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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A264541 a(n) = numerator(Jtilde3(n)).

Original entry on oeis.org

0, 1, 65, 13247, 704707, 660278641, 357852111131, 309349386395887, 240498440880062263, 148443546307725010253, 61760947097005048531, 13658972396318235617977, 723464275788899734058353751, 489812222050789870424202126629, 2614176630672654770175367214389, 204702102697072009862200307064701369
Offset: 0

Views

Author

Michel Marcus, Nov 17 2015

Keywords

Comments

Jtilde3(n) are Apéry-like rational numbers that arise in the calculation of zetaQ(3), the spectral zeta function for the non-commutative harmonic oscillator using a Gaussian hypergeometric function.

Crossrefs

Cf. A002117 (zeta(3)), A260832 (Jtilde2), A264542 (denominators).
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.)

Programs

  • Mathematica
    Numerator[Table[-2*Sum[(-1)^k*Binomial[-1/2, k]^2*Binomial[n, k]*Sum[ 1/(Binomial[-1/2, j]^2*(2*j + 1)^3), {j, 0, k - 1}], {k, 0, n}], {n, 0, 50}]] (* G. C. Greubel, Oct 24 2017 *)
  • PARI
    a(n) = numerator(-2*sum(k=0, n, (-1)^k*binomial(-1/2, k)^2*binomial(n, k)*sum(j=0, k-1, 1/(binomial(-1/2,j)^2*(2*j+1)^3))));

Formula

Jtilde3(n) = J3(n) - J3(0)*Jtilde2(n) (normalization).
4n^2*J3(n) - (8n^2-8n+3)*J3(n-1) + 4(n-1)^2*J3(n-2) = 2^n*(n-1)!/(2n-1)!! with J3(0)=7*zeta(3) and J3(1)=21*zeta(3)/4 + 1/2.

A264542 a(n) = denominator(Jtilde3(n)).

Original entry on oeis.org

1, 2, 96, 17280, 860160, 774144000, 408748032000, 347163328512000, 266621436297216000, 163172319013896192000, 67488959156767948800, 14865958099336613068800, 785345441564243189248819200, 530893518497428395932201779200, 2831432098652951444971742822400, 221701133324526098141287462993920000
Offset: 0

Views

Author

Michel Marcus, Nov 17 2015

Keywords

Comments

Jtilde3(n) are Apéry-like rational numbers that arise in the calculation of zetaQ(3), the spectral zeta function for the non-commutative harmonic oscillator using a Gaussian hypergeometric function.

Crossrefs

Cf. A002117 (zeta(3)), A260832 (Jtilde2), A264541 (numerators).
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.)

Programs

  • Mathematica
    Denominator[Table[-2*Sum[(-1)^k*Binomial[-1/2, k]^2*Binomial[n, k]* Sum[1/(Binomial[-1/2, j]^2*(2*j + 1)^3), {j, 0, k - 1}], {k, 0, n}], {n, 0, 50}]] (* G. C. Greubel, Oct 23 2017 *)
  • PARI
    a(n) = denominator(-2*sum(k=0, n, (-1)^k*binomial(-1/2, k)^2*binomial(n, k)*sum(j=0, k-1, 1/(binomial(-1/2,j)^2*(2*j+1)^3))));

Formula

Jtilde3(n) = J3(n) - J3(0)*Jtilde2(n) (normalization).
4n^2*J3(n) - (8n^2-8n+3)*J3(n-1) + 4(n-1)^2*J3(n-2) = 2^n*(n-1)!/(2n-1)!! with J3(0)=7*zeta(3) and J3(1)=21*zeta(3)/4 + 1/2.

A143414 Apéry-like numbers for the constant 1/e: a(n) = (1/(n-1)!)*Sum_{k = 0..n-1} binomial(n-1,k)*(2*n-k)!.

Original entry on oeis.org

0, 2, 30, 492, 9620, 222630, 5989242, 184139480, 6377545512, 245868202890, 10446648201110, 485126443539012, 24449173476952380, 1329144227959100462, 77535552689576436210, 4831278674685354629040, 320262424087652686405712
Offset: 0

Views

Author

Peter Bala, Aug 14 2008

Keywords

Comments

This sequence satisfies the recursion (n-1)^2*a(n) - n^2*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1), which leads to a rapidly converging series for the constant 1/e: 1/e = 1/2 - 2 * Sum_{n >= 2} (-1)^n * n^2/(a(n)*a(n-1)).
Notice the striking resemblance to the theory of the Apéry numbers A(n) = A005258(n), which satisfy a similar recurrence relation n^2*A(n) - (n-1)^2*A(n-2) = (11*n^2-11*n+3)*A(n-1) and which appear in the series acceleration formula zeta(2) = 5*Sum_{n>=1} 1/(n^2*A(n)*A(n-1)). Compare with A143413 and A143415.

Crossrefs

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.)

Programs

  • Maple
    a := n -> 1/(n-1)!*add (binomial(n-1,k)*(2*n-k)!,k = 0..n-1): seq(a(n),n = 0..19);
    # Alternative:
    A143414 := n -> `if`(n=0, 0, ((2*n)!/(n-1)!)*hypergeom([1-n], [-2*n], 1)):
    seq(simplify(A143414(n)), n = 0..16); # Peter Luschny, May 14 2020
  • Mathematica
    Table[(1/(n-1)!)*Sum[Binomial[n-1,k]*(2*n-k)!, {k,0,n-1}], {n,0,50}] (* G. C. Greubel, Oct 24 2017 *)
  • PARI
    for(n=0,25, print1((1/(n-1)!)*sum(k=0,n-1, binomial(n-1,k)*(2*n-k)!), ", ")) \\ G. C. Greubel, Oct 24 2017

Formula

a(n) = (1/(n-1)!)*Sum_{k = 0..n-1} binomial(n-1,k)*(2*n-k)!.
Recurrence relation: a(0) = 0, a(1) = 2, (n-1)^2*a(n) - n^2*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1), n >= 2.
Let b(n) denote the solution to this recurrence with initial conditions b(0) = -1, b(1) = 1. Then b(n) = A143413(n) = (1/(n-1)!)*Sum_{k = 0..n+1} (-1)^k*binomial(n+1,k)*(2*n-k)!.
The rational number b(n)/a(n) is equal to the Padé approximation to exp(x) of degree (n+1,n-1) evaluated at x = -1 and b(n)/a(n) -> 1/e very rapidly. For example, |b(100)/a(100) - 1/e| is approximately 2.177 * 10^(-437).
The identity a(n)*b(n-1) - a(n-1)*b(n) = (-1)^n *2*n^2 leads to rapidly converging series for the constants 1/e and e: 1/e = 1/2 - 2*Sum_{n >= 2} (-1)^n * n^2/(a(n)*a(n-1)) = 1/2 - 2*(2^2/(2*30) - 3^2/(30*492) + 4^2/(492*9620) - ...); e = 2 * Sum_{n >= 1} (-1)^n * n^2/(b(n)*b(n-1)) = 2*(1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...).
a(n) = (BesselK(n-1/2,1/2)-(1-2*n)*BesselK(n+1/2,1/2)) * exp(1/2)/(2*Pi^(1/2)). - Mark van Hoeij, Nov 12 2009
a(n) = ((2*n)!/(n-1)!)*hypergeom([1-n], [-2*n], 1) for n > 0. - Peter Luschny, May 14 2020
a(n) ~ 2^(2*n + 1/2) * n^(n+1) / exp(n - 1/2). - Vaclav Kotesovec, Jul 11 2021

A143415 Another sequence of Apery-like numbers for the constant 1/e: a(n) = 1/(n+1)!*Sum_{k = 0..n-1} C(n-1,k)*(2*n-k)!.

Original entry on oeis.org

0, 1, 5, 41, 481, 7421, 142601, 3288205, 88577021, 2731868921, 94969529101, 3675200329841, 156725471006105, 7302990263511541, 369216917569411601, 20130327811188977621, 1177435382675193700021, 73546210385434763486705
Offset: 0

Views

Author

Peter Bala, Aug 14 2008

Keywords

Comments

This sequence is a modified version of A143414.

Crossrefs

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.)

Programs

  • Maple
    a := n -> 1/(n+1)!*add (binomial(n-1,k)*(2*n-k)!,k = 0..n-1): seq(a(n),n = 0..19);
    # Alternative:
    A143415 := n -> `if`(n=0, 0, ((2*n)!/(n+1)!)*hypergeom([1-n], [-2*n], 1)):
    seq(simplify(A143415(n)), n = 0..17); # Peter Luschny, May 14 2020
  • Mathematica
    Table[(1/(n+1)!)*Sum[Binomial[n-1,k]*(2*n-k)!, {k,0,n-1}], {n,0,50}] (* G. C. Greubel, Oct 24 2017 *)
  • PARI
    for(n=0,25, print1((1/(n+1)!)*sum(k=0,n-1, binomial(n-1,k)*(2*n-k)!), ", ")) \\ G. C. Greubel, Oct 24 2017

Formula

a(n) = 1/(n+1)!*sum {k = 0..n-1} C(n-1,k)*(2*n-k)!.
a(n) = 1/(n*(n+1))*A143414(n) for n > 0.
Recurrence relation: a(0) = 0, a(1) = 1, (n-1)*(n+1)*a(n) - (n-2)*n*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1) for n >= 2. 1/e = 1/2 - 2 * Sum_{n = 1..inf} (-1)^(n+1)/(n*(n+2)*a(n)*a(n+1)) = 1/2 - 2*[1/(3*1*5) - 1/(8*5*41) + 1/(15*41*481) - 1/(24*481*7421) + ...] .
Conjectural congruences: for r >= 0 and prime p, calculation suggests the congruences a(p^r*(p+1)) == a(p^r) (mod p^(r+1)) may hold.
a(n) = ((2*n)!/(n+1)!)*hypergeom([1-n], [-2*n], 1) for n > 0. - Peter Luschny, May 14 2020

A219692 a(n) = Sum_{j=0..floor(n/3)} (-1)^j C(n,j) * C(2j,j) * C(2n-2j,n-j) * (C(2n-3j-1,n) + C(2n-3j,n)).

Original entry on oeis.org

2, 6, 54, 564, 6390, 76356, 948276, 12132504, 158984694, 2124923460, 28877309604, 398046897144, 5554209125556, 78328566695736, 1114923122685720, 15999482238880464, 231253045986317814, 3363838379489630916
Offset: 0

Views

Author

Jason Kimberley, Nov 25 2012

Keywords

Comments

This sequence is s_18 in Cooper's paper.
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Every prime eventually divides some term of this sequence. - Amita Malik, Aug 20 2017

Crossrefs

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.)

Programs

  • Magma
    s_18 := func where C is Binomial;
    
  • Mathematica
    Table[Sum[(-1)^j*Binomial[n,j]*Binomial[2j,j]*Binomial[2n-2j, n-j]* (Binomial[2n-3j-1,n] +Binomial[2n-3j,n]), {j,0,Floor[n/3]}], {n,0,20}] (* G. C. Greubel, Oct 24 2017 *)
  • PARI
    {a(n) = sum(j=0,floor(n/3), (-1)^j*binomial(n,j)*binomial(2*j,j)* binomial(2*n-2*j,n-j)*(binomial(2*n-3*j-1,n) +binomial(2*n-3*j,n)))}; \\ G. C. Greubel, Apr 02 2019
    
  • Sage
    [sum((-1)^j*binomial(n,j)*binomial(2*j,j)*binomial(2*n-2*j,n-j)* (binomial(2*n-3*j-1,n)+binomial(2*n-3*j,n)) for j in (0..floor(n/3))) for n in (0..20)] # G. C. Greubel, Apr 02 2019

Formula

1/Pi
= 2*3^(-5/2) Sum {k>=0} (n a(n)/18^n) [Cooper, equation (42)]
= 2*3^(-5/2) Sum {k>=0} (n a(n)/A001027(n)).
G.f.: 1+hypergeom([1/8, 3/8],[1],256*x^3/(1-12*x)^2)^2/sqrt(1-12*x). - Mark van Hoeij, May 07 2013
Conjecture D-finite with recurrence: n^3*a(n) -2*(2*n-1)*(7*n^2-7*n+3)*a(n-1) +12*(4*n-5)*(n-1)* (4*n-3)*a(n-2)=0. - R. J. Mathar, Jun 14 2016
a(n) ~ 3 * 2^(4*n + 1/2) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 08 2023

A260667 a(n) = (1/n^2) * Sum_{k=0..n-1} (2k+1)*S(k,n)^2, where S(k,x) denotes the polynomial Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).

Original entry on oeis.org

1, 37, 1737, 102501, 6979833, 523680739, 42129659113, 3572184623653, 315561396741609, 28807571694394593, 2701627814373536601, 259121323945378645947, 25330657454041707496017, 2516984276442279642274311, 253667099464270541534450025, 25884030861250181046253181349, 2670255662315910532447096232073
Offset: 1

Views

Author

Zhi-Wei Sun, Nov 14 2015

Keywords

Comments

Conjecture: For k = 0,1,2,... define S(k,x):= Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).
(i) For any integer n > 0, the polynomial (1/n^2) * Sum_{k=0..n-1}(2k+1)*S(k,x)^2 is integer-valued (and hence a(n) is always integral).
(ii) Let r be 0 or 1, and let x be any integer. Then, for any positive integers m and n, we have the congruence
Sum_{k=0..n-1} (-1)^(k*r)*(2k+1)*S(k,x)^(2m) == 0 (mod n).
(iii) For any odd prime p, we have Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p)(1-7*p^3*B_{p-3}) (mod p^4), where (a/p) is the Legendre symbol, and B_0,B_1,B_2,... are Bernoulli numbers. Also, for any prime p > 3 we have Sum_{k=0..p-1} S(k,-1/3)^2 == p - (14/3)*(p/3)*p^3*B_{p-2}(1/3) (mod p^4), where B_n(x) denotes the Bernoulli polynomial of degree n; Sum_{k=0..p-1} S(k,-1/4)^2 == (2/p)*p - 26*(-2/p)*p^3*E_{p-3} (mod p^4), where E_0,E_1,E_2,... are Euler numbers; Sum_{k=0..p-1} S(k,-1/6)^2 == (3/p)*p - (155/12)*(-1/p)*p^3*B_{p-2}(1/3) (mod p^4).
Our conjecture is motivated by a conjecture of Kimoto and Wakayama which states that Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p) (mod p^3) for any odd prime p. The Kimoto-Wakayama conjecture was confirmed by Long, Osburn and Swisher in 2014.
For more related conjectures, see Sun's paper arXiv.1512.00712. - Zhi-Wei Sun, Dec 03 2015

Examples

			a(2) = 37 since (1/2^2) * Sum_{k=0..1} (2k+1)*S(k,2)^2 = (S(0,2)^2 + 3*S(1,2)^2)/4 = (1^2 + 3*7^2)/4 = 148/4 = 37.
G.f. = x + 37*x^2 + 1737*x^3 + 102501*x^4 + 6979833*x^5 + 523680739*x^6 + ...
		

Crossrefs

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.)

Programs

  • Maple
    # Implementing Mark van Hoeij's formula.
    c := n -> binomial(2*n, n)/(n + 1):
    h := n -> simplify(hypergeom([-n,-n,-n], [1,-2*n], 1)):
    b := n -> c(n)^2*((n+11)*(2+4*n)^2*h(n+1)^2-2*(n+1)*(11*n+16)*(1+2*n)*h(n)*h(n+1)-h(n)^2*(n+1)^3)/(25*(n+2)):
    a := n -> b(n-1): seq(a(n), n = 1..17);  # Peter Luschny, Nov 11 2022
  • Mathematica
    S[k_,x_]:=S[k,x]=Sum[Binomial[k,j]Binomial[x,j]Binomial[x+j,j],{j,0,k}]
    a[n_]:=a[n]=Sum[(2k+1)*S[k,n]^2,{k,0,n-1}]/n^2
    Do[Print[n," ",a[n]],{n,1,17}]

Formula

a(n) ~ phi^(10*n + 3) / (10 * Pi^2 * n^3), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Nov 06 2021
Conjecture: a(p-1) == 1 (mod p^3) for all primes p >= 5. - Peter Bala, Aug 15 2022
a(n) = ((n+10)*A005258(n)^2 - (11*n+5)*A005258(n)*A005258(n-1) - n*A005258(n-1)^2)/(25*(n+1)). - Mark van Hoeij, Nov 11 2022

A262177 Decimal expansion of Q_5 = zeta(5) / (Sum_{k>=1} (-1)^(k+1) / (k^5 * binomial(2k, k))), a conjecturally irrational constant defined by an Apéry-like formula.

Original entry on oeis.org

2, 0, 9, 4, 8, 6, 8, 6, 2, 2, 0, 1, 0, 0, 3, 6, 9, 9, 3, 8, 5, 0, 2, 4, 9, 2, 9, 3, 7, 3, 2, 9, 4, 1, 6, 3, 0, 2, 9, 6, 7, 5, 8, 7, 4, 8, 5, 6, 7, 7, 8, 1, 8, 2, 7, 4, 0, 1, 2, 7, 5, 8, 7, 8, 3, 7, 4, 3, 8, 0, 0, 7, 8, 7, 6, 8, 4, 6, 8, 1, 5, 6, 3, 2, 0, 6, 0, 4, 4, 2, 3, 2, 0, 9, 0, 4, 3, 1, 3, 6, 9, 3, 1
Offset: 1

Views

Author

Jean-François Alcover, Sep 14 2015

Keywords

Comments

The similar constant Q_3 = zeta(3) / (Sum_{k>=1} (-1)^(k+1) / (k^3 * binomial(2k, k))) evaluates to 5/2.

Examples

			2.09486862201003699385024929373294163029675874856778182740127587837438...
		

Crossrefs

Cf. A013663.
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.)

Programs

  • Mathematica
    Q5 = Zeta[5]/Sum[(-1)^(k+1)/(k^5*Binomial[2k, k]), {k, 1, Infinity}]; RealDigits[Q5, 10, 103] // First
  • PARI
    zeta(5)/suminf(k=1, (-1)^(k+1)/(k^5*binomial(2*k,k))) \\ Michel Marcus, Sep 14 2015

Formula

Equals 2*zeta(5)/6F5(1,1,1,1,1,1; 3/2,2,2,2,2; -1/4).

A005261 a(n) = Sum_{k = 0..n} C(n,k)^5.

Original entry on oeis.org

1, 2, 34, 488, 9826, 206252, 4734304, 113245568, 2816649826, 72001228052, 1883210876284, 50168588906768, 1357245464138656, 37198352117916992, 1030920212982957184, 28847760730478655488, 814066783370083977826
Offset: 0

Views

Author

Keywords

Comments

a(n) is the constant term in the expansion of ((1 + w) * (1 + x) * (1 + y) * (1 + z) + (1 + 1/w) * (1 + 1/x) * (1 + 1/y) * (1 + 1/z))^n. - Seiichi Manyama, Oct 27 2019

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=5 of A309010.
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

Programs

  • Maple
    a := n -> hypergeom([seq(-n, i=1..5)],[seq(1, i=1..4)], -1):
    seq(simplify(a(n)),n=0..16); # Peter Luschny, Jul 27 2016
  • Mathematica
    RecurrenceTable[{32*(55n^2+33n+6)*(n-1)^4*a[n-2]-(19415n^6-27181n^5+7453n^4+3289n^3-956n^2-276n+96)*a[n-1]-(1155n^6+693n^5-732n^4-715n^3+45n^2+210n+56)*a[n]+(55n^2-77n+28)*(n+1)^4*a[n+1]==0,a[0]==1,a[1]==2,a[2]==34},a,{n,0,25}]
    (* or directly *)
    Table[Sum[Binomial[n,k]^5,{k,0,n}],{n,0,25}] (* Vaclav Kotesovec, Apr 27 2012 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)^5); \\ Michel Marcus, Mar 09 2016
    
  • Python
    def A005261(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m = m*(n-k)**5//(k+1)**5
        return g # Chai Wah Wu, Oct 04 2022

Formula

a(n) ~ 4*5^(-1/2)*Pi^-2*n^-2*2^(5*n). - Joe Keane (jgk(AT)jgk.org), Jun 21 2002
Recurrence (M. A. Perlstadt, 1987): 32*(55*n^2 + 33*n + 6)*(n - 1)^4*a(n-2) - (19415*n^6 - 27181*n^5 + 7453*n^4 + 3289*n^3 - 956*n^2 - 276*n + 96)*a(n-1) - (1155*n^6 + 693*n^5 - 732*n^4 - 715*n^3 + 45*n^2 + 210*n + 56)*a(n) + (55*n^2 - 77*n + 28)*(n + 1)^4*a(n+1) = 0. [Vaclav Kotesovec, Apr 27 2012]
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^5*C(n,k)^5 = C(n,r)^5*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
Sum_{n>=0} a(n) * x^n / (n!)^5 = (Sum_{n>=0} x^n / (n!)^5)^2. - Ilya Gutkovskiy, Jul 17 2020
From Peter Bala, Nov 01 2024: (Start)
For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n, k)^4 * binomial(n-1, k).
For n >= 1, a(n) = 2 * hypergeom([-n, -n, -n, -n, -n + 1], [1, 1, 1, 1], -1). (End)

Extensions

More terms from Matthew Conroy, Mar 16 2006

A069865 a(n) = Sum_{k = 0..n} C(n,k)^6.

Original entry on oeis.org

1, 2, 66, 1460, 54850, 2031252, 86874564, 3848298792, 180295263810, 8709958973540, 433617084579316, 22071658807720392, 1145600816547477316, 60423221241495866600, 3231675487858598367240, 174928470621208572186960
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org), Jun 21 2002

Keywords

Crossrefs

Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.

Programs

  • Maple
    a := n -> hypergeom([seq(-n, i=1..6)],[seq(1, i=1..5)],1):
    seq(simplify(a(n)),n=0..15); # Peter Luschny, Jul 27 2016
  • Mathematica
    RecurrenceTable[{24(6n-7)(2n-1)(6n-5)(91n^3 + 91n^2 + 35n + 5)(n-1)^3*a[n-2] -(153881n^9-307762n^8 + 185311n^7 + 2960n^6-31631n^5-88n^4 + 5239n^3-610n^2-440n + 100)*a[n-1] -n(3458n^8 + 1729n^7-2947n^6-2295n^5 + 901n^4 + 1190n^3 + 52n^2-228n-60)*a[n] + n(91n^3-182n^2 + 126n-30)(n + 1)^5*a[n + 1]==0, a[0]==1,a[1]==2,a[2]==66},a,{n,0,25}] (* Vaclav Kotesovec, Apr 27 2012 *)
    Table[Sum[Binomial[n, k]^6, {k, 0, n}], {n, 0, 25}] (* Vincenzo Librandi, May 03 2013 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n, k)^6); \\ Michel Marcus, Mar 09 2016
    
  • Python
    def A069865(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m = m*(n-k)**6//(k+1)**6
        return g # Chai Wah Wu, Oct 04 2022

Formula

a(n) ~ 4*3^(-1/2)*Pi^(-5/2)*n^(-5/2)*2^(6*n).
Recurrence (M. A. Perlstadt, 1987): 24*(6*n - 7)*(2*n - 1)*(6*n - 5)*(91*n^3 + 91*n^2 + 35*n + 5)*(n - 1)^3*a(n-2) - (153881*n^9 - 307762*n^8 + 185311*n^7 + 2960*n^6 - 31631*n^5 - 88*n^4 + 5239*n^3 - 610*n^2 - 440*n + 100)*a(n-1) - n*(3458*n^8 + 1729*n^7 - 2947*n^6 - 2295*n^5 + 901*n^4 + 1190*n^3 + 52*n^2 - 228*n - 60)*a(n) + n*(91*n^3 - 182*n^2 + 126*n - 30)*(n + 1)^5*a(n+1) = 0. [Vaclav Kotesovec, Apr 27 2012]
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^6*C(n,k)^6 = C(n,r)^6*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
Sum_{n>=0} a(n) * x^n / (n!)^6 = (Sum_{n>=0} x^n / (n!)^6)^2. - Ilya Gutkovskiy, Jul 17 2020
From Peter Bala, Nov 01 2024: (Start)
For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n, k)^5 * binomial(n-1, k).
For n >= 1, a(n) = 2 * hypergeom([-n, -n, -n, -n, -n, -n + 1], [1, 1, 1, 1, 1], 1).
(End)

A245086 Central values of the n-th discrete Chebyshev polynomials of order 2n.

Original entry on oeis.org

1, 0, -6, 0, 90, 0, -1680, 0, 34650, 0, -756756, 0, 17153136, 0, -399072960, 0, 9465511770, 0, -227873431500, 0, 5550996791340, 0, -136526995463040, 0, 3384731762521200, 0, -84478098072866400, 0, 2120572665910728000, 0, -53494979785374631680, 0
Offset: 0

Views

Author

Nikita Gogin, Jul 11 2014

Keywords

Comments

In the general case the n-th discrete Chebyshev polynomial of order N is D(N,n;x) = Sum_{i = 0..n} (-1)^i*C(n,i)*C(N-x,n-i)*C(x,i). For N = 2*n , x = n, one gets a(n) = D(2n,n;n) = Sum_{i = 0..n} (-1)^i*C(n,i)^3 that equals (due to Dixon's formula) 0 for odd n and (-1)^m*(3m)!/(m!)^3 for n = 2*m. (Riordan, 1968) So, a(2*m) = (-1)^m*A006480(m).

References

  • John Riordan, Combinatorial Identities, John Willey&Sons Inc., 1968.

Crossrefs

Programs

  • Mathematica
    Table[Coefficient[Simplify[JacobiP[n,0,-(2*n+1),(1+t^2)/(1-t^2)]*(1-t^2)^n],t,n],{n,0,20}]
  • Python
    from math import factorial
    def A245086(n): return 0 if n&1 else (-1 if (m:=n>>1)&1 else 1)*factorial(3*m)//factorial(m)**3 # Chai Wah Wu, Oct 04 2022

Formula

a(n) is a coefficient at t^n in (1-t^2)^n*P(0,-(2*n+1);n;(1+t^2)/(1-t^2)), where P(a,b;k;x) is the k-th Jacobi polynomial (Gogin and Hirvensalo, 2007).
G.f.: Hypergeometric2F1[1/3,2/3,1,-27*x^2].
a(2*m+1) = 0, a(2*m) = (-1)^m*A006480(m).
From Peter Bala, Aug 04 2016: (Start)
a(n) = Sum_{k = 0..n} (-1)^k*binomial(n,k)*binomial(2*n - k,n)*binomial(n + k,n) (Sun and Wang).
a(n) = Sum_{k = 0..n} (-1)^(n + k)*binomial(n + k, n - k)*binomial(2*k, k)*binomial(2*n - k, n) (Gould, Vol.5, 9.23).
a(n) = -1/(n + 1)^3 * A273630(n+1). (End)
From Peter Bala, Mar 22 2022: (Start)
a(n) = - (3*(3*n-2)*(3*n-4)/n^2)*a(n-2).
a(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A002894(n) = binomial(2*n,n)^2 = [x^n] (1 - x)^(2*n) * P(2*n,(1 + x)/(1 - x)). Cf. A103882. (End)
From Peter Bala, Jul 23 2023: (Start)
a(n) = [x^n] G(x)^(3*n), where the power series G(x) = 1 - x^2 + 2*x^4 - 14*x^6 + 127*x^8 - 1364*x^10 + ... appears to have integer coefficients.
exp(Sum_{n >= 1} a(n)*x^n/n) = F(x)^3, where the power series F(x) = 1 - x^2 + 8*x^4 - 101*x^6 + 1569*x^8 - 27445*x^10 + ..., appears to have integer coefficients. See A229452.
Row 1 of A364303. (End)
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(n+k, k)^2 * binomial(3*n+1, n-k). Cf. A183204.- Peter Bala, Sep 20 2024
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