 2)!*(2*n + 3)!*((6*n + 5)!)^2),n=0..9);

Formula

G.f.: 14*hypergeometric8F7([7/12, 2/3, 5/6, 11/12, 13/12, 17/12, 13/6, 7/3], [1, 7/6, 4/3, 3/2, 3/2, 5/3, 11/6], 186624*z).
E.g.f.: 14*hypergeometric8F8([7/12, 2/3, 5/6, 11/12, 13/12, 17/12, 13/6, 7/3], [1, 1, 7/6, 4/3, 3/2, 3/2, 5/3, 11/6], 186624*z).
a(n) = Integral_{x=0..186624} x^n*W(x) dx, n>=0, where W(x) = (1/(20736*Pi))*MeijerG([[], [0, 0, 1/6, 1/3, 1/2, 1/2, 2/3, 5/6]], [[-5/12, -1/3, -1/6, -1/12, 1/12, 5/12, 7/6, 4/3], []], x/186624). MeijerG is the Meijer G - function. W(x) can be represented as an expression containing the sum of 4 generalized hypergeometric functions of type 8F7. W(x) is a positive function in the interval [0, 186624], is singular at x=0 and monotonically decreases to zero at x = 186624. This integral representation as the n-th power moment of the positive function W(x) in the interval [0, 186624] is unique, as W(x) is the solution of the Hausdorff moment problem.
Let b(n) = Gamma(7+ 12*n)/(6*Gamma(2 + 2*n)*Gamma(3 + 4*n)*Gamma(6 + 6*n)), then a(n) = b(n) * A272399(n+2). - Peter Luschny, Jan 06 2024

A368875 a(n) = 24*(3*n + 1)!/(n!*((n + 2)!)^2).

Original entry on oeis.org

6, 16, 105, 1008, 12012, 164736, 2494206, 40646320, 701149020, 12655450080, 237026033790, 4577828250240, 90739095674400, 1838979005667840, 37993593597567210, 798259862714284080, 17022152442879594780, 367791659430639444000, 8040845154302354844450
Offset: 0

Views

Author

Karol A. Penson, Jan 08 2024

Keywords

Comments

According to A. Adolphson and S. Sperber (see Links), see page 14, second equation after Eq.(7.4): for any two integers K, L, the ratios (3*K+1)!*(3*L+1)!/(K!*L!*((K+L+1)!)^2) are proven to be integers. Here a(n) results from K = 1 and L = n, n >= 0.

Links

Crossrefs

Cf. A368650, A304126, A368545, A082368, A113424, A368692.

Programs

  • Maple
    seq(24*(3*n + 1)!/(n!*((n + 2)!)^2),n=0..17);
  • Mathematica
    Table[24*(3*n + 1)!/(n!*((n + 2)!)^2),{n,0,16}] (* James C. McMahon, Jan 08 2024 *)
  • SageMath
    def a(n): return (24 * (n + 1) * (n + 2) * gamma(3*n + 2)) / gamma(n + 3)^3
print([a(n) for n in range(19)])  # Peter Luschny, Jan 09 2024

Formula

G.f.: 6*hypergeometric3F2([2/3, 1, 4/3], [3, 3], 27*z).
G.f.: -(hypergeometric2F1([-4/3, -2/3], [1], 27*z) - 1)/(3*z^2) + 8/z.
E.g.f.: 6*hypergeometric3F3([2/3, 1, 4/3], [3, 3, 1], 27*z).
a(n) = Integral_{x=0..27} x^n*W(x) dx, n >= 0, where
W(x) = (243*2^(2/3)*Gamma(5/6)*Gamma(2/3)*hypergeometric2F1([-4/3, -4/3], [1/3], x/27)) / (16*Pi^(5/2)*x^(1/3)) - (3*sqrt(3)*2^(1/3)*x^(1/3)* hypergeometric2F1([-2/3, -2/3], [5/3], x/27))/(2*sqrt(Pi)*Gamma(5/6)* Gamma(2/3)).
W(x) is a positive function in the interval [0, 27], is singular at x = 0 with the singularity x^(-1/3), and monotonically decreases to zero at x = 27, with W'(x) tending to zero at x = 27. This integral representation as the n-th power moment of the positive function W(x) in the interval [0, 27] is unique, as W(x) is the solution of the Hausdorff moment problem.
Previous Showing 11-15 of 15 results.

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