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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A005259 Apery (Apéry) numbers: Sum_{k=0..n} (binomial(n,k)*binomial(n+k,k))^2.

Original entry on oeis.org

1, 5, 73, 1445, 33001, 819005, 21460825, 584307365, 16367912425, 468690849005, 13657436403073, 403676083788125, 12073365010564729, 364713572395983725, 11111571997143198073, 341034504521827105445, 10534522198396293262825, 327259338516161442321485
Offset: 0

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Comments

Conjecture: For each n = 1,2,3,... the Apéry polynomial A_n(x) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
The expansions of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + 232771*x^5 + ... and exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + 3*x + 27*x^2 + 390*x^3 + 7038*x^4 + 144550*x^5 + ... both appear to have integer coefficients. See A267220. - Peter Bala, Jan 12 2016
Diagonal of the rational function R(x, y, z, w) = 1 / (1 - (w*x*y*z + w*x*y + w*z + x*y + x*z + y + z)); also diagonal of rational function H(x, y, z, w) = 1/(1 - w*(1+x)*(1+y)*(1+z)*(x*y*z + y*z + y + z + 1)). - Gheorghe Coserea, Jun 26 2018
Named after the French mathematician Roger Apéry (1916-1994). - Amiram Eldar, Jun 10 2021

Examples

			G.f. = 1 + 5*x + 73*x^2 + 1445*x^3 + 33001*x^4 + 819005*x^5 + 21460825*x^6 + ...
a(2) = (binomial(2,0) * binomial(2+0,0))^2 + (binomial(2,1) * binomial(2+1,1))^2 + (binomial(2,2) * binomial(2+2,2))^2 = (1*1)^2 + (2*3)^2 + (1*6)^2 = 1 + 36 + 36 = 73. - _Michael B. Porter_, Jul 14 2016
		

References

  • Julian Havil, The Irrationals, Princeton University Press, Princeton and Oxford, 2012, pp. 137-153.
  • Wolfram Koepf, Hypergeometric Identities. Ch. 2 in Hypergeometric Summation: An Algorithmic Approach to Summation and Special Function Identities. Braunschweig, Germany: Vieweg, pp. 55, 119 and 146, 1998.
  • Maxim Kontsevich and Don Zagier, Periods, pp. 771-808 of B. Engquist and W. Schmid, editors, Mathematics Unlimited - 2001 and Beyond, 2 vols., Springer-Verlag, 2001.
  • Leonard Lipshitz and Alfred van der Poorten, "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), pp. 339-358.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Apéry's number or Apéry's constant zeta(3) is A002117. - N. J. A. Sloane, Jul 11 2023
Related to diagonal of rational functions: A268545-A268555.
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.
Cf. A092826 (prime terms).

Programs

  • GAP
    List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*Binomial(n+k,k)^2)); # Muniru A Asiru, Sep 28 2018
    
  • Haskell
    a005259 n = a005259_list !! n
    a005259_list = 1 : 5 : zipWith div (zipWith (-)
       (tail $ zipWith (*) a006221_list a005259_list)
       (zipWith (*) (tail a000578_list) a005259_list)) (drop 2 a000578_list)
    -- Reinhard Zumkeller, Mar 13 2014
    
  • Magma
    [&+[Binomial(n, k) ^2 *Binomial(n+k, k)^2: k in [0..n]]:n in  [0..17]]; // Marius A. Burtea, Jan 20 2020
    
  • Maple
    a := proc(n) option remember; if n=0 then 1 elif n=1 then 5 else (n^(-3))* ( (34*(n-1)^3 + 51*(n-1)^2 + 27*(n-1) +5)*a((n-1)) - (n-1)^3*a((n-1)-1)); fi; end;
    # Alternative:
    a := n -> hypergeom([-n, -n, 1+n, 1+n], [1, 1, 1], 1):
    seq(simplify(a(n)), n=0..17); # Peter Luschny, Jan 19 2020
  • Mathematica
    Table[HypergeometricPFQ[{-n, -n, n+1, n+1}, {1,1,1}, 1],{n,0,13}] (* Jean-François Alcover, Apr 01 2011 *)
    Table[Sum[(Binomial[n,k]Binomial[n+k,k])^2,{k,0,n}],{n,0,30}] (* Harvey P. Dale, Oct 15 2011 *)
    a[ n_] := SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ SeriesCoefficient[ 1 / (1 - t (1 + x ) (1 + y ) (1 + z ) (x y z + (y + 1) (z + 1))), {t, 0, n}], {x, 0, n}], {y, 0, n}], {z, 0, n}]; (* Michael Somos, May 14 2016 *)
  • PARI
    a(n)=sum(k=0,n,(binomial(n,k)*binomial(n+k,k))^2) \\ Charles R Greathouse IV, Nov 20 2012
    
  • Python
    def A005259(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m *= ((n+k+1)*(n-k))**2
            m //=(k+1)**4
        return g # Chai Wah Wu, Oct 02 2022

Formula

D-finite with recurrence (n+1)^3*a(n+1) = (34*n^3 + 51*n^2 + 27*n + 5)*a(n) - n^3*a(n-1), n >= 1.
Representation as a special value of the hypergeometric function 4F3, in Maple notation: a(n)=hypergeom([n+1, n+1, -n, -n], [1, 1, 1], 1), n=0, 1, ... - Karol A. Penson Jul 24 2002
a(n) = Sum_{k >= 0} A063007(n, k)*A000172(k). A000172 = Franel numbers. - Philippe Deléham, Aug 14 2003
G.f.: (-1/2)*(3*x - 3 + (x^2-34*x+1)^(1/2))*(x+1)^(-2)*hypergeom([1/3,2/3],[1],(-1/2)*(x^2 - 7*x + 1)*(x+1)^(-3)*(x^2 - 34*x + 1)^(1/2)+(1/2)*(x^3 + 30*x^2 - 24*x + 1)*(x+1)^(-3))^2. - Mark van Hoeij, Oct 29 2011
Let g(x, y) = 4*cos(2*x) + 8*sin(y)*cos(x) + 5 and let P(n,z) denote the Legendre polynomial of degree n. Then G. A. Edgar posted a conjecture of Alexandru Lupas that a(n) equals the double integral 1/(4*Pi^2)*int {y = -Pi..Pi} int {x = -Pi..Pi} P(n,g(x,y)) dx dy. (Added Jan 07 2015: Answered affirmatively in Math Overflow question 178790) - Peter Bala, Mar 04 2012; edited by G. A. Edgar, Dec 10 2016
a(n) ~ (1+sqrt(2))^(4*n+2)/(2^(9/4)*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Nov 01 2012
a(n) = Sum_{k=0..n} C(n,k)^2 * C(n+k,k)^2. - Joerg Arndt, May 11 2013
0 = (-x^2+34*x^3-x^4)*y''' + (-3*x+153*x^2-6*x^3)*y'' + (-1+112*x-7*x^2)*y' + (5-x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
From Peter Bala, Jan 18 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j) * C(n,k)^2 * C(n+k,k)^2 * C(n,j) * C(n+k+j,k+j).
a(n) = Sum_{0 <= j, k <= n} C(n,k) * C(n+k,k) * C(k,j)^3 (see Koepf, p. 55).
a(n) = Sum_{0 <= j, k <= n} C(n,k)^2 * C(n,j)^2 * C(3*n-j-k,2*n) (see Koepf, p. 119).
Diagonal coefficients of the rational function 1/((1 - x - y)*(1 - z - t) - x*y*z*t) (Straub, 2014). (End)
a(n) = [x^n] 1/(1 - x)*( Legendre_P(n,(1 + x)/(1 - x)) )^m at m = 2. At m = 1 we get the Apéry numbers A005258. - Peter Bala, Dec 22 2020
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*A108625(n, k). - Peter Bala, Jul 18 2024
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n,k)^2 * C(n,j)^2 * C(k+j,k), see Labelle et al. link. - Max Alekseyev, Mar 12 2025

A006480 De Bruijn's S(3,n): (3n)!/(n!)^3.

Original entry on oeis.org

1, 6, 90, 1680, 34650, 756756, 17153136, 399072960, 9465511770, 227873431500, 5550996791340, 136526995463040, 3384731762521200, 84478098072866400, 2120572665910728000, 53494979785374631680, 1355345464406015082330, 34469858696831179429500, 879619727485803060256500, 22514366432046593564460000
Offset: 0

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Author

Keywords

Comments

Number of paths of length 3n in an n X n X n grid from (0,0,0) to (n,n,n), using steps (0,0,1), (0,1,0), and (1,0,0).
Appears in Ramanujan's theory of elliptic functions of signature 3.
S(s,n) = Sum_{k=0..2n} (-1)^(k+n) * binomial(2n, k)^s. The formula S(3,n) = (3n)!/(n!)^3 is due to Dixon (according to W. N. Bailey 1935). - Charles R Greathouse IV, Dec 28 2011
a(n) is the number of ballot results that end in a 3-way tie when 3n voters each cast two votes for two out of three candidates vying for 2 slots on a county board; in such a tie, each of the three candidates receives 2n votes. Note there are C(3n,2n) ways to choose the voters who cast a vote for the youngest candidate. The n voters who did note vote for the youngest candidate voted for the two older candidates. Then there are C(2n,n) ways to choose the other n voters who voted for both the youngest and the second youngest candidate. The remaining voters vote for the oldest candidate. Hence there are C(3n,2n)*C(2n,n)=(3n)!/(n!)^3 ballot results. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X+Y+1/(X*Y))^(3*n). - Mark van Hoeij, May 07 2013
For n > 2 a(n) is divisible by (n+2)*(n+1)^2, a(n) = (n+1)^2*(n+2)*A161581(n). - Alexander Adamchuk, Dec 27 2013
a(n) is the number of permutations of the multiset {1^n, 2^n, 3^n}, the number of ternary words of length 3*n with n of each letters. - Joerg Arndt, Feb 28 2016
Diagonal of the rational function 1/(1 - x - y - z). - Gheorghe Coserea, Jul 06 2016
At least two families of elliptic curves, x = 2*H1 = (p^2+q^2)*(1-q) and x = 2*H2 = p^2+q^2-3*p^2*q+q^3 (0Bradley Klee, Feb 25 2018
The ordinary generating function also determines periods along a family of tetrahedral-symmetric sphere curves ("du troisième ordre"). Compare links to Goursat "Étude des surfaces..." and "Proof Certificate". - Bradley Klee, Sep 28 2018

Examples

			G.f.: 1 + 6*x + 90*x^2 + 1680*x^3 + 34650*x^4 + 756756*x^5 + 17153136*x^6 + ...
		

References

  • L. A. Aizenberg and A. P. Yuzhakov, "Integral representations and residues in multidimensional complex analysis", American Mathematical Society, 1983, p. 194.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 174.
  • N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row 3 of A187783.
Related to diagonal of rational functions: A268545-A268555. Elliptic Integrals: A002894, A113424, A000897. Factors: A005809, A000984. Integrals: A007004, A024486. Sphere Curves: A318245, A318495.

Programs

  • GAP
    List([0..20],n->Factorial(3*n)/Factorial(n)^3); # Muniru A Asiru, Mar 31 2018
    
  • Magma
    [Factorial(3*n)/(Factorial(n))^3: n in [0..20] ]; // Vincenzo Librandi, Aug 20 2011
    
  • Maple
    seq((3*n)!/(n!)^3, n=0..16); # Zerinvary Lajos, Jun 28 2007
  • Mathematica
    Sum [ (-1)^(k+n) Binomial[ 2n, k ]^3, {k, 0, 2n} ]
    a[ n_] := If[ n < 0, 0, (-1)^n HypergeometricPFQ[ {-2 n, -2 n, -2 n}, {1, 1}, 1]]; (* Michael Somos, Oct 22 2014 *)
    Table[Multinomial[n, n, n], {n, 0, 100}] (* Emanuele Munarini, Oct 25 2016 *)
    CoefficientList[Series[Hypergeometric2F1[1/3,2/3,1,27*x],{x,0,5}],x] (* Bradley Klee, Feb 28 2018 *)
    Table[(3n)!/(n!)^3,{n,0,20}] (* Harvey P. Dale, Mar 09 2025 *)
  • Maxima
    makelist(multinomial_coeff(n,n,n),n,0,24); /* Emanuele Munarini, Oct 25 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, (3*n)! / n!^3)}; /* Michael Somos, Dec 03 2002 */
    
  • PARI
    {a(n) = my(A, m); if( n<1, n==0, m=1; A = 1 + O(x); while( m<=n, m*=3; A = subst( (1 + 2*x) * subst(A, x, (x/3)^3), x, serreverse(x * (1 + x + x^2) / (1 + 2*x)^3 / 3 + O(x^m)))); polcoeff(A, n))}; /* Michael Somos, Dec 03 2002 */
    
  • Python
    from math import factorial
    def A006480(n): return factorial(3*n)//factorial(n)**3 # Chai Wah Wu, Oct 04 2022

Formula

Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 1/2 * sqrt(3) * 27^n / (Pi*n) - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
From Karol A. Penson, Nov 21 2001: (Start)
O.g.f.: hypergeom([1/3, 2/3], [1], 27*x).
E.g.f.: hypergeom([1/3, 2/3], [1, 1], 27*x).
Integral representation as n-th moment of a positive function on [0, 27]:
a(n) = int( x^n*(-1/24*(3*sqrt(3)*hypergeom([2/3, 2/3], [4/3], 1/27*x)* Gamma(2/3)^6*x^(1/3) - 8*hypergeom([1/3, 1/3], [2/3], 1/27*x)*Pi^3)/Pi^3 /x^(2/3)/Gamma(2/3)^3), x=0..27). This representation is unique. (End)
a(n) = Sum_{k=-n..n} (-1)^k*binomial(2*n, n+k)^3. - Benoit Cloitre, Mar 02 2005
a(n) = C(2n,n)*C(3n,n) = A104684(2n,n). - Paul Barry, Mar 14 2006
G.f. satisfies: A(x^3) = A( x*(1+3*x+9*x^2)/(1+6*x)^3 )/(1+6*x). - Paul D. Hanna, Oct 29 2010
D-finite with recurrence: n^2*a(n) - 3*(3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = (n+1)^2*(n+2)*A161581(n) for n>2. - Alexander Adamchuk, Dec 27 2013
0 = a(n)^2*(472392*a(n+1)^2 - 83106*a(n+1)*a(n+2) + 3600*a(n+2)^2) + a(n)*a(n+1)*(-8748*a(n+1)^2 + 1953*a(n+1)*a(n+2) - 120*a(n+2)^2) + a(n+1)^2*(36*a(n+1)^2 - 12*a(n+1)*a(n+2) + a(n+2)^2) for all n in Z. - Michael Somos, Oct 22 2014
0 = x*(27*x-1)*y'' + (54*x-1)*y' + 6*y, where y is g.f. - Gheorghe Coserea, Jul 06 2016
From Peter Bala, Jul 15 2016: (Start)
a(n) = 3*binomial(2*n - 1,n)*binomial(3*n - 1,n) = 3*[x^n] 1/(1 - x)^n * [x^n] 1/(1 - x)^(2*n) for n >= 1.
a(n) = binomial(2*n,n)*binomial(3*n,n) = ([x^n](1 + x)^(2*n)) *([x^n](1 + x)^(3*n)) = [x^n](F(x)^(6*n)), where F(x) = 1 + x + 2*x^2 + 14*x^3 + 127*x^4 + 1364*x^5 + 16219*x^6 + ... appears to have integer coefficients. Cf. A002894.
This sequence occurs as the right-hand side of several binomial sums:
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)^3 = a(n) (Dixon's identity).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n,n - k)*binomial(3*n + k,k) = a(n) (Gould, Vol. 4, 6.86)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*binomial(2*n + k,n)*binomial(3*n + k,n) = a(n).
Sum_{k = 0..n} binomial(n,k)*binomial(2*n + k,k)*binomial(3*n,n - k) = a(n).
Sum_{k = 0..n} (-1)^(k)*binomial(n,k)*binomial(3*n - k,n)*binomial(4*n - k,n) = a(n).
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n + k,2*n - k)*binomial(2*k,k)*binomial(4*n - k,2*n) = a(n) (see Gould, Vol.5, 9.23).
Sum_{k = 0..2*n} (-1)^k*binomial(3*n,k)*binomial(3*n - k,n)^3 = a(n) (Sprugnoli, Section 2.9, Table 10, p. 123). (End)
From Bradley Klee, Feb 28 2018: (Start)
a(n) = A005809(n)*A000984(n).
G.f.: F(x) = 1/(2*Pi) Integral_{z=0..2*Pi} 2F1(1/3,2/3; 1/2; 27*x*sin^2(z)) dz.
With G(x) = x*2F1(1/3,2/3; 2; 27*x): F(x) = d/dx G(x). (Cf. A007004) (End)
F(x)*G(1/27-x) + F(1/27-x)*G(x) = 1/(4*Pi*sqrt(3)). - Bradley Klee, Sep 29 2018
Sum_{n>=0} 1/a(n) = A091683. - Amiram Eldar, Nov 15 2020
From Peter Bala, Sep 20 2021: (Start)
a(n) = Sum_{k = n..2*n} binomial(2*n,k)^2 * binomial(k,n). Cf. A001459.
a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for any prime p >= 5 and any positive integers n and k (write a(n) as C(3*n,2*n)*C(2*n,n) and apply Mestrovic, equation 39, p. 12). (End)
a(n) = 6*A060542(n). - R. J. Mathar, Jun 21 2023
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^3 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^3 * binomial(2*n, n+k)^3 = x*(x + n)*(x + 2*n)*a(n) (x arbitrary). Compare with Dixon's identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^3 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Aug 14 2023: (Start)
a(n) = (-1)^n * [x^(2*n)] ( (1 - x)^(4*n) * Legendre_P(2*n, (1 + x)/(1 - x)) ).
Row 1 of A364509. (End)
From Peter Bala, Oct 10 2024: (Start)
The following hold for n >= 1:
a(n) = Sum_{k = 0.. 2*n} (-1)^(n+k) * k/n * binomial(2*n, k)^3 = 3/2 * Sum_{k = 0.. 2*n} (-1)^(n+k) * (k/n)^2 * binomial(2*n, k)^3.
a(n) = 3/2 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)^2*binomial(2*n-1, k).
a(n) = 3 * Sum_{0..2*n-1} (-1)^(n+k) * k/n * binomial(2*n, k)*binomial(2*n-1, k)^2. (End)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(2*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

Extensions

a(14)-a(16) from Eric W. Weisstein
Terms a(17) and beyond from T. D. Noe, Jun 29 2008

A268545 From the diagonal of 1/(1 - (y + z + x w + x z w + x y w)).

Original entry on oeis.org

1, 10, 246, 7540, 255430, 9163980, 341237820, 13042646760, 508236930630, 20101587623260, 804500381097556, 32508382071448920, 1324112273705453596, 54296281503438398200, 2239266766596344681400, 92809720054802928741840, 3863305447624183692730950, 161427619265399264526790140
Offset: 0

Views

Author

N. J. A. Sloane, Feb 29 2016

Keywords

Comments

From Gheorghe Coserea, Jul 03 2016: (Start)
Also diagonal of R(x,y,z) = 1/(1 - x - y - z - x*y - y*z).
Annihilating differential operator: x*(4*x+3)*(16*x^2+44*x-1)*Dx^2 + (128*x^3+320*x^2+264*x-3)*Dx + 16*x^2+12*x+30.
(End)

Crossrefs

Programs

  • Maple
    A268545 := proc(n)
        1/(1-y-z-x*w-x*z*w-x*y*w) ;
        coeftayl(%, x=0, n) ;
        coeftayl(%, y=0, n) ;
        coeftayl(%, z=0, n) ;
        coeftayl(%, w=0, n) ;
    end proc:
    seq(A268545(n), n=0..40) ; # R. J. Mathar, Apr 15 2016
    #alternative program
    with(combinat): seq(add(binomial(n,j)*add(binomial(j,k)*binomial(n+k,j)*binomial(2*n+k,n), k = 0..j), j = 0..n), n = 0..20); # Peter Bala, Jan 26 2018
  • Mathematica
    a[n_] := a[n] = 1/(1 - y - z - x*w - x*z*w - x*y*w) // SeriesCoefficient[#, {x, 0, n}]& // SeriesCoefficient[#, {y, 0, n}]& // SeriesCoefficient[#, {z, 0, n}]& // SeriesCoefficient[#, {w, 0, n}]&;
    Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 0, 40}]
    (* or: *)
    HypergeometricPFQ[{1/12, 5/12}, {1}, (6912*x^3*(-16*x^2 - 44*x + 1))/ (16*x^2 - 40*x + 1)^3]/(16*x^2 - 40*x + 1)^(1/4) + O[x]^41 // CoefficientList[#, x]& (* Jean-François Alcover, Nov 12 2017, after Gheorghe Coserea *)
  • PARI
    \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
    read("hypergeom.gpi");
    N = 21; x = 'x + O('x^N);
    Vec(hypergeom([1/12, 5/12],[1],6912*x^3*(1-44*x-16*x^2)/(1-40*x+16*x^2)^3, N)/(1-40*x+16*x^2)^(1/4)) \\ Gheorghe Coserea, Jul 03 2016
    
  • PARI
    diag(expr, N=22, var=variables(expr)) = {
      my(a = vector(N));
      for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
      for (n = 1, N, a[n] = expr;
        for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
      return(a);
    };
    diag(1/(1 - x - y - z - x*y - y*z), 18)
    \\ test: diag(1/(1-x-y-z-x*y-y*z)) == diag(1/(1-(y+z+x*w + x*z*w + x*y*w)))
    \\ Gheorghe Coserea, Jun 16 2018

Formula

D-finite with recurrence: n^2*(10*n-13)*a(n) +2*(-220*n^3+506*n^2-334*n+63)*a(n-1) -4*(10*n-3)*(-3+2*n)^2*a(n-2)=0. - R. J. Mathar, Apr 15 2016
a(n) ~ (1+sqrt(5))^(5*n+2) / (5^(1/4) * Pi * n * 2^(3*n+3)). - Vaclav Kotesovec, Jul 01 2016
G.f.: hypergeom([1/12, 5/12],[1],6912*x^3*(1-44*x-16*x^2)/(1-40*x+16*x^2)^3)/(1-40*x+16*x^2)^(1/4). - Gheorghe Coserea, Jul 01 2016
0 = x*(4*x+3)*(16*x^2+44*x-1)*y'' + (128*x^3+320*x^2+264*x-3)*y' + (16*x^2+12*x+30)*y, where y is the g.f. - Gheorghe Coserea, Jul 03 2016
a(n) = Sum_{j = 0..n} C(n,j)*Sum_{k = 0..j} C(j,k)*C(n+k,j)*C(2*n+k,n) (apply Eger, Theorem 3 to the set of column vectors S = {[1,0,0], [0,1,0], [0,0,1], [1,1,0], [0,1,1]}. - Peter Bala, Jan 26 2018

A005260 a(n) = Sum_{k = 0..n} binomial(n,k)^4.

Original entry on oeis.org

1, 2, 18, 164, 1810, 21252, 263844, 3395016, 44916498, 607041380, 8345319268, 116335834056, 1640651321764, 23365271704712, 335556407724360, 4854133484555664, 70666388112940818, 1034529673001901732, 15220552520052960516, 224929755893153896200, 3337324864503769353060
Offset: 0

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Author

Keywords

Comments

This sequence is s_10 in Cooper's paper. - Jason Kimberley, Nov 25 2012
Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*x*z + w*y*z + x*y*z + w*x + y*z)). - Gheorghe Coserea, Jul 13 2016
This is one of the Apéry-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Every prime eventually divides some term of this sequence. - Amita Malik, Aug 20 2017
Two walkers, A and B, stand on the South-West and North-East corners of an n X n grid, respectively. A walks by either North or East steps while B walks by either South or West steps. Sequence values a(n) < binomial(2*n,n)^2 count the simultaneous walks where A and B meet after exactly n steps and change places after 2*n steps. - Bradley Klee, Apr 01 2019
a(n) is the constant term in the expansion of ((1 + x) * (1 + y) * (1 + z) + (1 + 1/x) * (1 + 1/y) * (1 + 1/z))^n. - Seiichi Manyama, Oct 27 2019

Examples

			G.f. = 1 + 2*x + 18*x^2 + 164*x^3 + 1810*x^4 + 21252*x^5 + 263844*x^6 + ...
		

References

  • H. W. Gould, Combinatorial Identities, Morgantown, 1972, (X.14), p. 79.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=4 of A309010.
Related to diagonal of rational functions: A268545-A268555.
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.)
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Row sums of A202750.

Programs

  • Maple
    A005260 := proc(n)
            add( (binomial(n,k))^4,k=0..n) ;
    end proc:
    seq(A005260(n),n=0..10) ; # R. J. Mathar, Nov 19 2012
  • Mathematica
    Table[Sum[Binomial[n, k]^4, {k, 0, n}], {n, 0, 20}] (* Wesley Ivan Hurt, Mar 09 2014 *)
  • PARI
    {a(n) = sum(k=0, n, binomial(n, k)^4)};
    
  • Python
    def A005260(n):
        m, g = 1, 0
        for k in range(n+1):
            g += m
            m = m*(n-k)**4//(k+1)**4
        return g # Chai Wah Wu, Oct 04 2022

Formula

a(n) ~ 2^(1/2)*Pi^(-3/2)*n^(-3/2)*2^(4*n). - Joe Keane (jgk(AT)jgk.org), Jun 21 2002
D-finite with recurrence: n^3*a(n) = 2*(2*n - 1)*(3*n^2 - 3*n + 1)*a(n-1) + (4*n - 3)*(4*n - 4)*(4*n - 5)*a(n-2). [Yuan]
G.f.: 5*hypergeom([1/8, 3/8],[1], (4/5)*((1-16*x)^(1/2)+(1+4*x)^(1/2))*(-(1-16*x)^(1/2)+(1+4*x)^(1/2))^5/(2*(1-16*x)^(1/2)+3*(1+4*x)^(1/2))^4)^2/(2*(1-16*x)^(1/2)+3*(1+4*x)^(1/2)). - Mark van Hoeij, Oct 29 2011
1/Pi = sqrt(15)/18 * Sum_{n >= 0} a(n)*(4*n + 1)/36^n (Cooper, equation (5)) = sqrt(15)/18 * Sum_{n >= 0} a(n)*A016813(n)/A009980(n). - Jason Kimberley, Nov 26 2012
0 = (-x^2 + 12*x^3 + 64*x^4)*y''' + (-3*x + 54*x^2 + 384*x^3)*y'' + (-1 + 40*x + 444*x^2)*y' + (2 + 60*x)*y, where y is g.f. - Gheorghe Coserea, Jul 13 2016
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^4*C(n,k)^4 = C(n,r)^4*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
a(n) = hypergeom([-n, -n, -n, -n], [1, 1, 1], 1). - Peter Luschny, Jul 27 2016
Sum_{n>=0} a(n) * x^n / (n!)^4 = (Sum_{n>=0} x^n / (n!)^4)^2. - Ilya Gutkovskiy, Jul 17 2020
a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k)*C(2k,k)*C(2n-2k,n-k)*(-1)^(n-k). This can be proved via the Zeilberger algorithm. - Zhi-Wei Sun, Aug 23 2020
a(n) = (-1)^n*binomial(2*n, n)*hypergeom([1/2, -n, -n, n + 1], [1, 1, 1/2 - n], 1). - Peter Luschny, Aug 24 2020
a(n) = Sum_{k=0..n} binomial(n,k)^2*binomial(2*k,n)*binomial(2*n-k,n) [Theorem 1 in Belbachir and Otmani]. - Michel Marcus, Dec 06 2020
a(n) = [x^n] (1 - x)^(2*n) P(n,(1 + x)/(1 - x))^2, where P(n,x) denotes the n-th Legendre polynomial. See Gould, p. 66. This formula is equivalent to the binomial sum identity of Zhi-Wei Sun given above. - Peter Bala, Mar 24 2022
From Peter Bala, Oct 31 2024: (Start)
For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n, k)^3 * binomial(n-1, k).
For n >= 1, a(n) = 2 * hypergeom([-n, -n, -n, -n + 1], [1, 1, 1], 1). (End)
G.f.: Sum_{k>=0} Sum_{l=0..p*k} Sum_{m=0..l} (-1)^m*binomial(p*k+1,m)*binomial(l+k-m,k)^p*x^(l+k)/(1-x)^(p*k+1), where p = 4. - Miles Wilson, Apr 12 2025

Extensions

Edited by Michael Somos, Aug 09 2002
Minor edits by Vaclav Kotesovec, Aug 28 2014

A006077 (n+1)^2*a(n+1) = (9n^2+9n+3)*a(n) - 27*n^2*a(n-1), with a(0) = 1 and a(1) = 3.

Original entry on oeis.org

1, 3, 9, 21, 9, -297, -2421, -12933, -52407, -145293, -35091, 2954097, 25228971, 142080669, 602217261, 1724917221, 283305033, -38852066421, -337425235479, -1938308236731, -8364863310291, -24286959061533, -3011589296289, 574023003011199, 5028616107443691
Offset: 0

Views

Author

Keywords

Comments

This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Conjecture: Let W(n) be the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to a(i+j) for all i,j = 0,...,n. If n == 1 (mod 3) then W(n) = 0. When n == 0 or 2 (mod 3), W(n)*(-1)^(floor((n+1)/3))/6^n is always a positive odd integer. - Zhi-Wei Sun, Aug 21 2013
Conjecture: Let p == 1 (mod 3) be a prime, and write 4*p = x^2 + 27*y^2 with x, y integers and x == 1 (mod 3). Then W(p-1) == (-1)^{(p+1)/2}*(x-p/x) (mod p^2), where W(n) is defined as the above. - Zhi-Wei Sun, Aug 23 2013
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Diagonal of rational functions 1/(1 - (x^2*y + y^2*z - z^2*x + 3*x*y*z)), 1/(1 - (x^3 + y^3 - z^3 + 3*x*y*z)), 1/(1 + x^3 + y^3 + z^3 - 3*x*y*z). - Gheorghe Coserea, Aug 04 2018

Examples

			G.f. = 1 + 3*x + 9*x^2 + 21*x^3 + 9*x^4 - 297*x^5 - 2421*x^6 - 12933*x^7 - ...
		

References

  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • D. Zagier, Integral solutions of Apery-like recurrence equations, in: Groups and Symmetries: from Neolithic Scots to John McKay, CRM Proc. Lecture Notes 47, Amer. Math. Soc., Providence, RI, 2009, pp. 349-366.

Crossrefs

Related to diagonal of rational functions: A268545-A268555.
Cf. A091401.
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

  • Maple
    a := n -> 3^n*hypergeom([-n/3, (1-n)/3, (2-n)/3], [1, 1], 1):
    seq(simplify(a(n)), n=0..24); # Peter Luschny, Nov 01 2017
  • Mathematica
    Table[Sum[(-1)^k*3^(n - 3*k)*Binomial[n, 3*k]*Binomial[2*k, k]* Binomial[3*k, k], {k, 0, Floor[n/3]}], {n, 0, 50}] (* G. C. Greubel, Oct 24 2017 *)
    a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/3, 2/3}, {1}, x^3 / (x - 1/3)^3 ] / (1 - 3 x), {x, 0, n}]; (* Michael Somos, Nov 01 2017 *)
  • PARI
    subst(eta(q)^3/eta(q^3), q, serreverse(eta(q^9)^3/eta(q)^3*q)) \\ (generating function) Helena Verrill (verrill(AT)math.lsu.edu), Apr 20 2009 [for (-1)^n*a(n)]
    
  • PARI
    diag(expr, N=22, var=variables(expr)) = {
      my(a = vector(N));
      for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
      for (n = 1, N, a[n] = expr;
        for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
      return(a);
    };
    diag(1/(1 + x^3 + y^3 + z^3 - 3*x*y*z), 25)
    
  • PARI
    seq(N) = {
      my(a = vector(N)); a[1] = 3; a[2] = 9;
      for (n = 2, N-1, a[n+1] = ((9*n^2+9*n+3)*a[n] - 27*n^2*a[n-1])/(n+1)^2);
      concat(1,a);
    };
    seq(24)
    \\ test: y=subst(Ser(seq(202)), 'x, -'x/27); 0 == x*(x^2+9*x+27)*y'' + (3*x^2+18*x+27)*y' + (x+3)*y
    \\ Gheorghe Coserea, Nov 09 2017
    
  • PARI
    {a(n) = my(A); if( n<0, 0, A = x * O(x^n); (-1)^n * polcoeff( subst(eta(x + A)^3 / eta(x^3 + A), x, serreverse( x * eta(x^9 + A)^3 / eta(x + A)^3)), n))}; /* Michael Somos, Nov 01 2017 */

Formula

G.f.: hypergeom([1/3, 2/3], [1], x^3/(x-1/3)^3) / (1-3*x). - Mark van Hoeij, Oct 25 2011
a(n) = Sum_{k=0..floor(n/3)}(-1)^k*3^(n-3k)*C(n,3k)*C(2k,k)*C(3k,k). - Zhi-Wei Sun, Aug 21 2013
0 = x*(x^2+9*x+27)*y'' + (3*x^2 + 18*x + 27)*y' + (x + 3)*y, where y(x) = A(x/-27). - Gheorghe Coserea, Aug 26 2016
a(n) = 3^n*hypergeom([-n/3, (1-n)/3, (2-n)/3], [1, 1], 1). - Peter Luschny, Nov 01 2017
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=3*(-1 + 9*x)*T(x) + (-1 + 9*x)^2*T'(x) + x*(1 - 9*x + 27*x^2)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = 3*(-8 + 9*(1 - 9*x)^3)*(1 - 9*x);
g3 = 8 - 36*(1 - 9*x)^3 + 27*(1 - 9*x)^6;
which determine an elliptic surface with four singular fibers. (End)

Extensions

More terms from Kok Seng Chua (chuaks(AT)ihpc.nus.edu.sg), Jun 20 2000

A008977 a(n) = (4*n)!/(n!)^4.

Original entry on oeis.org

1, 24, 2520, 369600, 63063000, 11732745024, 2308743493056, 472518347558400, 99561092450391000, 21452752266265320000, 4705360871073570227520, 1047071828879079131681280, 235809301462142612780721600, 53644737765488792839237440000, 12309355935372581458927646400000
Offset: 0

Views

Author

Keywords

Comments

Number of paths of length 4*n in an n X n X n X n grid from (0,0,0,0) to (n,n,n,n).
a(n) occurs in Ramanujan's formula 1/Pi = (sqrt(8)/9801) * Sum_{n>=0} (4*n)!/(n!)^4 * (1103 + 26390*n)/396^(4*n). - Susanne Wienand, Jan 05 2013
a(n) is the number of ballot results that lead to a 4-way tie when 4*n voters each cast three votes for three out of four candidates vying for 3 slots on a county commission; each of these ballot results give 3*n votes to each of the four candidates. - Dennis P. Walsh, May 02 2013
a(n) is the constant term of (X + Y + Z + 1/(X*Y*Z))^(4*n). - Mark van Hoeij, May 07 2013
In Narumiya and Shiga on page 158 the g.f. is given as a hypergeometric function. - Michael Somos, Aug 12 2014
Diagonal of the rational function R(x,y,z,w) = 1/(1-(w+x+y+z)). - Gheorghe Coserea, Jul 15 2016

Examples

			a(13)=52!/(13!)^4=53644737765488792839237440000 is the number of ways of dealing the four hands in Bridge or Whist. - _Henry Bottomley_, Oct 06 2000
a(1)=24 since, in a 4-voter 3-vote election that ends in a four-way tie for candidates A, B, C, and D, there are 4! ways to arrange the needed vote sets {A,B,C}, {A,B,D}, {A,C,D}, and {B,C,D} among the 4 voters. - _Dennis P. Walsh_, May 02 2013
G.f. = 1 + 24*x + 2520*x^2 + 369600*x^3 + 63063000*x^4 + 11732745024*x^5 + ...
		

Crossrefs

Row 4 of A187783.
Related to diagonal of rational functions: A268545-A268555.

Programs

  • Magma
    [Factorial(4*n)/Factorial(n)^4: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014
    
  • Maple
    A008977 := n->(4*n)!/(n!)^4;
  • Mathematica
    Table[(4n)!/(n!)^4,{n,0,16}] (* Harvey P. Dale, Oct 24 2011 *)
    a[ n_] := If[ n < 0, 0, (4 n)! / n!^4]; (* Michael Somos, Aug 12 2014 *)
    a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/4, 2/4, 3/4}, {1, 1}, 256 x], {x, 0, n}]; (* Michael Somos, Aug 12 2014 *)
  • Maxima
    A008977(n):=(4*n)!/(n!)^4$ makelist(A008977(n),n,0,20); /* Martin Ettl, Nov 15 2012 */
    
  • PARI
    a(n) = (4*n)!/n!^4; \\ Gheorghe Coserea, Jul 15 2016
    
  • Python
    from math import factorial
    def A008977(n): return factorial(n<<2)//factorial(n)**4 # Chai Wah Wu, Mar 15 2023

Formula

a(n) = A139541(n)*(A001316(n)/A049606(n))^3. - Reinhard Zumkeller, Apr 28 2008
Self-convolution of A178529, where A178529(n) = (4^n/n!^2) * Product_{k=0..n-1} (8*k + 1)*(8*k + 3).
G.f.: hypergeom([1/8, 3/8], [1], 256*x)^2. - Mark van Hoeij, Nov 16 2011
a(n) ~ 2^(8*n - 1/2) / (Pi*n)^(3/2). - Vaclav Kotesovec, Mar 07 2014
G.f.: hypergeom([1/4, 2/4, 3/4], [1, 1], 256*x). - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(24*n)), where F(x) = 1 + x + 29*x^2 + 2246*x^3 + 239500*x^4 + 30318701*x^5 + 4271201506*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008978, A008979, A186420 and A188662. (End)
0 = (x^2-256*x^3)*y''' + (3*x-1152*x^2)*y'' + (1-816*x)*y' - 24*y, where y is the g.f. - Gheorghe Coserea, Jul 15 2016
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..3*n} (-1)^(n+k)*binomial(4*n,n + k)* binomial(n + k,k)^4.
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(4*n,k)*binomial(n + k,k)^4. (End)
E.g.f.: 3F3(1/4,1/2,3/4; 1,1,1; 256*x). - Ilya Gutkovskiy, Jan 23 2018
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.
a(n) = [(x*y*z)^n] (1 + x + y + z)^(4*n). (End)
D-finite with recurrence n^3*a(n) -8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Aug 01 2022
a(n) = 24*A082368(n). - R. J. Mathar, Jun 21 2023

A004987 a(n) = (3^n/n!)*Product_{k=0..n-1} (3*k + 1). 3-central binomial coefficients.

Original entry on oeis.org

1, 3, 18, 126, 945, 7371, 58968, 480168, 3961386, 33011550, 277297020, 2344420260, 19927572210, 170150808870, 1458435504600, 12542545339560, 108179453553705, 935434098376155, 8107095519260010, 70403724246205350, 612512400941986545, 5337608065351597035, 46582761297613937760
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

Diagonal of rational function R(x,y) = (1 - 9*x*y) / (1 - 2*x - 3*y + 3*y^2 + 9*x^2*y). - Gheorghe Coserea, Jul 01 2016
This is the k = 3 variant of the k-central binomial coefficients c(n,k) with g.f. (1 - k^2*x)^(-1/k), which yield the usual central binomial coefficients A001405 for k = 2. - M. F. Hasler, Nov 12 2024

Examples

			G.f.: 1 + 3*x + 18*x^2 + 126*x^3 + 945*x^4 + 7371*x^5 + 58968*x^6 + 480168*x^7 + ...
		

Crossrefs

Related to diagonal of rational functions: A268545-A268555.

Programs

  • GAP
    List([0..25], n-> 3^n*Product([0..n-1], k-> 3*k+1)/Factorial(n) ); # G. C. Greubel, Aug 22 2019
  • Magma
    [1] cat [3^n*&*[3*k+1: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019
    
  • Maple
    a:= n-> (3^n/n!)*mul(3*k+1, k=0..n-1); seq(a(n), n=0..25); # G. C. Greubel, Aug 22 2019
  • Mathematica
    Table[(-9)^n Binomial[-1/3, n], {n, 0, 25}] (* Jean-François Alcover, Sep 28 2016, after Peter Luschny *)
  • PARI
    a(n) = prod(k=0, n-1, 3*k + 1)*3^n/n! \\ Michel Marcus, Jun 30 2013
    
  • PARI
    my(x='x, y='y);
    R = (1 - 9*x*y) / (1 - 2*x - 3*y + 3*y^2 + 9*x^2*y);
    diag(n, expr, var) = {
      my(a = vector(n));
      for (i = 1, #var, expr = taylor(expr, var[#var - i + 1], n));
      for (k = 1, n, a[k] = expr;
           for (i = 1, #var, a[k] = polcoeff(a[k], k-1)));
      return(a);
    };
    diag(20, R, [x,y])  \\ Gheorghe Coserea, Jul 01 2016
    
  • PARI
    Vec((1-9*x+O(x^25))^(-1/3)) \\ yields the same as:
    apply( {A004987(n)=prod(k=0, n-1, 9*k+3)\n!}, [0..24]) \\ M. F. Hasler, Nov 12 2024
    
  • Sage
    [9^n*rising_factorial(1/3, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019
    

Formula

G.f.: (1 - 9*x)^(-1/3).
a(n) = (3^n/n!)*A007559(n), n >= 1, a(0) := 1.
a(n) ~ Gamma(1/3)^-1*n^(-2/3)*3^(2*n)*{1 - 1/9*n^-1 + ...}.
Representation as n-th moment of a positive function on (0, 9): a(n) = Integral_{x=0..9} ( x^n*(1/(Pi*sqrt(3)*6*(x/9)^(2/3)*(1-x/9)^(1/3))) ), n >= 0. This function is the solution of the Hausdorff moment problem on (0, 9) with moments equal to a(n). As a consequence this representation is unique. - Karol A. Penson, Jan 30 2003
D-finite with recurrence: n*a(n) + 3*(2-3*n)*a(n-1)=0. - R. J. Mathar, Jun 07 2013
0 = a(n) * (81*a(n+1) - 15*a(n+2)) + a(n+1) * (-3*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Jan 27 2014
G.f. A(x)=:y satisfies 0 = y'' * y - 4 * y' * y'. - Michael Somos, Jan 27 2014
a(n) = (-9)^n*binomial(-1/3, n). - Peter Luschny, Mar 23 2014
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([1/3], [1], 9*x). - Karol A. Penson, Dec 19 2015
Sum_{n>=0} 1/a(n) = (sqrt(3)*Pi + 3*(12 + log(3)))/32 = 1.3980385924595932... - Ilya Gutkovskiy, Jul 01 2016
Binomial transform of A216316. - Peter Bala, Jul 02 2023
From Peter Bala, Mar 31 2024: (Start)
a(n) = (9^n)*Sum_{k = 0..2*n} (-1)^k*binomial(-1/3, k)* binomial(-1/3, 2*n - k).
(9^n)*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
Sum_{k = 0..n} a(k)*a(n-k) = A004988(n).
Sum_{k = 0..2*n} a(k)*a(2*n-k) = 18^n/(2*n)! * Product_{k = 1..n} (6*k - 1)*(3*k - 2). (End)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^3). - Seiichi Manyama, Jun 20 2025

Extensions

More terms from Ralf Stephan, Mar 13 2004
More terms from Benoit Cloitre, Jun 05 2004

A183204 Central terms of triangle A181544.

Original entry on oeis.org

1, 4, 48, 760, 13840, 273504, 5703096, 123519792, 2751843600, 62659854400, 1451780950048, 34116354472512, 811208174862904, 19481055861877120, 471822589361293680, 11511531876280913760, 282665135367572129040
Offset: 0

Views

Author

Paul D. Hanna, Dec 30 2010

Keywords

Comments

The g.f. for row n of triangle A181544 is (1-x)^(3n+1)*Sum_{k>=0}C(n+k-1,k)^3*x^k.
This sequence is s_7 in Cooper's paper. - Jason Kimberley, Nov 06 2012
Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*x*z + w*y*z + x*y + x*z + y + z)). - Gheorghe Coserea, Jul 14 2016
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

Examples

			Triangle A181544 begins:
(1);
1, (4), 1;
1, 20, (48), 20, 1;
1, 54, 405, (760), 405, 54, 1;
1, 112, 1828, 8464, (13840), 8464, 1828, 112, 1; ...
		

Crossrefs

Related to diagonal of rational functions: A268545-A268555.
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
    P:=PolynomialRing(Integers()); C:=Binomial;
    A183204:=func; // or directly:
    A183204:=func;
    [A183204(n):n in[0..16]]; // Jason Kimberley, Oct 29 2012
  • Mathematica
    Table[Sum[Binomial[n,j]^2 * Binomial[2*j,n] * Binomial[j+n,j],{j,0,n}],{n,0,20}] (* Vaclav Kotesovec, Apr 05 2015 *)
  • PARI
    {a(n)=polcoeff((1-x)^(3*n+1)*sum(j=0, 2*n, binomial(n+j, j)^3*x^j), n)}
    

Formula

a(n) = [x^n] (1-x)^(3n+1) * Sum_{k>=0} C(n+k-1,k)^3*x^k.
a(n) = Sum_{j = 0..n} C(n,j)^2 * C(2*j,n) * C(j+n,j). [Formula of Wadim Zudilin provided by Jason Kimberley, Nov 06 2012]
1/Pi = sqrt(7) Sum_{n>=0} (-1)^n a(n) (11895n + 1286)/22^(3n+3). [Cooper, equation (41)] - Jason Kimberley, Nov 06 2012
G.f.: sqrt((1-13*x+(1-26*x-27*x^2)^(1/2))/(1-21*x+8*x^2+(1-8*x)*(1-26*x-27*x^2)^(1/2)))*hypergeom([1/12,5/12],[1],13824*x^7/(1-21*x+8*x^2+(1-8*x)*(1-26*x-27*x^2)^(1/2))^3)^2. - Mark van Hoeij, May 07 2013
a(n) ~ 3^(3*n+3/2) / (4 * (Pi*n)^(3/2)). - Vaclav Kotesovec, Apr 05 2015
G.f. A(x) satisfies 1/(1+4*x)^2 * A( x/(1+4*x)^3 ) = 1/(1+2*x)^2 * A( x^2/(1+2*x)^3 ) [see Cooper, Guillera, Straub, Zudilin]. - Joerg Arndt, Apr 08 2016
a(n) = (-1)^n*binomial(3n+1,n)* 4F3({-n,n+1,n+1,n+1};{1,1,2(n+1)}; 1). - M. Lawrence Glasser, May 15 2016
Conjecture D-finite with recurrence: n^3*a(n) - (2*n-1)*(13*n^2-13*n+4)*a(n-1) - 3*(n-1)*(3*n-4)*(3*n-2)*a(n-2) = 0. - R. J. Mathar, May 15 2016
0 = (-x^2+26*x^3+27*x^4)*y''' + (-3*x+117*x^2+162*x^3)*y'' + (-1+86*x+186*x^2)*y' + (4+24*x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
From Jeremy Tan, Mar 14 2024: (Start)
The conjectured D-finite recurrence can be proved by Zeilberger's algorithm.
a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n+k,n) * binomial(2*n-k,n) = [(w*x*y*z)^n] ((w+y)*(x+z)*(y+z)*(w+x+y+z))^n. (End)
a(n) = Sum_{0 <= j, k <= n} binomial(n, k)^2 * binomial(n, j)^2 * binomial(k+j, n) = Sum_{k = 0..n} binomial(n, k)^2 * A108625(n, k). - Peter Bala, Jul 08 2024
From Peter Bala, Sep 18 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n+k, k)^3*binomial(3*n+1, n-k). Cf A245086.
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*A143007(n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)

A002897 a(n) = binomial(2n,n)^3.

Original entry on oeis.org

1, 8, 216, 8000, 343000, 16003008, 788889024, 40424237568, 2131746903000, 114933031928000, 6306605327953216, 351047164190381568, 19774031697705428416, 1125058699232216000000, 64561313052442296000000
Offset: 0

Views

Author

Keywords

Comments

Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*z + x + y + z)). - Gheorghe Coserea, Jul 14 2016
Conjecture: The g.f. is also the diagonal of the rational function 1/(1 - (x + y)*(1 - 4*z*t) - z - t) = 1/det(I - M*diag(x, y, z, t)), I the 4 x 4 unit matrix and M the 4 x 4 matrix [1, 1, 1, 1; 1, 1, 1, 1; 1, 1, 1, -1; 1 , 1, -1, 1]. If true, then a(n) = [(x*y*z)^n] (1 + x + y + z)^(2*n)*(1 + x + y - z)^n*(1 + x - y + z)^n. - Peter Bala, Apr 10 2022

References

  • S. Ramanujan, Modular Equations and Approximations to pi, pp. 23-39 of Collected Papers of Srinivasa Ramanujan, Ed. G. H. Hardy et al., AMS Chelsea 2000. See page 36, equation (25).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Related to diagonal of rational functions: A268545-A268555.

Programs

  • Magma
    [Binomial(2*n, n)^3: n in [0..20]]; // Vincenzo Librandi, Nov 18 2011
  • Mathematica
    a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/2, 1/2, 1/2}, {1, 1}, 64x], {x, 0, n}];
    Table[Binomial[2n,n]^3,{n,0,20}] (* Harvey P. Dale, Dec 06 2017 *)
  • PARI
    {a(n) = binomial(2*n, n)^3}; /* Michael Somos, Jan 31 2007 */
    
  • Sage
    [binomial(2*n, n)**3 for n in range(21)] # Zerinvary Lajos, Apr 21 2009
    

Formula

Expansion of (K(k)/(Pi/2))^2 in powers of (kk'/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at modulus k. - Michael Somos, Jan 31 2007
G.f.: F(1/2, 1/2, 1/2; 1, 1; 64x) where F() is a hypergeometric function. - Michael Somos, Jan 31 2007
G.f.: hypergeom([1/4,1/4],[1],64*x)^2. - Mark van Hoeij, Nov 17 2011
D-finite with recurrence n^3*a(n) - 8*(2*n - 1)^3*a(n-1) = 0. - R. J. Mathar, Mar 08 2013
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)^3 = ( [x^n](1 + x)^(2*n) )^3 = [x^n](F(x)^(8*n)), where F(x) = 1 + x + 6*x^2 + 111*x^3 + 2806*x^4 + 84456*x^5 + 2832589*x^6 + 102290342*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002894, A006480, A008977, A186420 and A188662. (End)
a(n) ~ 64^n/(Pi*n)^(3/2). - Ilya Gutkovskiy, Jul 13 2016
0 = (-x^2 + 64*x^3)*y''' + (-3*x + 288*x^2)*y'' + (-1 + 208*x)*y' + 8*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
a(n) = Sum_{k = 0..n} (2*n + k)!/(k!^3*(n - k)!^2). Cf. A001850(n) = Sum_{k = 0..n} (n + k)!/(k!^2*(n - k)!). - Peter Bala, Jul 27 2016
It appears that a(n) is the coefficient of (x*y*z)^(2*n) in the expansion of (1 + x*y + x*z - y*z)^(2*n) * (1 + x*y - x*z + y*z)^(2*n) * (1 - x*y + x*z + y*z)^(2*n). Cf. A000172. - Peter Bala, Sep 21 2021
From Peter Bala, Sep 24 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)*binomial(2*n+k,n).
a(n) = the coefficient of (x*y*z*t^2)^n in the expansion of 1/(1 - x - y)*(1 - z - t) - x*y*z*t) (a(n) = A(n,n,n,2*n) in the notation of Straub, Theorem 1.2). (End)
a(n) = (8/5) * Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)*binomial(2*n+k-1,n) for n >= 1. - Peter Bala, Jul 09 2024
a(n) = Sum_{k = 0..n} binomial(n, k)^2 * A108625(2*n, k). Cf. A183204. - Peter Bala, Oct 12 2024
From Peter Bala, Oct 16 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k)*binomial(2*n+k, k)*A108625(n, k) = 8 * Sum_{k = 0..n} (-1)^(n+k+1) * binomial(n-1, k)*binomial(2*n+k-1, k)*A108625(n, k) = (8/5) * Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k)*binomial(2*n+k-1, k)*A108625(n, k) for n >= 1. Cf. A176285. (End)

A002896 Number of 2n-step polygons on cubic lattice.

Original entry on oeis.org

1, 6, 90, 1860, 44730, 1172556, 32496156, 936369720, 27770358330, 842090474940, 25989269017140, 813689707488840, 25780447171287900, 825043888527957000, 26630804377937061000, 865978374333905289360, 28342398385058078078010, 932905175625150142902300
Offset: 0

Views

Author

Keywords

Comments

Number of walks with 2n steps on the cubic lattice Z^3 beginning and ending at (0,0,0).
If A is a random matrix in USp(6) (6 X 6 complex matrices that are unitary and symplectic) then a(n) is the 2n-th moment of tr(A^k) for all k >= 7. - Andrew V. Sutherland, Mar 24 2008
Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*x*z + w*y + x*z + y + z)). - Gheorghe Coserea, Jul 14 2016
Constant term in the expansion of (x + 1/x + y + 1/y + z + 1/z)^(2n). - Harry Richman, Apr 29 2020

Examples

			1 + 6*x + 90*x^2 + 1860*x^3 + 44730*x^4 + 1172556*x^5 + 32496156*x^6 + ...
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

C(2n, n) times A002893.
Related to diagonal of rational functions: A268545-A268555.
Row k=3 of A287318.

Programs

  • Maple
    a := proc(n) local k; binomial(2*n,n)*add(binomial(n,k)^2 *binomial(2*k,k), k=0..n); end;
    # second Maple program
    a:= proc(n) option remember; `if`(n<2, 5*n+1,
          (2*(2*n-1)*(10*n^2-10*n+3) *a(n-1)
           -36*(n-1)*(2*n-1)*(2*n-3) *a(n-2)) /n^3)
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Nov 02 2012
    A002896 := n -> binomial(2*n,n)*hypergeom([1/2, -n, -n], [1, 1], 4):
    seq(simplify(A002896(n)), n=0..16); # Peter Luschny, May 23 2017
  • Mathematica
    Table[Binomial[2n,n] Sum[Binomial[n,k]^2 Binomial[2k,k],{k,0,n}],{n,0,20}] (* Harvey P. Dale, Jan 24 2012 *)
    a[ n_] := If[ n < 0, 0, HypergeometricPFQ[ {-n, -n, 1/2}, {1, 1}, 4] Binomial[ 2 n, n]] (* Michael Somos, May 21 2013 *)
  • PARI
    a(n)=binomial(2*n,n)*sum(k=0,n,binomial(n, k)^2*binomial(2*k, k)) \\ Charles R Greathouse IV, Oct 31 2011
    
  • Sage
    def A002896():
        x, y, n = 1, 6, 1
        while True:
            yield x
            n += 1
            x, y = y, ((4*n-2)*((10*(n-1)*n+3)*y-18*(n-1)*(2*n-3)*x))//n^3
    a = A002896()
    [next(a) for i in range(17)]  # Peter Luschny, Oct 09 2013

Formula

a(n) = C(2*n, n)*Sum_{k=0..n} C(n, k)^2*C(2*k, k).
a(n) = (4^n*p(1/2, n)/n!)*hypergeom([-n, -n, 1/2], [1, 1], 4), where p(a, k) = Product_{i=0..k-1} (a+i).
E.g.f.: Sum_{n>=0} a(n)*x^(2*n)/(2*n)! = BesselI(0, 2*x)^3. - Corrected by Christopher J. Smyth, Oct 29 2012
D-finite with recurrence: n^3*a(n) = 2*(2*n-1)*(10*n^2-10*n+3)*a(n-1) - 36*(n-1)*(2*n-1)*(2*n-3)*a(n-2). - Vladeta Jovovic, Jul 16 2004
An asymptotic formula follows immediately from an observation of Bruce Richmond and me in SIAM Review - 31 (1989, 122-125). We use Hayman's method to find the asymptotic behavior of the sum of squares of the multinomial coefficients multi(n, k_1, k_2, ..., k_m) with m fixed. From this one gets a_n ~ (3/4)*sqrt(3)*6^(2*n)/(Pi*n)^(3/2). - Cecil C Rousseau (ccrousse(AT)memphis.edu), Mar 14 2006
G.f.: (1/sqrt(1+12*z)) * hypergeom([1/8,3/8],[1],64/81*z*(1+sqrt(1-36*z))^2*(2+sqrt(1-36*z))^4/(1+12*z)^4) * hypergeom([1/8, 3/8],[1],64/81*z*(1-sqrt(1-36*z))^2*(2-sqrt(1-36*z))^4/(1+12*z)^4). - Sergey Perepechko, Jan 26 2011
a(n) = binomial(2*n,n)*A002893(n). - Mark van Hoeij, Oct 29 2011
G.f.: (1/2)*(10-72*x-6*(144*x^2-40*x+1)^(1/2))^(1/2)*hypergeom([1/6, 1/3],[1],54*x*(108*x^2-27*x+1+(9*x-1)*(144*x^2-40*x+1)^(1/2)))^2. - Mark van Hoeij, Nov 12 2011
PSUM transform is A174516. - Michael Somos, May 21 2013
0 = (-x^2+40*x^3-144*x^4)*y''' + (-3*x+180*x^2-864*x^3)*y'' + (-1+132*x-972*x^2)*y' + (6-108*x)*y, where y is the g.f. - Gheorghe Coserea, Jul 14 2016
a(n) = [(x y z)^0] (x + 1/x + y + 1/y + z + 1/z)^(2*n). - Christopher J. Smyth, Sep 25 2018
a(n) = (1/Pi)^3*Integral_{0 <= x, y, z <= Pi} (2*cos(x) + 2*cos(y) + 2*cos(z))^(2*n) dx dy dz. - Peter Bala, Feb 10 2022
a(n) = Sum_{i+j+k=n, 0<=i,j,k<=n} multinomial(2n [i,i,j,j,k,k]). - Shel Kaphan, Jan 16 2023
Sum_{k>=0} a(k)/36^k = A086231 = (sqrt(3)-1) * (Gamma(1/24) * Gamma(11/24))^2 / (32*Pi^3). - Vaclav Kotesovec, Apr 23 2023
G.f.: HeunG(1/9,1/12,1/4,3/4,1,1/2,4*x)^2 (see Hassani et al.). - Stefano Spezia, Feb 16 2025
Showing 1-10 of 73 results. Next