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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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

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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).

A036829 a(n) = Sum_{k=0..n-1} C(3*k,k)*C(3*n-3*k-2,n-k-1).

Original entry on oeis.org

0, 1, 7, 48, 327, 2221, 15060, 102012, 690519, 4671819, 31596447, 213633696, 1444131108, 9760401756, 65957919496, 445671648228, 3011064814455, 20341769686311, 137412453018933, 928188965638464, 6269358748632207, 42343731580741821
Offset: 0

Views

Author

Keywords

References

  • M. Petkovsek et al., A=B, Peters, 1996, p. 97.

Crossrefs

Programs

  • Haskell
    a036829 n = sum $ map
       (\k -> (a007318 (3*k) k) * (a007318 (3*n-3*k-2) (n-k-1))) [0..n-1]
    -- Reinhard Zumkeller, May 24 2012
  • Mathematica
    Table[Sum[Binomial[3k,k]Binomial[3n-3k-2,n-k-1],{k,0,n-1}],{n,0,30}] (* Harvey P. Dale, Jan 10 2012 *)

Formula

G.f.: (g-g^2)/(3*g-1)^2 where g*(1-g)^2 = x. - Mark van Hoeij, Nov 09 2011
Recurrence: 8*(n-1)*(2*n-1)*a(n) = 6*(36*n^2-81*n+49)*a(n-1) - 81*(3*n-5)*(3*n-4)*a(n-2). - Vaclav Kotesovec, Nov 19 2012
a(n) ~ 3^(3*n-1)/2^(2*n+1). - Vaclav Kotesovec, Dec 29 2012
L.g.f.: Sum_{k>=1} a(k)*x^k/k = (1/3) * log( Sum_{k>=0} binomial(3*k,k)*x^k ). - Seiichi Manyama, Jul 19 2025
G.f.: (g-1)/(3-2*g)^2 where g=1+x*g^3. - Seiichi Manyama, Jul 26 2025

A036916 a(n) = Sum_{k=0..n} binomial(2*n-2*k,n-k)^2 * binomial(n,k)^2.

Original entry on oeis.org

1, 5, 53, 761, 12661, 229705, 4410665, 88127485, 1813270645, 38158684745, 817458330553, 17767242718285, 390819348043369, 8683822363169933, 194618212789162733, 4394243766346694161, 99862206804817230965, 2282427331053360624713
Offset: 0

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Author

Keywords

References

  • Cf. M. Petkovsek et al., A=B, Peters, p. ix.

Crossrefs

Cf. A007318.
Row n=4 of A275784.

Programs

  • Haskell
    a036916 n = sum $ map
       (\k -> (a007318 (2*n-2*k) (n-k))^2 * (a007318 n k)^2) [0..n]
    -- Reinhard Zumkeller, May 24 2012
  • Mathematica
    Table[Sum[Binomial[2n-2k,n-k]^2 Binomial[n,k]^2,{k,0,n}],{n,0,20}] (* Harvey P. Dale, Mar 31 2013 *)

Formula

(n - 1)*(144*n^3 - 864*n^2 + 1693*n - 1075)*n^3*a(n) - 2*(n - 1)*(2592*n^6 - 19440*n^5 + 56322*n^4 - 80296*n^3 + 60004*n^2 - 23017*n + 3580)*a(n - 1) + (42336*n^7 - 423360*n^6 + 1769838*n^5 - 4006912*n^4 + 5293968*n^3 - 4062414*n^2 + 1661406*n - 274520)*a(n - 2) - 2*(34848*n^5 - 261360*n^4 + 741842*n^3 - 984642*n^2 + 598948*n - 127215)*(n - 2)^2*a(n - 3) + 225*(144*n^3 - 432*n^2 + 397*n - 102)*(n - 2)^2*(n - 3)^2*a(n - 4) = 0 - Vladeta Jovovic, Jul 15 2004
a(n) ~ 5^(2*n+2) / (16 * (Pi*n)^(3/2)). - Vaclav Kotesovec, Mar 02 2014

A036915 a(n) = Sum_{k=0..n} C(2*n-2*k,n-k)^2 * C(2*n,k)^2.

Original entry on oeis.org

1, 8, 136, 2996, 76168, 2121808, 62886484, 1947164948, 62250464648, 2038917457664, 68056394358736, 2306354482651568, 79138758509239636, 2743929765899780336, 95983445574557632948, 3383168407446955236196, 120039826640626408744328
Offset: 0

Views

Author

Keywords

References

  • There's an erroneous assertion (caused by a typographical error) in the first printing of M. Petkovsek et al., "A=B", Peters, p. ix. that this is the same as A036917; it is not.

Programs

  • Mathematica
    Table[Sum[Binomial[2n-2k,n-k]^2 Binomial[2n,k]^2,{k,0,n}],{n,0,20}] (* Harvey P. Dale, May 31 2015 *)

A172390 G.f. satisfies: A(x) = G(x/A(x))^2 and G(x)^2 = A(x*G(x)^2) where G(x) = Sum_{n>=0} C(2*n,n)^2*x^n.

Original entry on oeis.org

1, 8, 24, 0, -168, 0, 2112, 0, -32040, 0, 536256, 0, -9542976, 0, 177126912, 0, -3390361128, 0, 66436117440, 0, -1326185205696, 0, 26872637815296, 0, -551301904867392, 0, 11428295231789568, 0, -239010764560888320, 0
Offset: 0

Views

Author

Paul D. Hanna, Feb 04 2010

Keywords

Examples

			G.f.: A(x) = 1 + 8*x + 24*x^2 - 168*x^4 + 2112*x^6 - 32040*x^8 + ...
A(x) = G(x/A(x))^2 where G(x) = 1/AGM(1, (1-16*x)^(1/2)) is the power series:
G(x) = 1 + 2^2*x + 6^2*x^2 + 20^2*x^3 + 70^2*x^4 + 252^2*x^5 + ... + C(2*n,n)^2*x^n + ...
The square root of g.f. A(x) begins:
A(x)^(1/2) = 1 + 4*x + 4*x^2 - 16*x^3 - 28*x^4 + 176*x^5 + 336*x^6 + ... + A158101(n)*x^n + ...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(G=sum(m=0,n,binomial(2*m,m)^2*x^m)+x*O(x^n));polcoeff(x/serreverse(x*G^2),n)}
    
  • PARI
    {a(n)=if(n==1,8,polcoeff(agm(1,sqrt(1-16*x +x^2*O(x^n)))^(2*n-2),n)/(1-n))} \\ Paul D. Hanna, Mar 20 2010

Formula

G.f.: A(x) = x/Series_Reversion(x*G(x)^2) where G(x) = Sum_{n>=0} C(2*n,n)^2*x^n = 1/agm(1, (1-16*x)^(1/2)) = g.f. of A002894 and G(x)^2 is the g.f. of A036917.
Self-convolution of A158101, which is a bisection of A158100; A158100 has g.f. F(x) that satisfies: F(x) = 1/AGM(1, 1 - 8*x/F(x) ).
a(n) = [x^n] AGM(1,(1-16*x)^(1/2))^(2*n-2)/(1-n) for n>1 where AGM is the arithmetic-geometric mean of Gauss. - Paul D. Hanna, Mar 20 2010

A300116 a(n) = Sum_{k=0..n} binomial(2k,k)^3 * binomial(2n-2k,n-k) * 2^(4*(n-k)).

Original entry on oeis.org

1, 40, 2008, 109120, 6173656, 357903040, 21090174400, 1257411781120, 75630327895000, 4580277582101440, 278915640538355008, 17061127317021130240, 1047543937631077672384, 64523332938885758410240, 3985152917145136901283328, 246717298245058901071237120
Offset: 0

Views

Author

Seiichi Manyama, Feb 25 2018

Keywords

Crossrefs

Programs

  • PARI
    {a(n) = sum(k=0, n, binomial(2*k, k)^3*binomial(2*n-2*k, n-k)*2^(4*(n-k)))}

Formula

n^3 * a(n) = 8 * (2*n-1) * (8*n^2-8*n+5) * a(n-1) - 4096 * (n-1)^3 * a(n-2) for n > 1.
a(n) ~ Gamma(1/4)^4 * 2^(6*n - 2) / (Pi^(7/2) * sqrt(n)). - Vaclav Kotesovec, Jul 10 2021

A089624 Expansion of sqrt(2/Pi*EllipticK(4*sqrt(x))).

Original entry on oeis.org

1, 2, 16, 168, 1986, 25092, 330816, 4492560, 62352720, 879956000, 12583279360, 181872982400, 2652039363240, 38959845007440, 575974743052800, 8561706637619520, 127874111328349890, 1917875205285147780
Offset: 0

Views

Author

D. G. Rogers and Vladeta Jovovic, Dec 31 2003

Keywords

Comments

When convolved with itself gives A002894.

Crossrefs

Cf. A036917.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sqrt[Sum[Binomial[2*k, k]^2*x^k, {k, 0, nmax}]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Apr 10 2018 *)
    nmax = 20; CoefficientList[Series[Sqrt[2*EllipticK[16*x]/Pi], {x, 0, nmax}], x] (* Vaclav Kotesovec, Apr 10 2018 *)
  • PARI
    {a(n) = if(n<0, 0, polcoeff( sqrt( sum(k=0, n, binomial(2*k, k)^2 * x^k, x*O(x^n)) ), n))} /* Michael Somos, Aug 17 2007 */
    
  • PARI
    {a(n) = local(A); if(n<0, 0, A = x*O(x^n); polcoeff( subst( sum(k = 1, sqrtint(n), 2*x^k^2, 1+A), x, serreverse(x * (eta(x+A) * eta(x^4+A)^2 / eta(x^2+A)^3)^8 )), n))} /* Michael Somos, Aug 17 2007 */

Formula

Expansion of theta_3(q) in powers of (m/16) where q = exp(-Pi K'/K) and m = k^2 is the elliptic modulus. - Michael Somos, Aug 17 2007
a(n) ~ 2^(4*n-1) / (n*sqrt(Pi*log(n))) * (1 - (gamma/2 + 2*log(2)) / log(n) + (3*gamma^2/8 + 3*log(2)*gamma + 6*log(2)^2 - Pi^2/16) / log(n)^2), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Sep 29 2019
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