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 1-4 of 4 results.

A293490 a(n) = Sum_{k=0..n} binomial(2*k, k)*binomial(2*n-k, n).

Original entry on oeis.org

1, 4, 18, 84, 400, 1932, 9436, 46512, 231066, 1155660, 5813808, 29396952, 149305884, 761282032, 3894953640, 19987999696, 102847396416, 530446714812, 2741576339716, 14196136939600, 73631851898220, 382483602131400, 1989514312826400, 10361255764532400, 54020655931542300, 281933439875693424
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 10 2017

Keywords

Comments

Main diagonal of iterated partial sums array of central binomial coefficients (starting with the first partial sums).

Crossrefs

Cf. A000984, A002426, A006134, A026375, A270447.

Programs

  • GAP
    A293490 := Concatenation([1], List([1..3*10^2],n -> Sum([0..n],k -> Binomial(2*k, k)*(Binomial(2*n - k, n))))); # Muniru A Asiru, Oct 15 2017
  • Mathematica
    Table[Sum[Binomial[2 k, k] Binomial[2 n - k, n], {k, 0, n}], {n, 0, 25}]
Table[SeriesCoefficient[1/(Sqrt[1 - 4 x] (1 - x)^(n + 1)), {x, 0, n}], {n, 0, 25}]
Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) 1/(1 - 2 x/(1 + ContinuedFractionK[-x, 1, {k, 1, n}])), {x, 0, n}], {n, 0, 25}]
CoefficientList[Series[1/(Sqrt[2 Sqrt[1-4 x]-1] Sqrt[1-4 x]),{x,0,25}],x] (* Alexander M. Haupt, Jun 24 2018 *)
  • PARI
    a(n) = sum(k=0, n, binomial(2*k, k)*binomial(2*n-k, n)); \\ Michel Marcus, Oct 15 2017

Formula

a(n) = [x^n] 1/(sqrt(1 - 4*x)*(1 - x)^(n+1)).
a(n) = [x^n] 1/((1 - x)^(n+1)*(1 - 2*x/(1 - x/(1 - x/(1 - x/(1 - ...)))))), a continued fraction.
a(n) = 4^n*Gamma(n+1/2)*2F1(-n,n+1; 1/2-n; 1/4)/(sqrt(Pi)*Gamma(n+1)).
From Vaclav Kotesovec, Oct 16 2017: (Start)
D-finite with recurrence: 3*(n-1)*n*a(n) = 14*(n-1)*(2*n-1)*a(n-1) - 4*(4*n-5)*(4*n-3)*a(n-2).
a(n) ~ 2^(4*n + 3/2) / (3^(n + 1/2) * sqrt(Pi*n)).
(End)
G.f.: 1/(sqrt(2*sqrt(1-4*x)-1)*sqrt(1-4*x)). - Alexander M. Haupt, Jun 24 2018

A293468 a(n) = Sum_{k=0..n} k!*binomial(2*n-k, n).

Original entry on oeis.org

1, 3, 11, 44, 189, 880, 4542, 26712, 182793, 1461368, 13477650, 140564536, 1627370146, 20621925504, 283161372284, 4182215376240, 66065933347425, 1111053154779720, 19814069772086730, 373435157945506680, 7415765258637418950, 154751460071567005920, 3385387828167428482020
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 09 2017

Keywords

Links

Crossrefs

Cf. A000142, A000522, A003422, A014144, A092392, A270447, A293469.

Programs

  • Maple
    seq(simplify( GAMMA(n+1/2)*4^n*hypergeom([1,1,-n],[-2*n],1)/(sqrt(Pi)*n!)),n=0..30); # Robert Israel, Oct 09 2017
  • Mathematica
    Table[Sum[k! Binomial[2 n - k, n], {k, 0, n}], {n, 0, 22}]
Table[Sum[Gamma[k + 1] Gamma[2 n - k  + 1]/(Gamma[n + 1] Gamma[n - k + 1]), {k, 0, n}], {n, 0, 22}]
Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) 1/(1 + ContinuedFractionK[-Floor[(k + 1)/2] x, 1, {k, 1, n}]), {x, 0, n}], {n, 0, 22}]
Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) Sum[k! x^k, {k, 0, n}], {x, 0, n}], {n, 0, 22}]
A293468[n_] := DifferenceRoot[Function[{a,k}, {(k+1)(k-n)a[k] + (k(n-2)-k^2+3n)
a[k+1] + (k-2n) a[k+2] == 0, a[0] == 0, a[1] == Binomial[2n, n]}]][1+n];
Table[A293468[n], {n, 0, 22}] (* Peter Luschny, Oct 09 2017 *)

Formula

a(n) = [x^n] 1/((1 - x)^(n+1)*(1 - x/(1 - x/(1 - 2*x/(1 - 2*x/(1 - 3*x/(1 - 3*x/(1 - ...)))))))), a continued fraction.
a(n) = Gamma(n+1/2)*4^n*hypergeom([1,1,-n],[-2n],1)/(sqrt(Pi)*n!). - Robert Israel, Oct 09 2017
a(n) ~ exp(1) * n!. - Vaclav Kotesovec, Oct 18 2017

A293469 a(n) = Sum_{k=0..n} (2*k-1)!!*binomial(2*n-k, n).

Original entry on oeis.org

1, 3, 12, 57, 330, 2436, 23226, 277389, 3966534, 65517210, 1220999208, 25279328958, 575024187192, 14247595540542, 381846383109030, 11004598454925405, 339324532631899110, 11146022446431209490, 388535338484934710040, 14324570939127320452350, 556887682690152668745660
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 09 2017

Keywords

Links

Crossrefs

Cf. A001147, A076795, A084262, A092392, A270447, A293468.

Programs

  • Maple
    seq(add(doublefactorial(2*k-1)*binomial(2*n-k,n),k=0..n),n=0..40); # Robert Israel, Oct 09 2017
  • Mathematica
    Table[Sum[(2 k - 1)!! Binomial[2 n - k, n], {k, 0, n}], {n, 0, 20}]
Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) 1/(1 + ContinuedFractionK[-k x, 1, {k, 1, n}]), {x, 0, n}], {n, 0, 20}]
Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) Sum[(2 k - 1)!! x^k, {k, 0, n}], {x, 0, n}], {n, 0, 20}]

Formula

a(n) = [x^n] 1/((1 - x)^(n+1)*(1 - x/(1 - 2*x/(1 - 3*x/(1 - 4*x/(1 - 5*x/(1 - 6*x/(1 - ...)))))))), a continued fraction.
a(n) = Gamma(n+1/2)*hypergeom([1/2, 1, -n], [-2*n], 2)*4^n/(n!*sqrt(Pi)). - Robert Israel, Oct 09 2017
a(n) ~ 2^(n + 1/2) * n^n / exp(n - 1/2). - Vaclav Kotesovec, Oct 18 2017

A270530 a(n) = Sum_{k=0..n}((binomial(2*k,k)/(k+1)*binomial(2*n+2,n-k))).

Original entry on oeis.org

1, 5, 23, 105, 484, 2267, 10821, 52705, 262010, 1328768, 6867266, 36115455, 192954358, 1045481465, 5735154907, 31802349105, 178010615678, 1004542994462, 5709066033900, 32646940202200, 187701954810320
Offset: 0

Views

Author

Vladimir Kruchinin, Mar 18 2016

Keywords

Comments

Binomial transform of Catalan numbers.

Links

Crossrefs

Cf. A000108, A007317, A270447.

Programs

  • Maple
    A270530 := proc(n)
    add(binomial(2*k,k)/(k+1)*binomial(2*n+2,n-k),k=0..n) ;
end proc: # R. J. Mathar, Jun 07 2016
  • Mathematica
    CoefficientList[Series[1/(2*x*Sqrt[1 - 4*x]) + (-Sqrt[((5*x + 2*Sqrt[1 - 4*x] - 2))/(x^3*(4 - 16*x))]), {x,0,50}], x] (* G. C. Greubel, Apr 09 2017 *)
  • Maxima
    a(n):=sum((binomial(2*k,k)/(k+1)*binomial(2*n+2,n-k)),k,0,n);
makelist(coeff(taylor(1/(2*x*sqrt(1-4*x))+(-sqrt(((5*x+2*sqrt(1-4*x)-2))/(x^3*(4-16*x)))),x,0,10),x,n),n,0,10);
  • PARI
    x='x+O('x^50); Vec(1/(2*x*sqrt(1-4*x))+(-sqrt(((5*x+2*sqrt(1-4*x)-2))/(x^3*(4-16*x))))) \\ G. C. Greubel, Apr 09 2017

Formula

G.f.: 1/(2*x*sqrt(1-4*x))+(-sqrt(((5*x+2*sqrt(1-4*x)-2))/(x^3*(4-16*x)))).
a(n) ~ 5^(2*n + 7/2) / (3^(3/2) * sqrt(Pi) * n^(3/2) * 2^(2*n+4)). - Vaclav Kotesovec, Mar 18 2016
Conjecture: 2*n*(2*n+3)*(n+1)*a(n) -n*(77*n^2+27*n-4)*a(n-1) +(549*n^3-987*n^2+686*n-168)*a(n-2) -20*(2*n-3)*(43*n^2-104*n+70)*a(n-3) +500*(2*n-5)*(n-2)*(2*n-3)*a(n-4)=0. - R. J. Mathar, Jun 07 2016
Conjecture: 2*n*(2*n+3)*(n+3)*(n+1)*a(n) -n*(57*n^3+228*n^2+107*n+8)*a(n-1) +4*(2*n-1) *(33*n^3+99*n^2-88*n+36)*a(n-2) -100*(n-1)*(2*n-1)*(2*n-3)*(n+4)*a(n-3)=0. - R. J. Mathar, Jun 07 2016
Showing 1-4 of 4 results.

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