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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A203517 a(n) = A203516(n)/A000178(n).

Original entry on oeis.org

1, 4, 96, 15360, 17203200, 138726604800, 8203736501452800, 3603868630142209228800, 11873738053102139590311936000, 295578185800614925763054760099840000, 55920479534877093093661639943174183976960000
Offset: 1

Views

Author

Clark Kimberling, Jan 03 2012

Keywords

Links

Crossrefs

Cf. A000178, A000217, A034910, A093883, A203516.

Programs

  • Magma
    [2^Binomial(n,2)*(&*[Binomial(2*k,k): k in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Feb 19 2024
  • Mathematica
    f[j_] := 2 j - 1; z = 15;
v[n_] := Product[Product[f[k] + f[j], {j, 1, k - 1}], {k, 2, n}]
d[n_] := Product[(i - 1)!, {i, 1, n}]   (* A000178 *)
Table[v[n], {n, 1, z}]                  (* A203516 *)
Table[v[n + 1]/(4 v[n]), {n, 1, z - 1}] (* A034910 *)
Table[v[n]/d[n], {n, 1, 20}]            (* A203517 *)
Table[2^(-1/24 - 3*n/2 + 3*n^2/2) * Glaisher^(3/2) * Pi^(1/4 - n/2) * BarnesG[1/2 + n] / E^(1/8) / BarnesG[1 + n], {n, 1, 12}] (* Vaclav Kotesovec, Sep 01 2023 *)
  • SageMath
    [2^binomial(n,2)*product(binomial(2*k,k) for k in range(n)) for n in range(1,21)] # G. C. Greubel, Feb 19 2024

Formula

a(n) ~ A^(3/2) * 2^(-7/24 - 3*n/2 + 3*n^2/2) * exp(-1/8 + n/2) * n^(1/8 - n/2) / Pi^(n/2), where A = A074962 is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Sep 01 2023
a(n) = 2^binomial(n,2) * Product_{j=0..n-1} binomial(2*j, j). - G. C. Greubel, Feb 19 2024

A254620 a(n) = 9^n*(2*n + 1)!/n!.

Original entry on oeis.org

1, 54, 4860, 612360, 99202320, 19642059360, 4596241890240, 1240985310364800, 379741504971628800, 129871594700297049600, 49091462796712284748800, 20323865597838885886003200, 9145739519027498648701440000, 4444829406247364343268899840000
Offset: 0

Views

Author

Peter Bala, Feb 03 2015

Keywords

Crossrefs

Cf. A002162, A002391, A034910, A073000, A105531, A254381, A254619.

Programs

  • Maple
    seq(9^n*(2*n + 1)!/n!, n = 0..14);
  • Mathematica
    Table[9^n (2n+1)!/n!,{n,0,20}] (* Harvey P. Dale, Aug 13 2019 *)

Formula

E.g.f.: 1/(1 - 36*x)^(3/2) = 1 + 54*x + 4860*x^2/2! + 612360*x^3/3! + ....
Recurrence equation: a(n) = 18*(2*n + 1)*a(n-1) with a(0) = 1.
2nd order recurrence equation: a(n) = (40*n + 16)*a(n-1) - 36*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 54.
Define a sequence b(n) := a(n)*sum {k = 0..n} 1/((2*k + 1)*9^k) beginning [1, 56, 5052, 636672, 103142544, 20422253952, 4778808090048, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). Using this observation we obtain the continued fraction expansion log(2) = 2/3*Sum {k >= 0} 1/((2*k + 1)*9^k) = 2/3*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))).
Alternative 2nd order recurrence equation: a(n) = (32*n + 20)*a(n-1) + 36*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 54.
Define now a sequence c(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*9^k) beginning [1, 52, 4692, 591072, 95755344, 18959527872, 4436530187328, ...], which, along with a(n), satisfies the alternative 2nd order recurrence equation. From this observation we find the continued fraction expansion arctan(1/3) = 1/3*Sum {k >= 0} (-1)^k/((2*k + 1)*9^k) = 1/3*(1 - 2/(54 + 36*3^2/(84 + 36*5^2/(116 + ... + 36*(2*n - 1)^2/((32*n + 20) + ... ))))). Cf. A254381 and A254619.

A092145 Numerator of I(n) = 2*(Integral_{x=0..1/2} (1+x^2)^n dx).

Original entry on oeis.org

1, 13, 283, 8667, 342969, 16671885, 962672355, 64467073755, 4917699360945, 421377918441165, 40104072098340075, 4200511400073848475, 480454695780380469225, 59617988532820945752525, 7980059238850231812954675, 1146519564522299271411982875
Offset: 0

Views

Author

Al Hakanson (hawkuu(AT)excite.com), Mar 31 2004

Keywords

Comments

The denominator is A034910(n+1) = 2^(n-1)*(2*n+2)!/(n+1)!.
The terms in the sequence are numerators of unreduced fractions. They equal the value of the integral multiplied by b(n). The reduced fractions are 1, 13/12, 283/240, 2889/2240, 114323/80640, 1111459/709632, 21392719/12300288 etc. - R. J. Mathar, Nov 24 2008

Examples

			I(3) = 8667/6720.

Links

Crossrefs

Cf. A034910.

Programs

  • Magma
    [Numerator(&+[Binomial(n,k)/(4^k*(2*k+1)): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Feb 05 2024
  • Maple
    f:= n -> simplify(hypergeom([1/2, -n], [3/2], -1/4)*(2*n+2)!*2^(n-1)/(n+1)!):
map(f, [$0..20]); # Robert Israel, Nov 07 2016
  • Mathematica
    a[n_]:= (2^(1+3*n)*Gamma[3/2+n]*Hypergeometric2F1[-n,1/2,3/2,-1/4] )/Sqrt[Pi];
Table[a[n], {n, 0, 20}] (* Gerry Martens, Aug 09 2015 *)
  • SageMath
    [numerator(sum(binomial(n,k)/(4^k*(2*k+1)) for k in range(n+1))) for n in range(31)] # G. C. Greubel, Feb 05 2024

Formula

a(n) = (2^(3*n+1)*Gamma(n+3/2)/sqrt(Pi))*Hypergeometric2F1([-n, 1/2], [3/2], -1/4). - Gerry Martens, Aug 09 2015
a(n) = Sum_{k=0..n} binomial(n,k)/(4^k*(2*k+1)). - G. C. Greubel, Feb 05 2024
a(n) ~ 2^(n + 1/2) * 5^(n+1) * n^n / exp(n). - Vaclav Kotesovec, Feb 05 2024

Extensions

More terms from Alois P. Heinz, Aug 09 2015
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