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.

Showing 1-3 of 3 results.

A254322 Expansion of e.g.f.: (1-11*x)^(-10/11).

Original entry on oeis.org

1, 10, 210, 6720, 288960, 15603840, 1014249600, 77082969600, 6706218355200, 657209398809600, 71635824470246400, 8596298936429568000, 1126115160672273408000, 159908352815462823936000, 24465977980765812062208000, 4012420388845593178202112000
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 28 2015

Keywords

Comments

Generally, for k > 1, if e.g.f. = (1-k*x)^(-(k-1)/k) then a(n) ~ n! * k^n / (n^(1/k) * Gamma((k-1)/k)).

Links

Crossrefs

Sequences of the form k^n*Pochhammer((k-1)/k, n): A000007 (k=1), A001147 (k=2), A008544 (k=3), A008545 (k=4), A008546 (k=5), A008543 (k=6), A049209 (k=7), A049210 (k=8), A049211 (k=9), A049212 (k=10), this sequence (k=11), A346896 (k=12).
Cf. A000108, A254282, A254286, A254287.

Programs

  • Magma
    m=11; [Round(m^n*Gamma(n +(m-1)/m)/Gamma((m-1)/m)): n in [0..20]]; // G. C. Greubel, Feb 08 2022
  • Mathematica
    CoefficientList[Series[(1-11*x)^(-10/11), {x, 0, 20}], x] * Range[0, 20]!
FullSimplify[Table[11^n * Gamma[n+10/11] / Gamma[10/11], {n, 0, 18}]]
  • Sage
    m=11; [m^n*rising_factorial((m-1)/m, n) for n in (0..20)] # G. C. Greubel, Feb 08 2022

Formula

D-finite with recurrence: a(0) = 1; a(n) = (11*n-1) * a(n-1) for n > 0. [corrected by Georg Fischer, Dec 23 2019]
a(n) = 11^n * Gamma(n+10/11) / Gamma(10/11).
a(n) ~ n! * 11^n / (n^(1/11) * Gamma(10/11)).
From Nikolaos Pantelidis, Jan 17 2021: (Start)
G.f.: 1/G(0) where G(k) = 1 - (22*k+10)*x - 11*(k+1)*(11*k+10)*x^2/G(k+1) (continued fraction).
G.f.: 1/(1-10*x-110*x^2/(1-32*x-462*x^2/(1-54*x-1056*x^2/(1-76*x-1892*x^2/(1-98*x-2970*x^2/(1-...)))))) (Jacobi continued fraction).
G.f.: 1/Q(0) where Q(k) = 1 - x*(11*k+10)/(1 - x*(11*k+11)/Q(k+1)) (continued fraction).
G.f.: 1/(1-10*x/(1-11*x/(1-21*x/(1-22*x/(1-32*x/(1-33*x/(1-43*x/(1-44*x/(1-54*x/(1-55*x/(1-...))))))))))) (Stieltjes continued fraction).
(End)
G.f.: hypergeometric2F0([1, 10/11], [--], 11*x). - G. C. Greubel, Feb 08 2022
Sum_{n>=0} 1/a(n) = 1 + (e/11)^(1/11)*(Gamma(10/11) - Gamma(10/11, 1/11)). - Amiram Eldar, Dec 22 2022

A254286 Expansion of (1 - (1-256*x)^(1/4)) / (64*x).

Original entry on oeis.org

1, 96, 14336, 2523136, 484442112, 98180268032, 20645907791872, 4459516083044352, 983075545417777152, 220208922173582082048, 49967406340478261526528, 11459191854083014643417088, 2651480699775516003646046208, 618173786004806016850049630208
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 27 2015

Keywords

Links

Crossrefs

Cf. A000108, A008545, A025749, A048779, A254282, A254287.

Programs

  • Magma
    [Round(2^(8*n)*Gamma(n+3/4)/(Gamma(3/4)*Gamma(n+2))): n in [0..30]]; // G. C. Greubel, Aug 10 2022
  • Mathematica
    CoefficientList[Series[(1-(1-256*x)^(1/4)) / (64*x),{x,0,20}],x]
CoefficientList[Series[Hypergeometric1F1[3/4,2,256*x],{x,0,20}],x] * Range[0,20]! (* Vaclav Kotesovec, Jan 28 2015 *)
  • SageMath
    [2^(8*n)*rising_factorial(3/4,n)/factorial(n+1) for n in (0..30)] # G. C. Greubel, Aug 10 2022

Formula

G.f.: (1 - (1-256*x)^(1/4)) / (64*x).
a(n) ~ 256^n / (Gamma(3/4) * n^(5/4)).
Recurrence: (n+1)*a(n) = 64*(4*n-1)*a(n-1).
a(n) = 256^n * Gamma(n+3/4) / (Gamma(3/4) * Gamma(n+2)).
E.g.f.: hypergeom([3/4], [2], 256*x). - Vaclav Kotesovec, Jan 28 2015
From Peter Bala, Sep 01 2017: (Start)
a(n) = (-1)^n*binomial(1/4, n+1)*4^(4*n+1). Cf. A000108(n) = (-1)^n*binomial(1/2, n+1)*2^(2*n+1).
a(n) = 16^n*A025749(n+1); a(n) = 32^n*A048779(n+1).
(End)

A254287 Expansion of (1 - (1 - 3125*x)^(1/5)) / (625*x).

Original entry on oeis.org

1, 1250, 2343750, 5126953125, 12176513671875, 30441284179687500, 78821182250976562500, 209368765354156494140625, 567040406167507171630859375, 1559361116960644721984863281250, 4341403109719976782798767089843750, 12210196246087434701621532440185546875
Offset: 0

Views

Author

Vaclav Kotesovec, Jan 27 2015

Keywords

Comments

In general, if k > 1 and g.f. = (1 - (1 - k^k * x)^(1/k)) / (k^(k-1) * x), then a(n) ~ k^(k*n) / (Gamma((k-1)/k) * n^((k+1)/k)).

Links

Crossrefs

Cf. A000108 (k=2), A254282 (k=3), A254286 (k=4).
Cf. A008546, A025748.

Programs

  • Magma
    [Round(5^(5*n)*Gamma(n+4/5)/(Gamma(4/5)*Gamma(n+2))): n in [0..30]]; // G. C. Greubel, Aug 10 2022
  • Mathematica
    CoefficientList[Series[(1-(1-3125*x)^(1/5)) / (625*x),{x,0,20}],x]
CoefficientList[Series[Hypergeometric1F1[4/5,2,3125*x],{x,0,20}],x] * Range[0,20]! (* Vaclav Kotesovec, Jan 28 2015 *)
  • SageMath
    [5^(5*n)*rising_factorial(4/5, n)/factorial(n+1) for n in (0..30)] # G. C. Greubel, Aug 10 2022

Formula

G.f.: (1 - (1 - 3125*x)^(1/5)) / (625*x).
a(n) ~ 3125^n / (Gamma(4/5) * n^(6/5)).
Recurrence: (n+1)*a(n) = 625*(5*n-1)*a(n-1).
a(n) = 5^(5*n) * Gamma(n+4/5) / (Gamma(4/5) * Gamma(n+2)).
E.g.f.: hypergeom([4/5], [2], 3125*x). - Vaclav Kotesovec, Jan 28 2015
From Peter Bala, Sep 01 2017: (Start)
a(n) = (-1)^n*binomial(1/5, n+1)*5^(5*n+1). Cf. A000108(n) = (-1)^n*binomial(1/2, n+1)*2^(2*n+1).
a(n) = 125^n*A025748(n+1). (End)
Showing 1-3 of 3 results.

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