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 1-7 of 7 results.

A005258 Apéry numbers: a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n+k,k).

Original entry on oeis.org

1, 3, 19, 147, 1251, 11253, 104959, 1004307, 9793891, 96918753, 970336269, 9807518757, 99912156111, 1024622952993, 10567623342519, 109527728400147, 1140076177397091, 11911997404064793, 124879633548031009, 1313106114867738897, 13844511065506477501
Offset: 0

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Author

Keywords

Comments

This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Equals the main diagonal of square array A108625. - Paul D. Hanna, Jun 14 2005
This sequence is t_5 in Cooper's paper. - Jason Kimberley, Nov 25 2012
Conjecture: For each n=1,2,3,... the polynomial a_n(x) = Sum_{k=0..n} C(n,k)^2*C(n+k,k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
Diagonal of rational functions 1/(1 - x - x*y - y*z - x*z - x*y*z), 1/(1 + y + z + x*y + y*z + x*z + x*y*z), 1/(1 - x - y - z + x*y + x*y*z), 1/(1 - x - y - z + y*z + x*z - x*y*z). - Gheorghe Coserea, Jul 07 2018

Examples

			G.f. = 1 + 3*x + 19*x^2 + 147*x^3 + 1251*x^4 + 11253*x^5 + 104959*x^6 + ...
		

References

  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • S. Melczer, An Invitation to Analytic Combinatorics, 2021; p. 129.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A007318.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

Programs

  • GAP
    a:=n->Sum([0..n],k->(-1)^(n-k)*Binomial(n,k)*Binomial(n+k,k)^2);;
    A005258:=List([0..20],n->a(n));; # Muniru A Asiru, Feb 11 2018
    
  • GAP
    List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*Binomial(n+k,k))); # Muniru A Asiru, Jul 29 2018
    
  • Haskell
    a005258 n = sum [a007318 n k ^ 2 * a007318 (n + k) k | k <- [0..n]]
    -- Reinhard Zumkeller, Jan 04 2013
    
  • Magma
    [&+[Binomial(n,k)^2 * Binomial(n+k,k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Nov 28 2018
    
  • Maple
    with(combinat): seq(add((multinomial(n+k,n-k,k,k))*binomial(n,k), k=0..n), n=0..18); # Zerinvary Lajos, Oct 18 2006
    a := n -> binomial(2*n, n)*hypergeom([-n, -n, -n], [1, -2*n], 1):
    seq(simplify(a(n)), n=0..20); # Peter Luschny, Feb 10 2018
  • Mathematica
    a[n_] := HypergeometricPFQ[ {n+1, -n, -n}, {1, 1}, 1]; Table[ a[n], {n, 0, 18}] (* Jean-François Alcover, Jan 20 2012, after Vladeta Jovovic *)
    Table[Sum[Binomial[n,k]^2 Binomial[n+k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Aug 25 2019 *)
  • PARI
    {a(n) = if( n<0, -(-1)^n * a(-1-n), sum(k=0, n, binomial(n, k)^2 * binomial(n+k, k)))} /* Michael Somos, Sep 18 2013 */
    
  • Python
    def A005258(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m *= (n+k+1)*(n-k)**2
            m //= (k+1)**3
        return g # Chai Wah Wu, Oct 02 2022

Formula

a(n) = hypergeom([n+1, -n, -n], [1, 1], 1). - Vladeta Jovovic, Apr 24 2003
D-finite with recurrence: (n+1)^2 * a(n+1) = (11*n^2+11*n+3) * a(n) + n^2 * a(n-1). - Matthijs Coster, Apr 28 2004
Let b(n) be the solution to the above recurrence with b(0) = 0, b(1) = 5. Then the b(n) are rational numbers with b(n)/a(n) -> zeta(2) very rapidly. The identity b(n)*a(n-1) - b(n-1)*a(n) = (-1)^(n-1)*5/n^2 leads to a series acceleration formula: zeta(2) = 5 * Sum_{n >= 1} 1/(n^2*a(n)*a(n-1)) = 5*(1/(1*3) + 1/(2^2*3*19) + 1/(3^2*19*147) + ...). Similar results hold for the constant e: see A143413. - Peter Bala, Aug 14 2008
G.f.: hypergeom([1/12, 5/12],[1], 1728*x^5*(1-11*x-x^2)/(1-12*x+14*x^2+12*x^3+x^4)^3) / (1-12*x+14*x^2+12*x^3+x^4)^(1/4). - Mark van Hoeij, Oct 25 2011
a(n) ~ ((11+5*sqrt(5))/2)^(n+1/2)/(2*Pi*5^(1/4)*n). - Vaclav Kotesovec, Oct 05 2012
1/Pi = 5*(sqrt(47)/7614)*Sum_{n>=0} (-1)^n a(n)*binomial(2n,n)*(682n+71)/15228^n. [Cooper, equation (4)] - Jason Kimberley, Nov 26 2012
a(-1 - n) = (-1)^n * a(n) if n>=0. a(-1 - n) = -(-1)^n * a(n) if n<0. - Michael Somos, Sep 18 2013
0 = a(n)*(a(n+1)*(+4*a(n+2) + 83*a(n+3) - 12*a(n+4)) + a(n+2)*(+32*a(n+2) + 902*a(n+3) - 147*a(n+4)) + a(n+3)*(-56*a(n+3) + 12*a(n+4))) + a(n+1)*(a(n+1)*(+17*a(n+2) + 374*a(n+3) - 56*a(n+4)) + a(n+2)*(+176*a(n+2) + 5324*a(n+3) - 902*a(n+4)) + a(n+3)*(-374*a(n+3) + 83*a(n+4))) + a(n+2)*(a(n+2)*(-5*a(n+2) - 176*a(n+3) + 32*a(n+4)) + a(n+3)*(+17*a(n+3) - 4*a(n+4))) for all n in Z. - Michael Somos, Aug 06 2016
a(n) = binomial(2*n, n)*hypergeom([-n, -n, -n],[1, -2*n], 1). - Peter Luschny, Feb 10 2018
a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*binomial(n+k,k)^2. - Peter Bala, Feb 10 2018
G.f. y=A(x) satisfies: 0 = x*(x^2 + 11*x - 1)*y'' + (3*x^2 + 22*x - 1)*y' + (x + 3)*y. - Gheorghe Coserea, Jul 01 2018
From Peter Bala, Jan 15 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(j+k)*C(n,k)*C(n+k,k)^2*C(n,j)* C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j)*C(n,k)^2*C(n+k,k)*C(n,j)* C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} (-1)^j*C(n,k)^2*C(n,j)*C(3*n-j-k,2*n). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 1. At m = 2 we get the Apéry numbers A005259. - Peter Bala, Dec 22 2020
a(n) = (-1)^n*Sum_{j=0..n} (1 - 5*j*H(j) + 5*j*H(n - j))*binomial(n, j)^5, where H(n) denotes the n-th harmonic number, A001008/A002805. (Paule/Schneider). - Peter Luschny, Jul 23 2021
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=(3 + x)*T(x) + (-1 + 22*x + 3*x^2)*T'(x) + x*(-1 + 11*x + x^2)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = 3*(1 - 12*x + 14*x^2 + 12*x^3 + x^4);
g3 = 1 - 18*x + 75*x^2 + 75*x^4 + 18*x^5 + x^6;
which determine an elliptic surface with four singular fibers. (End)
Conjecture: a(n)^2 = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A143007(n, k). - Peter Bala, Jul 08 2024

A111968 a(n) = Sum_{k=0..n} C(n,k)^2*C(n+k,k)^3, where C := binomial.

Original entry on oeis.org

1, 9, 325, 17577, 1152501, 84505509, 6664647781, 553268669865, 47710914870133, 4236909872278509, 385139801423145825, 35681384898462925125, 3358273513450241419125, 320308335005997679093125, 30900030366269721747776325, 3010365811746267487293617577
Offset: 0

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Author

N. J. A. Sloane, Nov 28 2005

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]^2*Binomial[n+k,k]^3,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Nov 04 2012 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)^2*binomial(n+k,k)^3); \\ Michel Marcus, Mar 10 2016

Formula

a(n) ~ (1+r)^(6*n+7/2)/r^(5*n+9/2)/(4*Pi^2*n^2)*sqrt((1-r)/(5-r)), where r is positive real root of the equation (1-r)^2*(1+r)^3=r^5, r = 0.77591859532439... - Vaclav Kotesovec, Nov 04 2012
Recurrence: (n-1)^2*n^4*(843719*n^6 - 10346391*n^5 + 52472779*n^4 - 140788713*n^3 + 210641238*n^2 - 166531044*n + 54330068)*a(n) = (n-1)^2*(109683470*n^10 - 1564397770*n^9 + 9692963299*n^8 - 34227043418*n^7 + 75994068609*n^6 - 110509975758*n^5 + 106422212572*n^4 - 67092633284*n^3 + 26619112256*n^2 - 6034674112*n + 596279040)*a(n-1) - (1736373702*n^12 - 31711114890*n^11 + 261518988565*n^10 - 1286766506127*n^9 + 4203065855621*n^8 - 9590033857111*n^7 + 15649936441072*n^6 - 18370855225904*n^5 + 15360506258964*n^4 - 8896962441876*n^3 + 3377234408016*n^2 - 751582555104*n + 73915071552)*a(n-2) - (n-2)^2*(36279917*n^10 - 590014481*n^9 + 4216923435*n^8 - 17398379754*n^7 + 45760527058*n^6 - 79915647314*n^5 + 93501944898*n^4 - 72055169071*n^3 + 34824212688*n^2 - 9481092472*n + 1100757336)*a(n-3) - (n-2)^2*(843719*n^6 - 5284077*n^5 + 13396609*n^4 - 17487127*n^3 + 12303648*n^2 - 4393232*n + 621656)*(n-3)^4*a(n-4). - Vaclav Kotesovec, Nov 04 2012

A112019 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+k,k)^2.

Original entry on oeis.org

1, 5, 55, 749, 11251, 178835, 2949115, 49906925, 860905315, 15071939255, 266982872905, 4774722189275, 86070844191775, 1561948324845095, 28507384046515555, 522867506128197869, 9631571375362268515, 178094411589895650815, 3304192479145474141741, 61487420580006795749999
Offset: 0

Views

Author

N. J. A. Sloane, Nov 28 2005

Keywords

Comments

Diagonal of rational function 1/(1 - x - y - z - x*y + x*z + x*y*z). - Gheorghe Coserea, Jul 01 2018
Diagonal of the rational function 1 / ((1-x)*(1-y)*(1-z) - z). - Seiichi Manyama, Apr 30 2025

Crossrefs

Programs

  • Maple
    seq(add((multinomial(n+k,n-k,k,k))*binomial(n+k,k),k=0..n),n=0..19); # Zerinvary Lajos, Oct 18 2006
    ogf := hypergeom([1/12,5/12],[1], -1728*(x^3+5*x^2+39*x-2)*x^4 / (x^4+4*x^3+30*x^2-20*x+1)^3 ) / (x^4+4*x^3+30*x^2-20*x+1)^(1/4);
    series(ogf, x=0, 30); # Mark van Hoeij, Jan 22 2013
  • Mathematica
    Table[HypergeometricPFQ[{-n, 1 + n, 1 + n}, {1, 1}, -1], {n, 0, 20}] (* Olivier Gérard, Apr 23 2009 *)
    Table[Sum[Binomial[n,k]*Binomial[n+k,k]^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Nov 04 2012 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*binomial(n+k,k)^2); \\ Michel Marcus, Mar 09 2016

Formula

a(n) = 3F2( {-n, 1 + n, 1 + n} ; {1, 1} )(-1). - Olivier Gérard, Apr 23 2009
a(n) ~ (1+r)^(4*n+5/2)/r^(3*n+5/2)/(2*Pi*n)*sqrt((1-r)/(3-r)), where r is positive real root of the equation (1-r)*(1+r)^2=r^3, r = 1/6*((44-3*sqrt(177))^(1/3)+(44+3*sqrt(177))^(1/3)-1) = 0.82948354095849... - Vaclav Kotesovec, Nov 04 2012
Recurrence: 2*n^2*(59*n - 83)*a(n) = (2301*n^3 - 5538*n^2 + 3797*n - 800)*a(n-1) + 5*(59*n^3 - 201*n^2 + 213*n - 64)*a(n-2) + (59*n - 24)*(n-2)^2*a(n-3). - Vaclav Kotesovec, Nov 04 2012
G.f. y=A(x) satisfies: 0 = x*(5*x + 8)*(x^3 + 5*x^2 + 39*x - 2)*y'' + (15*x^4 + 82*x^3 + 315*x^2 + 624*x - 16)*y' + (5*x^3 + 21*x^2 + 80)*y. - Gheorghe Coserea, Jul 01 2018

A014178 a(n) = Sum_{k = 0..n} binomial(n,k)^3*binomial(n+k,k).

Original entry on oeis.org

1, 3, 31, 399, 5871, 93753, 1577479, 27556623, 495001327, 9085988613, 169675769781, 3213444254133, 61573700137431, 1191526503165729, 23252920338835911, 457112339182896399, 9043566887755775727, 179928068420530483389, 3597714616543167088921
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Maple
    f := n->sum( 'binomial(n,k)^3*binomial(n+k,k)^1','k'=0..n);
  • Mathematica
    Table[Sum[Binomial[n,k]^3*Binomial[n+k,k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Nov 04 2012 *)
  • PARI
    a(n)=sum(k=0,n, binomial(n,k)^3*binomial(n+k,k) ); \\ Joerg Arndt, May 04 2013

Formula

a(n) ~ (1+r)^(2*n+3/2)/r^(4*n+7/2)/(4*Pi^(3/2)*n^(3/2))*sqrt((1-r)/(2+r)), where r is positive real root of the equation (1-r)^3*(1+r)=r^4, r = 0.58252220781047... - Vaclav Kotesovec, Nov 04 2012
Recurrence: (n-1)*n^3*(29412*n^4 - 246240*n^3 + 764259*n^2 - 1042332*n + 527381)*a(n) = 2*(n-1)*(235296*n^7 - 2322864*n^6 + 9245766*n^5 - 19022421*n^4 + 21621181*n^3 - 13561627*n^2 + 4459053*n - 605664)*a(n-1) + 2*(1647072*n^8 - 20377728*n^7 + 107506956*n^6 - 315721020*n^5 + 564159163*n^4 - 627527310*n^3 + 423779896*n^2 - 158592459*n + 25128864)*a(n-2) + 2*(n-2)*(4176504*n^7 - 49583844*n^6 + 243933522*n^5 - 641841009*n^4 + 971188553*n^3 - 841622632*n^2 + 385567572*n - 72023040)*a(n-3) - 4*(n-2)*(29412*n^4 - 128592*n^3 + 202011*n^2 - 134886*n + 32480)*(n-3)^3*a(n-4). - Vaclav Kotesovec, Nov 04 2012
The expansions exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 20*x^2 + 184*x^3 + 2060*x^4 + 26246*x^5 + ... and exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + x + 2*x^2 + 12*x^3 + 112*x^4 + 1296*x^5 + ... appear to have integer coefficients. Cf. A005258 and A005259. - Peter Bala, Jan 14 2016

A014180 Sum_{k = 0..n} binomial(n,k)^3*binomial(n+k,k)^2.

Original entry on oeis.org

1, 5, 109, 3533, 133501, 5629505, 254899765, 12129399245, 599084606845, 30455459491505, 1584249399505609, 83970120618566825, 4520585403820052581, 246592348286170615097, 13603606921687170927109, 757808346139996787715533, 42575668004558257371188605, 2410024012619343278147357297
Offset: 0

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Author

Keywords

Comments

Compare with the Apéry numbers A005258 and A005259.

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]^3*Binomial[n+k,k]^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Nov 04 2012 *)
  • PARI
    a(n)=sum(k=0,n, binomial(n,k)^3*binomial(n+k,k)^2 ); \\ Joerg Arndt, May 04 2013

Formula

a(n) ~ (1+r)^(4*n+5/2)/r^(5*n+9/2)/(4*Pi^2*n^2)*sqrt((1-r)/(5+r)), where r is positive real root of the equation (1-r)^3*(1+r)^2 = r^5, r = 0.65039847669867... - Vaclav Kotesovec, Nov 04 2012
The expansions exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 5*x + 67*x^2 + 1471*x^3 + 41456*x^4 + 1380268*x^5 + ... and exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + x + 3*x^2 + 39*x^3 + 924*x^4 + 27696*x^5 + ... appear to have integer coefficients. Cf. A005258 and A005259.- Peter Bala, Jan 14 2016

A218693 a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k)^3.

Original entry on oeis.org

1, 9, 271, 11193, 535251, 27854739, 1531656211, 87547358649, 5149886133907, 309721191497259, 18957806551405701, 1177134132932168739, 73964787438524189871, 4694347514292389411103, 300499277330710307643771, 19378727805024033594228153, 1257802636907811605342461587
Offset: 0

Views

Author

Vaclav Kotesovec, Nov 04 2012

Keywords

Crossrefs

Cf.
A001850 (p=1, q=1),
A112019 (p=1, q=2),
A005258 (p=2, q=1),
A005259 (p=2, q=2),
A111968 (p=2, q=3),
A014178 (p=3, q=1),
A014180 (p=3, q=2),
A092813 (p=3, q=3),
A218690 (p=4, q=2),
A092814 (p=4, q=4),
A092815 (p=5, q=5),
A218692 (p=6, q=3),
A218689 (p=6, q=6).

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]*Binomial[n+k,k]^3,{k,0,n}],{n,0,20}]
    a[n_] := HypergeometricPFQ[{-n, n+1, n+1 ,n+1},{1, 1, 1}, -1]; Table[a[n],{n,0,16}] (* Detlef Meya, May 25 2024 *)

Formula

Recurrence: 4*(n-1)*n^3*(29412*n^4 - 224352*n^3 + 632931*n^2 - 781692*n + 356309)*a(n) = 2*(n-1)*(4176504*n^7 - 38122740*n^6 + 140783586*n^5 - 270139161*n^4 + 288226505*n^3 - 170040251*n^2 + 51625509*n - 6283008)*a(n-1) + 2*(1647072*n^8 - 19152000*n^7 + 94636812*n^6 - 258386460*n^5 + 423728203*n^4 - 423743982*n^3 + 249392728*n^2 - 77793627*n + 9736704)*a(n-2) + 2*(n-2)*(235296*n^7 - 2618352*n^6 + 11905158*n^5 - 28432149*n^4 + 38188669*n^3 - 28610816*n^2 + 10954716*n - 1618272)*a(n-3) - (n-2)*(29412*n^4 - 106704*n^3 + 136347*n^2 - 71238*n + 12608)*(n-3)^3*a(n-4).
a(n) ~ (1+r)^(6*n+7/2)/r^(4*n+7/2)/(4*Pi^(3/2)*n^(3/2))*sqrt((1-r)/(2-r)), where r is positive real root of the equation (1-r)*(1+r)^3=r^4, r = 0.90340819201887...
Generally, Sum_{k=0..n} C(n,k)^p*C(n+k,k)^q is asymptotic to sqrt((r*(1-r^2))/((p+q+(p-q)*r)*(2*Pi*n)^(p+q-1))) * (1+r)^(q*n)/(1-r)^(p*n+p), where r is positive real root of the equation (1-r)^p*(1+r)^q=r^(p+q). - Vaclav Kotesovec, Nov 07 2012
a(n) = hypergeom([-n, n+1, n+1, n+1],[1, 1, 1], -1). - Detlef Meya, May 25 2024

A218690 Sum_{k=0..n} C(n,k)^4*C(n+k,k)^2.

Original entry on oeis.org

1, 5, 181, 9797, 616501, 45778505, 3678753205, 314352091205, 28154344881205, 2612340741982505, 249415797821914681, 24374175871569953225, 2428502737999745059381, 245952659136011117909897, 25260608441078125508484181, 2626023605592369669830217797
Offset: 0

Views

Author

Vaclav Kotesovec, Nov 04 2012

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k]^4*Binomial[n+k,k]^2,{k,0,n}],{n,0,20}]

Formula

a(n) ~ ((1+sqrt(5))/2)^(2*(5*n+4)-3/2)/(5^(1/4)*(2*Pi*n)^(5/2)*sqrt(2)).
Showing 1-7 of 7 results.