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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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  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.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

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

A008459 Square the entries of Pascal's triangle.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 9, 9, 1, 1, 16, 36, 16, 1, 1, 25, 100, 100, 25, 1, 1, 36, 225, 400, 225, 36, 1, 1, 49, 441, 1225, 1225, 441, 49, 1, 1, 64, 784, 3136, 4900, 3136, 784, 64, 1, 1, 81, 1296, 7056, 15876, 15876, 7056, 1296, 81, 1, 1, 100, 2025, 14400, 44100, 63504, 44100, 14400, 2025, 100, 1
Offset: 0

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Comments

Number of lattice paths from (0, 0) to (n, n) with steps (1, 0) and (0, 1), having k right turns. - Emeric Deutsch, Nov 23 2003
Product of A007318 and A105868. - Paul Barry, Nov 15 2005
Number of partitions that fit in an n X n box with Durfee square k. - Franklin T. Adams-Watters, Feb 20 2006
From Peter Bala, Oct 23 2008: (Start)
Narayana numbers of type B. Row n of this triangle is the h-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p. 60]. See A063007 for the corresponding f-vectors for associahedra of type B_n. See A001263 for the h-vectors for associahedra of type A_n. The Hilbert transform of this triangular array is A108625 (see A145905 for the definition of this term).
Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i, j <= n + 1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A063007 is the corresponding array of f-vectors for these type A_n polytopes. See A086645 for the array of h-vectors for type C_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
(End)
The n-th row consists of the coefficients of the polynomial P_n(t) = Integral_{s = 0..2*Pi} (1 + t^2 - 2*t*cos(s))^n/Pi/2 ds. For example, when n = 3, we get P_3(t) = t^6 + 9*t^4 + 9*t^2 + 1; the coefficients are 1, 9, 9, 1. - Theodore Kolokolnikov, Oct 26 2010
Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!^2 as defined in Wang and Wang. - Peter Bala, Jul 24 2013
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. T(n,k) is the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 231, and 321. T(n,k) is also the number of permutations pi of [n+1] with k descents such that s(pi) avoids the patterns 132, 312, and 321.
T(n,k) is the number of permutations of [n+1] with k descents that avoid the patterns 1342, 3142, 3412, and 3421. (End)
The number of convex polyominoes whose smallest bounding rectangle has size (k+1)*(n+1-k) and which contain the lower left corner of the bounding rectangle (directed convex polyominoes). - Günter Rote, Feb 27 2019
Let P be the poset [n] X [n] ordered by the product order. T(n,k) is the number of antichains in P containing exactly k elements. Cf. A063746. - Geoffrey Critzer, Mar 28 2020

Examples

			Pascal's triangle begins
  1
  1  1
  1  2   1
  1  3   3   1
  1  4   6   4   1
  1  5  10  10   5   1
  1  6  15  20  15   6   1
  1  7  21  35  35  21   7   1
...
so the present triangle begins
  1
  1   1
  1   4    1
  1   9    9     1
  1  16   36    16     1
  1  25  100   100    25    1
  1  36  225   400   225   36   1
  1  49  441  1225  1225  441  49   1
...
		

References

  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Chapter 12.
  • J. Riordan, An introduction to combinatorial analysis, Dover Publications, Mineola, NY, 2002, page 191, Problem 15. MR1949650
  • P. G. Tait, On the Linear Differential Equation of the Second Order, Proceedings of the Royal Society of Edinburgh, 9 (1876), 93-98 (see p. 97) [From Tom Copeland, Sep 09 2010, vol number corrected Sep 10 2010]

Crossrefs

Row sums are in A000984. Columns 0-3 are A000012, A000290, A000537, A001249.
Family of polynomials (see A062145): this sequence (c=1), A132813 (c=2), A062196 (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Cf. A007318, A055133, A116647, A001263, A086645, A063007, A108558, A108625 (Hilbert transform), A145903, A181543, A086645 (logarithmic derivative), A105868 (inverse binomial transform), A093118.

Programs

  • GAP
    Flat(List([0..10],n->List([0..n],k->Binomial(n,k)^2))); # Muniru A Asiru, Mar 30 2018
    
  • Magma
    /* As triangle */ [[Binomial(n, k)^2: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 15 2016
    
  • Maple
    seq(seq(binomial(n, k)^2, k=0..n), n=0..10);
  • Mathematica
    Table[Binomial[n, k]^2, {n, 0, 11}, {k, 0, n}]//Flatten (* Alonso del Arte, Dec 08 2013 *)
  • Maxima
    create_list(binomial(n,k)^2,n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • Maxima
    T(n,k):=if n=k then 1 else if k=0 then 1 else T(n-1,k)*(n+k)/(n-k)+T(n-1,k-1); /* Vladimir Kruchinin, Oct 18 2014 */
    
  • Maxima
    A(x,y):=1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2);
    taylor(x*A(x,y)+x*y*A(x,y)+sqrt(1+4*x^2*y*A(x,y)^2),x,0,7,y,0,7); /* Vladimir Kruchinin, Oct 23 2020 */
    
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, binomial(n, k)^2)}; /* Michael Somos, May 03 2004 */
    
  • PARI
    {T(n,k)=polcoeff(polcoeff(sum(m=0,n,(2*m)!/m!^2*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(2*m+1)),n,x),k,y)} \\ Paul D. Hanna, Oct 31 2010
    
  • Python
    def A008459(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2))**2 # Chai Wah Wu, Nov 12 2024

Formula

T(n,k) = A007318(n,k)^2. - Sean A. Irvine, Mar 29 2018
E.g.f.: exp((1+y)*x)*BesselI(0, 2*sqrt(y)*x). - Vladeta Jovovic, Nov 17 2003
G.f.: 1/sqrt(1-2*x-2*x*y+x^2-2*x^2*y+x^2*y^2); g.f. for row n: (1-t)^n P_n[(1+t)/(1-t)] where the P_n's are the Legendre polynomials. - Emeric Deutsch, Nov 23 2003 [The original version of the bivariate g.f. has been modified with the roles of x and y interchanged so that now x corresponds to n and y to k. - Petros Hadjicostas, Oct 22 2017]
G.f. for column k is Sum_{j = 0..k} C(k, j)^2*x^(k+j)/(1 - x)^(2*k+1). - Paul Barry, Nov 15 2005
Column k has g.f. (x^k)*Legendre_P(k, (1+x)/(1-x))/(1 - x)^(k+1) = (x^k)*Sum_{j = 0..k} C(k, j)^2*x^j/(1 - x)^(2*k+1). - Paul Barry, Nov 19 2005
Let E be the operator D*x*D, where D denotes the derivative operator d/dx. Then (1/n!^2) * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(2*n+1) = Sum_{k >= 0} binomial(n+k, k)^2*x^k. For example, when n = 3 we have (1/3!)^2*E^3(1/(1 - x)) = (1 + 9*x + 9*x^2 + x^3)/(1 - x)^7 = (1/3!)^2 * Sum_{k >= 0} ((k+1)*(k+2)*(k+3))^2*x^k. - Peter Bala, Oct 23 2008
G.f.: A(x, y) = Sum_{n >= 0} (2*n)!/n!^2 * x^(2*n)*y^n/(1 - x - x*y)^(2*n+1). - Paul D. Hanna, Oct 31 2010
From Peter Bala, Jul 24 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0, 2*sqrt(-y)). Generating function: E(y)*E(x*y) = 1 + (1 + x)*y + (1 + 4*x + x^2)*y^2/2!^2 + (1 + 9*x + 9*x^2 + x^3)*y^3/3!^2 + .... Cf. the unsigned version of A021009 with generating function exp(y)*E(x*y).
The n-th power of this array has the generating function E(y)^n*E(x*y). In particular, the matrix inverse A055133 has the generating function E(x*y)/E(y). (End)
T(n,k) = T(n-1,k)*(n+k)/(n-k) + T(n-1,k-1), T(n,0) = T(n,n) = 1. - Vladimir Kruchinin, Oct 18 2014
Observe that the recurrence T(n,k) = T(n-1,k)*(n+k)/(n-k) - T(n-1,k-1), for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1 gives Pascal's triangle A007318. - Peter Bala, Dec 21 2014
n-th row polynomial R(n, x) = [z^n] (1 + (1 + x)*z + x*z^2)^n. Note that 1/n*[z^(n-1)] (1 + (1 + x)*z + x*z^2)^n gives the row polynomials of A001263. - Peter Bala, Jun 24 2015
Binomial transform of A105868. If G(x,t) = 1/sqrt(1 - 2*(1 + t)*x + (1 - t)^2*x^2) denotes the o.g.f. of this array then 1 + x*d/dx log(G(x,t)) = 1 + (1 + t)*x + (1 + 6*t + t^2)*x^2 + ... is the o.g.f. for A086645. - Peter Bala, Sep 06 2015
T(n,k) = Sum_{i=0..n} C(n-i,k)*C(n,i)*C(n+i,i)*(-1)^(n-i-k). - Vladimir Kruchinin, Jan 14 2018
G.f. satisfies A(x,y) = x*A(x,y)+x*y*A(x,y)+sqrt(1+4*x^2*y*A(x,y)^2). - Vladimir Kruchinin, Oct 23 2020
G.f. satisfies the differential equation y * d^2(A(x,y))/dy^2 - x^2 * d^2(x*A(x,y))/dx^2 + 2*x^2* A(x,y)^3 = 0. - Sergii Voloshyn, Mar 07 2025
T(n,k) = Sum_{i=0..n} C(2*n+1,i)*C(n+k-i,n)^2*(-1)^i. - Natalia L. Skirrow, Apr 14 2025

A063007 T(n,k) = binomial(n,k)*binomial(n+k,k), 0 <= k <= n, triangle read by rows.

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 12, 30, 20, 1, 20, 90, 140, 70, 1, 30, 210, 560, 630, 252, 1, 42, 420, 1680, 3150, 2772, 924, 1, 56, 756, 4200, 11550, 16632, 12012, 3432, 1, 72, 1260, 9240, 34650, 72072, 84084, 51480, 12870, 1, 90, 1980, 18480, 90090, 252252, 420420, 411840, 218790, 48620
Offset: 0

Views

Author

Henry Bottomley, Jul 02 2001

Keywords

Comments

T(n,k) is the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's Cluster algebra of finite type B_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of triangle A008459 (squares of binomial coefficients). For example, x^2+6*x+6 = y^2+4*y+1. - Paul Boddington, Mar 07 2003
T(n,k) is the number of lattice paths from (0,0) to (n,n) using steps E=(1,0), N=(0,1) and D=(1,1) (i.e., bilateral Schroeder paths), having k N=(0,1) steps. E.g. T(2,0)=1 because we have DD; T(2,1) = 6 because we have NED, NDE, EDN, END, DEN and DNE; T(2,2)=6 because we have NNEE, NENE, NEEN, EENN, ENEN and ENNE. - Emeric Deutsch, Apr 20 2004
Another version of [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] = 1; 1, 0; 1, 2, 0; 1, 6, 6, 0; 1, 12, 30, 20, 0; ..., where DELTA is the operator defined in A084938. - Philippe Deléham Apr 15 2005
Terms in row n are the coefficients of the Legendre polynomial P(n,2x+1) with increasing powers of x.
From Peter Bala, Oct 28 2008: (Start)
Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p.60]. See A008459 for the corresponding h-vectors for associahedra of type B_n and A001263 and A033282 respectively for the h-vectors and f-vectors for associahedra of type A_n.
An alternative description of this triangle in terms of f-vectors is as follows. Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i,j <= n+1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the f-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A008459 is the corresponding array of h-vectors for these type A_n polytopes. See A127674 (without the signs) for the array of f-vectors for type C_n polytopes and A108556 for the array of f-vectors associated with type D_n polytopes.
The S-transform on the ring of polynomials is the linear transformation of polynomials that is defined on the basis monomials x^k by S(x^k) = binomial(x,k) = x(x-1)...(x-k+1)/k!. Let P_n(x) denote the S-transform of the n-th row polynomial of this array. In the notation of [Hetyei] these are the Stirling polynomials of the type B associahedra. The first few values are P_1(x) = 2*x + 1, P_2(x) = 3*x^2 + 3*x + 1 and P_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials P_n(-x) satisfy a Riemann hypothesis. See A142995 for further details. The sequence of values P_n(k) for k = 0,1,2,3, ... produces the n-th row of A108625. (End)
This is the row reversed version of triangle A104684. - Wolfdieter Lang, Sep 12 2016
T(n, k) is also the number of (n-k)-dimensional faces of a convex n-dimensional Lipschitz polytope of real functions f defined on the set X = {1, 2, ..., n+1} which satisfy the condition f(n+1) = 0 (see Gordon and Petrov). - Stefano Spezia, Sep 25 2021
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial ((x+1)*(x+2)*(x+3)*...*(x+n) / n!)^2 in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 09 2022
Chapoton's observation above is correct: the precise expansion is ((x+1)*(x+2)*(x+3)*...*(x+n)/ n!)^2 = Sum_{k = 0..n} (-1)^k*T(n,n-k)*binomial(x+2*n-k, 2*n-k), as can be verified using the WZ algorithm. For example, n = 3 gives ((x+1)*(x+2)*(x+3)/3!)^2 = 20*binomial(x+6,6) - 30*binomial(x+5,5) + 12*binomial(x+4,4) - binomial(x+3,3). - Peter Bala, Jun 24 2023

Examples

			The triangle T(n, k) starts:
  n\k 0  1    2     3     4      5      6      7      8     9
  0:  1
  1:  1  2
  2:  1  6    6
  3:  1 12   30    20
  4:  1 20   90   140    70
  5:  1 30  210   560   630    252
  6:  1 42  420  1680  3150   2772    924
  7:  1 56  756  4200 11550  16632  12012   3432
  8:  1 72 1260  9240 34650  72072  84084  51480  12870
  9:  1 90 1980 18480 90090 252252 420420 411840 218790 48620
... reformatted by _Wolfdieter Lang_, Sep 12 2016
From _Petros Hadjicostas_, Jul 11 2020: (Start)
Its inverse (from Table II, p. 92, in Ser's book) is
   1;
  -1/2,  1/2;
   1/3, -1/2,    1/6;
  -1/4,  9/20,  -1/4,   1/20;
   1/5, -2/5,    2/7,  -1/10,  1/70;
  -1/6,  5/14, -25/84,  5/36, -1/28,  1/252;
   1/7, -9/28,  25/84, -1/6,   9/154, -1/84, 1/924;
   ... (End)
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 366.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, Table I, p. 92.
  • 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

See A331430 for an essentially identical triangle, except with signed entries.
Columns include A000012, A002378, A033487 on the left and A000984, A002457, A002544 on the right.
Main diagonal is A006480.
Row sums are A001850. Alternating row sums are A033999.
Cf. A033282 (f-vectors type A associahedra), A108625, A080721 (f-vectors type D associahedra).
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

  • Haskell
    a063007 n k = a063007_tabl !! n !! k
    a063007_row n = a063007_tabl !! n
    a063007_tabl = zipWith (zipWith (*)) a007318_tabl a046899_tabl
    -- Reinhard Zumkeller, Nov 18 2014
    
  • Magma
    /* As triangle: */ [[Binomial(n,k)*Binomial(n+k,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 03 2015
  • Maple
    p := (n,x) -> orthopoly[P](n,1+2*x): seq(seq(coeff(p(n,x),x,k), k=0..n), n=0..9);
  • Mathematica
    Flatten[Table[Binomial[n, k]Binomial[n + k, k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Dec 24 2011 *)
    Table[CoefficientList[Hypergeometric2F1[-n, n + 1, 1, -x], x], {n, 0, 9}] // Flatten
    (* Peter Luschny, Mar 09 2018 *)
  • PARI
    {T(n, k) = local(t); if( n<0, 0, t = (x + x^2)^n; for( k=1, n, t=t'); polcoeff(t, k) / n!)} /* Michael Somos, Dec 19 2002 */
    
  • PARI
    {T(n, k) = binomial(n, k) * binomial(n+k, k)} /* Michael Somos, Sep 22 2013 */
    
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, (n+k)! / (k!^2 * (n-k)!))} /* Michael Somos, Sep 22 2013 */
    

Formula

T(n, k) = (n+k)!/(k!^2*(n-k)!) = T(n-1, k)*(n+k)/(n-k) = T(n, k-1)*(n+k)*(n-k+1)/k^2 = T(n-1, k-1)*(n+k)*(n+k-1)/k^2.
binomial(x, n)^2 = Sum_{k>=0} T(n,k) * binomial(x, n+k). - Michael Somos, May 11 2012
T(n, k) = A109983(n, k+n). - Michael Somos, Sep 22 2013
G.f.: G(t, z) = 1/sqrt(1-2*z-4*t*z+z^2). Row generating polynomials = P_n(1+2z), i.e., T(n, k) = [z^k] P_n(1+2*z), where P_n are the Legendre polynomials. - Emeric Deutsch, Apr 20 2004
Sum_{k>=0} T(n, k)*A000172(k) = Sum_{k>=0} T(n, k)^2 = A005259(n). - Philippe Deléham, Jun 08 2005
1 + z*d/dz(log(G(t,z))) = 1 + (1 + 2*t)*z + (1 + 8*t + 8*t^2)*z^2 + ... is the o.g.f. for a signed version of A127674. - Peter Bala, Sep 02 2015
If R(n,t) denotes the n-th row polynomial then x^3 * exp( Sum_{n >= 1} R(n,t)*x^n/n ) = x^3 + (1 + 2*t)*x^4 + (1 + 5*t + 5*t^2)*x^5 + (1 + 9*t + 21*t^2 + 14*t^3)*x^6 + ... is an o.g.f for A033282. - Peter Bala, Oct 19 2015
P(n,x) := 1/(1 + x)*Integral_{t = 0..x} R(n,t) dt are (modulo differences of offset) the row polynomials of A033282. - Peter Bala, Jun 23 2016
From Peter Bala, Mar 09 2018: (Start)
R(n,x) = Sum_{k = 0..n} binomial(2*k,k)*binomial(n+k,n-k)*x^k.
R(n,x) = Sum_{k = 0..n} binomial(n,k)^2*x^k*(1 + x)^(n-k).
n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x).
R(n,x) = (-1)^n*R(n,-1 - x).
R(n,x) = 1/n! * (d/dx)^n ((x^2 + x)^n). (End)
The row polynomials are R(n,x) = hypergeom([-n, n + 1], [1], -x). - Peter Luschny, Mar 09 2018
T(n,k) = C(n+1,k)*A009766(n,k). - Bob Selcoe, Jan 18 2020 (Connects this triangle with the Catalan triangle. - N. J. A. Sloane, Jan 18 2020)
If we let A(n,k) = (-1)^(n+k)*(2*k+1)*(n*(n-1)*...*(n-(k-1)))/((n+1)*...*(n+(k+1))) for n >= 0 and k = 0..n, and we consider both T(n,k) and A(n,k) as infinite lower triangular arrays, then they are inverses of one another. (Empty products are by definition 1.) See the example below. The rational numbers |A(n,k)| appear in Table II on p. 92 in Ser's (1933) book. - Petros Hadjicostas, Jul 11 2020
From Peter Bala, Nov 28 2021: (Start)
Row polynomial R(n,x) = Sum_{k >= n} binomial(k,n)^2 * x^(k-n)/(1+x)^(k+1) for x > -1/2.
R(n,x) = 1/(1 + x)^(n+1) * hypergeom([n+1, n+1], [1], x/(1 + x)).
R(n,x) = (1 + x)^n * hypergeom([-n, -n], [1], x/(1 + x)).
R(n,x) = hypergeom([(n+1)/2, -n/2], [1], -4*x*(1 + x)).
If we set R(-1,x) = 1, we can run the recurrence n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x) backwards to give R(-n,x) = R(n-1,x).
R(n,x) = [t^n] ( (1 + t)*(1 + x*(1 + t)) )^n. (End)
n*T(n,k) = (2*n-1)*T(n-1,k) + (4*n-2)*T(n-1,k-1) - (n-1)*T(n-2,k). - Fabián Pereyra, Jun 30 2022
From Peter Bala, Oct 07 2024: (Start)
n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n, k) * x^k o (1 + x)^(n-k), where o denotes the black diamond product of power series as defined by Dukes and White (see Bala, Section 4.4, exercise 3).
Denote this triangle by T. Then T * transpose(T) = A143007, the square array of crystal ball sequences for the A_n X A_n lattices.
Let S denote the triangle ((-1)^(n+k)*T(n, k))n,k >= 0, a signed version of this triangle. Then S^(-1) * T = A007318, Pascal's triangle; it appears that T * S^(-1) = A110098.
T = A007318 * A115951. (End)

A143007 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the 2*n-dimensional lattice A_n x A_n.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 13, 13, 1, 1, 25, 73, 25, 1, 1, 41, 253, 253, 41, 1, 1, 61, 661, 1445, 661, 61, 1, 1, 85, 1441, 5741, 5741, 1441, 85, 1, 1, 113, 2773, 17861, 33001, 17861, 2773, 113, 1, 1, 145, 4873, 46705, 142001, 142001, 46705, 4873, 145, 1
Offset: 0

Views

Author

Peter Bala, Jul 22 2008

Keywords

Comments

The A_n lattice consists of all vectors v = (x_1,...,x_(n+1)) in Z^(n+1) such that x_1 + ... + x_(n+1) = 0. The lattice is equipped with the norm ||v|| = 1/2*(|x_1| + ... + |x_(n+1)|). Pairs of lattice points (v,w) in the product lattice A_n x A_n have norm ||(v,w)|| = ||v|| + ||w||. Then the k-th term in the crystal ball sequence for the A_n x A_n lattice gives the number of such pairs (v,w) for which ||(v,w)|| is less than or equal to k.
This array has a remarkable relationship with Apery's constant zeta(3). The row (or column) and main diagonal entries of the array occur in series acceleration formulas for zeta(3). For row n entries there holds zeta(3) = (1+1/2^3+...+1/n^3) + Sum_{k >= 1} 1/(k^3*T(n,k-1)*T(n,k)). Also, as consequence of Apery's proof of the irrationality of zeta(3), we have a series acceleration formula along the main diagonal of the table: zeta(3) = 6 * sum {n >= 1} 1/(n^3*T(n-1,n-1)*T(n,n)). Apery's result appears to generalize to the other diagonals of the table. Calculation suggests the following result may hold: zeta(3) = 1 + 1/2^3 + ... + 1/k^3 + Sum_{n >= 1} (2*n+k)*(3*n^2 +3*n*k +k^2)/(n^3*(n+k)^3*T(n-1,n+k-1)*T(n,n+k)).
For the corresponding results for the constant zeta(2), related to the crystal ball sequences of the lattices A_n, see A108625. For corresponding results for log(2), coming from either the crystal ball sequences of the hypercubic lattices A_1 x ... x A_1 or the lattices of type C_n, see A008288 and A142992 respectively.

Examples

			The table begins
n\k|0...1.....2......3.......4.......5
======================================
0..|1...1.....1......1.......1.......1
1..|1...5....13.....25......41......61 A001844
2..|1..13....73....253.....661....1441 A143008
3..|1..25...253...1445....5741...17861 A143009
4..|1..41...661...5741...33001..142001 A143010
5..|1..61..1441..17861..142001..819005 A143011
........
Example row 1 [1,5,13,...]:
The lattice A_1 x A_1 is equivalent to the square lattice of all integer lattice points v = (x,y) in Z x Z equipped with the taxicab norm ||v|| = (|x| + |y|). There are 4 lattice points (marked with a 1 on the figure below) satisfying ||v|| = 1 and 8 lattice points (marked with a 2 on the figure) satisfying ||v|| = 2. Hence the crystal ball sequence for the A_1 x A_1 lattice begins 1, 1+4 = 5, 1+4+8 = 13, ... .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . 2 . . . . .
. . . . 2 1 2 . . . .
. . . 2 1 0 1 2 . . .
. . . . 2 1 2 . . . .
. . . . . 2 . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
Row 1 = [1,5,13,...] is the sequence of partial sums of A008574; row 2 = [1,13,73,...] is the sequence of partial sums of A008530, so row 2 is the crystal ball sequence for the lattice A_2 x A_2 (the 4-dimensional di-isohexagonal orthogonal lattice).
Read as a triangle the array begins
n\k|0...1....2....3...4...5
===========================
0..|1
1..|1...1
2..|1...5....1
3..|1..13...13....1
4..|1..25...73...25...1
5..|1..41..253..253..41...1
		

Crossrefs

Cf. A001844 (row 1), A005259 (main diagonal), A008288, A008530 (first differences of row 2), A008574 (first differences of row 1), A085478, A108625, A142992, A143003, A143004, A143005, A143006, A143008 (row 2), A143009 (row 3), A142010 (row 4), A143011 (row 5).
Cf. A227845 (antidiagonal sums), A246464.
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
    A:= func< n,k | (&+[(Binomial(n,j)*Binomial(n+k-j,k-j))^2: j in [0..n]]) >; // Array
    A143007:= func< n,k | A(n-k,k) >; // Antidiagonal triangle
    [A143007(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    with(combinat): T:= (n,k) -> add(binomial(n+j,2*j)*binomial(2*j,j)^2*binomial(k+j,2*j), j = 0..n): for n from 0 to 9 do seq(T(n,k),k = 0..9) end do;
  • Mathematica
    T[n_, k_]:= HypergeometricPFQ[{-k, k+1, -n, n+1}, {1, 1, 1}, 1]; Table[T[n-k, k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Mar 06 2013 *)
  • PARI
    /* Print as a square array: */
    {T(n, k)=sum(j=0, n, binomial(n+j, 2*j)*binomial(2*j, j)^2*binomial(k+j, 2*j))}
    for(n=0, 10, for(k=0,10, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (1) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m * y^m / (1-x +x*O(x^n))^(2*m+1) * sum(k=0, m, binomial(m, k)^2*x^k)^2 ); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (2) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m/(1-x*y +x*O(x^n))^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * x^k * y^k)^2 ); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (3) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m , k)^2 * y^k * sum(j=0, k, binomial(k, j)^2 * x^j)+x*O(x^n))); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (4) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * y^(m-k) * sum(j=0, k, binomial(k, j)^2 * x^j * y^j)+x*O(x^n))); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    /* End */
    
  • SageMath
    def A(n,k): return sum((binomial(n,j)*binomial(n+k-j,k-j))^2 for j in range(n+1)) # array
    def A143007(n,k): return A(n-k,k) # antidiagonal triangle
    flatten([[A143007(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n,k) = Sum_{j = 0..n} C(n+j,2*j)*C(2*j,j)^2*C(k+j,2*j).
The array is symmetric T(n,k) = T(k,n).
The main diagonal [1,5,73,1445,...] is the sequence of Apery numbers A005259.
The entries in the k-th column satisfy the Apery-like recursion n^3*T(n,k) + (n-1)^3*T(n-2,k) = (2*n-1)*(n^2-n+1+2*k^2+2*k)*T(n-1,k).
The LDU factorization of the square array is L * D * transpose(L), where L is the lower triangular array A085478 and D is the diagonal matrix diag(C(2n,n)^2). O.g.f. for row n: The generating function for the coordination sequence of the lattice A_n is [Sum_{k = 0..n} C(n,k)^2*x^k ]/(1-x)^n. Thus the generating function for the coordination sequence of the product lattice A_n x A_n is {[Sum_{k = 0..n} C(n,k)^2*x^k]/(1-x)^n}^2 and hence the generating function for row n of this array, the crystal ball sequence of the lattice A_n x A_n, equals [Sum_{k = 0..n} C(n,k)^2*x^k]^2/(1-x)^(2n+1) = 1/(1-x)*[Legendre_P(n,(1+x)/(1-x))]^2. See [Conway & Sloane].
Series acceleration formulas for zeta(3): Row n: zeta(3) = (1 + 1/2^3 + ... + 1/n^3) + Sum_{k >= 1} 1/(k^3*T(n,k-1)*T(n,k)), n = 0,1,2,... . For example, the fourth row of the table (n = 3) gives zeta(3) = (1 + 1/2^3 + 1/3^3) + 1/(1^3*1*25) + 1/(2^3*25*253) + 1/(3^3*253*1445) + ... . See A143003 for further details.
Main diagonal: zeta(3) = 6 * Sum_{n >= 1} 1/(n^3*T(n-1,n-1)*T(n,n)). Conjectural result for other diagonals: zeta(3) = 1 + 1/2^3 + ... + 1/k^3 + Sum_{n >= 1} (2*n+k)*(3*n^2+3*n*k+k^2)/(n^3*(n+k)^3*T(n-1,n+k-1)*T(n,n+k)).
Sum_{k=0..n} T(n-k,k) = A227845(n) (antidiagonal sums). - Paul D. Hanna, Aug 27 2014
The main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) == S(m*p^(r-1) - 1) (mod p^(3*r)) for prime p >= 5 and m, r in N (this is true: see A352653. - Peter Bala, Apr 16 2022).
From Paul D. Hanna, Aug 27 2014: (Start)
G.f. A(x,y) = Sum_{n>=0, k=0..n} T(n,k)*x^n*y^k can be expressed by:
(1) Sum_{n>=0} x^n * y^n / (1-x)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k]^2,
(2) Sum_{n>=0} x^n / (1 - x*y)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k * y^k]^2,
(3) Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * y^k * Sum_{j=0..k} C(k,j)^2 * x^j,
(4) Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * y^(n-k) * Sum_{j=0..k} C(k,j)^2 * x^j * y^j. (End)
From Peter Bala, Jun 23 2023: (Start)
T(n,k) = Sum_{j = 0..n} C(n,j)^2 * C(n+k-j,k-j)^2.
T(n,k) = binomial(n+k,k)^2 * hypergeom([-n, -n, -k, -k],[-n - k, -n - k, 1], 1).
T(n,k) = hypergeom([n+1, -n, k+1, -k], [1, 1, 1], 1). (End)
From Peter Bala, Jun 28 2023: (Start)
T(n,k) = the coefficient of (x*z)^n*(y*t)^k in the expansion of 1/( (1 - x - y)*(1 - z - t) - x*y*z*t ).
T(n,k) = A(n, k, n, k) in the notation of Straub, equation 7.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, k+1, -k], [1, 1, 1], 1) allows the table indexing to be extended to negative values of n and k; we have T(-n,k) = T(n-1,k) and T(n,-k) = T(n,k-1) leading to T(-n,-k) = T(n-1, k-1). (End)
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n, k) = T(n-k, k) be the antidiagonal triangle, then:
t(n, k) = t(n, n-k).
Sum_{k=0..floor(n/2)} t(n-k,k) = A246563(n).
t(2*n+1, n+1) = A352653(n+1). (End)
From Peter Bala, Sep 27 2024: (Start)
The square array = A063007 * transpose(A063007) (LU factorization).
Let L denote the lower triangular array (l(n,k))n,k >= 0, where l(n, k) = (-1)^(n+k) * binomial(n, k)*binomial(n+k, k). (L is a signed version of A063007 and L = A063007 * A007318 ^(-1).)
Then the square array = L * transpose(A108625).
L^2 * transpose(A108625) = the Hadamard product of A108625 with itself (both identities can be verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package to find recurrences for the double sums involved). (End)

Extensions

Spelling/notation corrections by Charles R Greathouse IV, Mar 18 2010

A143003 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).

Original entry on oeis.org

0, 1, 21, 1091, 114520, 21298264, 6410456640, 2923097201856, 1920450126458880, 1747596822651334656, 2133806329230225408000, 3405545462439659704320000, 6950705677729940374290432000, 17807686090745585163974737920000
Offset: 0

Views

Author

Peter Bala, Jul 19 2008

Keywords

Comments

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1)*a(n) - n^6*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k>=1} 1/k^3 for Apery's constant zeta(3). For other cases see A066989 (m=0), A143004 (m=2), A143005 (m=3) and A143006 (m=4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^3*p_m(n)*Sum_{k = 1..n} 1/(k^3*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..n} C(2*k,k)^2*C(n+k,2*k)*C(x+k,2*k) is a polynomial in x of degree 2*m.
The first few are p_0(x) = 1, p_1(x) = 2*x^2 + 2*x + 1, p_2(x) = (3*x^4 + 6*x^3 + 9*x^2 + 6*x + 2)/2 and p_3(x) = (10*x^6 + 30*x^5 + 85*x^4 + 120*x^3 + 121*x^2 + 66*x + 18)/18. For fixed n, the sequence [p_n(k)]k>=0 is the crystal ball sequence for the product lattice A_n x A_n. See A143007 for the table of values [p_n(k)] n,k >= 0. Observe that [p_n(n)] n >= 0 is the sequence of Apery numbers A005259.
The reciprocity law p_m(n) = p_n(m) holds for nonnegative integers m and n. In particular we have p_m(1) = 2*m^2 + 2*m + 1 and p_m(2) = (3*m^4 + 6*m^3 + 9*m^2 + 6*m + 2)/2.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^3*f(x+1) + x^3*f(x-1) = (2*x+1)*(x^2+x+2*m^2+2*m+1)*f(x), normalized so that f(0) = 1. The reciprocity law now yields the Apery-like recursion m^3*p_m(x) + (m-1)^3*p_(m-2)(x) = (2*m-1)*(m^2-m+1+2*x^2+2*x)*p_(m-1)(x).
The polynomial functions p_m(x) have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p. 4 of [BUMP et al.]).
The general recurrence in the first paragraph above has a second solution b(n) = n!^3*p_m(n) with initial conditions b(0) = 1, b(1) = 2*m^2+2*m+1. Hence the behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*p_m(k-1)*p_m(k)) = 1/((2*m^2+2*m+1) - 1^6/(3*(2*m^2+2*m+3) - 2^6/(5*(2*m^2+2*m+7) - 3^6/(7*(2*m^2+2*m+13) - ...)))) = Sum_{k>=1} 1/(m+k)^3. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii)].
For the corresponding results for the constant zeta(2) see A142995. For corresponding results for the constant log(2) see A142979 and A142992.

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

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
    p := n -> 2*n^2+2*n+1: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..14)
  • Mathematica
    RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n+1)(n^2+n+5)a[n]- n^6 a[n-1]}, a[n],{n,15}] (* Harvey P. Dale, Jun 20 2011 *)

Formula

a(n) = n!^3*p(n)*Sum_{k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n).
Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).
The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 5. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - (n-1)^6/((2*n-1)*(n^2-n+5)))))), for n >= 2. The behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*(4*k^4 + 1)) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - n^6/((2*n+1)*(n^2+n+5) - ...))))) = zeta(3) - 1, where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 1].

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)

A000897 a(n) = (4*n)! / ((2*n)!*n!^2).

Original entry on oeis.org

1, 12, 420, 18480, 900900, 46558512, 2498640144, 137680171200, 7735904619300, 441233078286000, 25467973278667920, 1484298740174927040, 87202550985276963600, 5157850293780050462400, 306839461354466267304000, 18344908596179023234548480
Offset: 0

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Author

Keywords

Comments

Appears in Ramanujan's theory of elliptic functions of signature 4.
H. A. Verrill proves that a(n) = Sum_{p + q + r = 3n} w^(p-q) * {(3n)!/(p! q! r!)}^2, with p, q, r >= 0 and w = primitive 3rd root of unity.
The family of elliptic curves "x=2*H1=p^2+q^2-(1/4)*q^4, 0sqrt(-1)*q" to H1 produces "x=2*H2=p^2-q^2-(1/4)*q^4, 0Bradley Klee, Feb 25 2018
Even-order terms in the diagonal of rational function 1/(1 - (x^2 + y^2 + z)). - Gheorghe Coserea, Aug 09 2018

Examples

			G.f.: 1 + 12*x + 420*x^2 + 18480*x^3 + 900900*x^4 + 46558512*x^5 + 2498640144*x^6 + ...
		

References

  • E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 96.

Crossrefs

Cf. A002897, A008977, A186420, A188662. Elliptic Integrals: A002894, A113424, A006480. Factors: A005809, A005810, A000984, A001448.

Programs

  • GAP
    a:=n->Sum([0..3*n],k->(-1)^k*Binomial(3*n,k)*Binomial(6*n-k,3*n)*
    Binomial(2*k,k));;
    A000897:=List([0..14],n->a(n)); # Muniru A Asiru, Feb 11 2018
  • Maple
    seq((4*n)!/(n!)^4/binomial(2*n,n), n=0..14); # Zerinvary Lajos, Jun 28 2007
  • Mathematica
    Table[(4n)!/((2n)! n!^2), {n, 0, 30}] (* Stefan Steinerberger, Apr 14 2006 *)
    a[ n_] := Binomial[ 4 n, 2 n] Binomial[ 2 n, n]; (* Michael Somos, Mar 24 2013 *)
    a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/4, 3/4, 1, 64 x], {x, 0, n}]; (* Michael Somos, Mar 24 2013 *)
    a[ n_] := If[ n < 0, 0, With[{m = 4 n}, (-1)^n m! SeriesCoefficient[ BesselI[ 0, 2 x] BesselJ[ 0, 2 x], {x, 0, m}]]]; (* Michael Somos, Aug 12 2014 *)
    a[ n_] := 64^n Pochhammer[1/4, n] Pochhammer[3/4, n] / n!^2; (* Michael Somos, Aug 12 2014 *)
  • PARI
    {a(n) = if( n<0, 0, (4*n)! / ((2*n)! * n!^2))}; /* Michael Somos, Oct 31 2005 */
    

Formula

E.g.f.: Sum_{k>=0} (-1)^k * a(k) * x^(4*k) / (4*k)! = BesselI(0, 2x) * BesselJ(0, 2x).
G.f.: F(1/4, 3/4; 1; 64*x). - Michael Somos, Oct 31 2005
a(n) = A008977(n)/A000984(n) - Zerinvary Lajos, Jun 28 2007
Sum_{k>=0} a(k) * x^(3k)/(3k)!^2 = f(x)*f(x*w)*f(x/w) where f(x) = BesselI(0, 2*sqrt(x)) and w = primitive 3rd root of unity. - Michael Somos, Jul 25 2007
In general, for (BesselI(b, 2x))*(BesselJ(b, 2x))=((x^(2*b))/((GAMMA(b+1))^2)*(1-(x^4)/(Q(0)+(x^4))); Q(k)=(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)-(x^4)+(x^4)*(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)/Q(k+1)) ; (continued fraction). - Sergei N. Gladkovskii, Nov 24 2011
D-finite with recurrence 0 = a(n)*4*(4*n + 1)*(4*n + 3) - a(n+1)*(n + 1)^2 for all n in Z. - Michael Somos, Aug 12 2014
0 = a(n)*(-4026531840*a(n+2) +2005401600*a(n+3) -103896576*a(n+4) +1251948*a(n+5)) + a(n+1)*(+41418752*a(n+2) -30435328*a(n+3) +1863228*a(n+4) -24604*a(n+5)) + a(n+2)*(-16896*a(n+2) +75608*a(n+3) -6740*a(n+4) +105*a(n+5)) for all n in Z. - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(3*n,n)*binomial(4*n,n) = A005809(n)*A005810(n) = ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(12*n)), where F(x) = 1 + x + 6*x^2 + 105*x^3 + 2448*x^4 + 67043*x^5 + 2028307*x^6 + ... appears to have integer coefficients. Cf. A002894, A002897, A006480, A008977, A186420 and A188662. (End)
a(n) ~ 2^(6*n-1/2)/(Pi*n). - Ilya Gutkovskiy, Jul 12 2016
G.f.: 2*EllipticK(sqrt((sqrt(1-64*x)-1)/(2*sqrt(1-64*x))))/(Pi*(1-64*x)^(1/4)) where EllipticK is the complete elliptic integral of the first kind (in Maple's notation). - Robert Israel, Jul 12 2016
a(n) = Sum_{k = 0..3*n} (-1)^k*C(3*n,k)*C(6*n-k,3*n)*C(2*k,k). - Peter Bala, Feb 10 2018
From Bradley Klee, Feb 27 2018: (Start)
a(n) = A000984(n)*A001448(n).
G.f.: (1/(sqrt(2)*Pi))*Integral_{q=-oo..oo} 1/sqrt(q^2+(1/4)*q^4+(1-64*x)) dq.
G.f.: (1/(2*Pi))*Integral_{phi=0..2*Pi} 1/sqrt(1-64*x*sin^4(phi)) dphi. (End)
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..2*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (4*n)!/((2*n)!*n!^2);
(3/2)*Sum_{k = 0..n} 2*n*(3*n+k-1)!/(k!*n!*(2*n)!) = (4*n)!/((2*n)!*n!^2).
Cf. A001451. (End)
a(n) = (4^n/n!^2)*Product_{k = 0..2*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(3*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

A142992 Square array, read by ascending antidiagonals, of the crystal ball sequences for the root lattices of type C_n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 5, 1, 1, 19, 25, 7, 1, 1, 33, 85, 49, 9, 1, 1, 51, 225, 231, 81, 11, 1, 1, 73, 501, 833, 489, 121, 13, 1, 1, 99, 985, 2471, 2241, 891, 169, 15, 1, 1, 129, 1765, 6321, 8361, 4961, 1469, 225, 17, 1
Offset: 0

Views

Author

Peter Bala, Jul 18 2008

Keywords

Comments

The lattice C_n consists of all integer lattice points v = (x_1,...,x_n) in Z^n such that the sum x_1 + ... + x_n is even. Let ||v|| = 1/2 * Sum_{i = 1..n} |x_i|; this defines a norm on C_n. The k-th term of the crystal ball sequence of C_n gives the number of lattice points v in C_n with ||v|| <= k [Bacher et al.]. The case n = 2 is illustrated in the Example section below.
This array has a remarkable relationship with the constant log(2). The row, column and (conjecturally) the diagonal entries of the array occur in series acceleration formulas for log(2) (see the Formula section below for some examples).
See A103884 for the table of coordination sequences of the C_n lattices. For the crystal ball sequences for the A_n and D_n lattices see A108625 and A108553 respectively. For the crystal ball sequences for the product lattices A_1 x ... x A_1(n copies) and A_n x A_n see A008288 and A143007 respectively.

Examples

			The square array begins
n\k|0...1....2.....3.....4......5
=================================
.0.|1...1....1.....1.....1......1
.1.|1...3....5.....7.....9.....11
.2.|1...9...25....49....81....121 A016754
.3.|1..19...85...231...489....891 A063496
.4.|1..33..225...833..2241...4961 A142993
.5.|1..51..501..2471..8361..22363 A142994
...
Triangular array begins
n\k|0...1...2...3...4...5
=========================
.0.|1
.1.|1...1
.2.|1...3...1
.3.|1...9...5...1
.4.|1..19..25...7...1
.5.|1..33..85..49...9...1
Case n = 2: The C_2 lattice consists of all integer lattice points v = (x,y) in Z x Z such that x + y is even, equipped with the taxicab type norm ||v|| = 1/2 * (|x| + |y|). There are 8 lattice points (marked with a 1 on the figure below) satisfying ||v|| = 1 and 16 lattice points (marked with a 2 on the figure) satisfying ||v|| = 2. Hence the crystal ball sequence for the C_2 lattice (row 2 of the table) begins 1, 1+8 = 9, 1+8+16 = 25, ... .
. . . . . . . . . . .
. . . . . 2 . . . . .
. . . . 2 . 2 . . . .
. . . 2 . 1 . 2 . . .
. . 2 . 1 . 1 . 2 . .
. 2 . 1 . 0 . 1 . 2 .
. . 2 . 1 . 1 . 2 . .
. . . 2 . 1 . 2 . . .
. . . . 2 . 2 . . . .
. . . . . 2 . . . . .
. . . . . . . . . . .
		

Crossrefs

Programs

  • Maple
    with combinat: T := (n,k) -> add(binomial(2n,2i)*binomial(k+i,n),i = 0..n): for n from 0 to 9 do seq(T(n,k), k = 0..9) end do;
  • Mathematica
    t[n_, k_] := Sum[ Binomial[2*n, 2*i]*Binomial[k+i, n], {i, 0, n}]; Table[t[n-k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

Formula

T(n,k) = Sum_{i = 0..n} C(2*n,2*i)*C(k+i,n).
O.g.f. for row n: 1/(1-x)^(n+1) * Sum_{k = 0..n} C(2*n,2*k)*x^k = 1/(1-x) * T(n,(1+x)/(1-x)), where T(n,x) denotes the Chebyshev polynomial of the first kind.
O.g.f. for the array: 1/(1-x) * {(1-t) - x*(1+t)}/{(1-t)^2 - x*(1+t)^2} = (1+x+x^2+x^3+...) + (1+3*x+5*x^2+7*x^3+...)*t + (1+9*x+25*x^2+49*x^3+...)*t^2 + ... .
Row n of the array has the form [p_n(0),p_n(1),p_n(2),...], where the polynomial function p_n(x) = Sum_{k = 0..n} C(2*n,2*k)*C(x+k,n). The first few are p_0(x) = 1, p_1(x) = 2*x+1, p_2(x) = (2*x+1)^2, p_3(x) = (2*x+1)*(8*x^2+8*x+3)/3 and p_4(x) = (2*x+1)^2*(4*x^2+4*x+3)/3.
Alternative expressions for p_n(x) include p_n(x) = Sum_{k = 0..n} 2^(2*k)*n/(n+k)*C(n+k,2*k)*C(x,k) and p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(2*x+1,k).
The polynomials p_n(x) satisfy the 3-term recurrence relation n*p_n(x) = 2*(2*x+1)*p_(n-1)(x)+(n-2)*p(n-2)(x) for n >= 2; their generating function is 1/2*((1+t)/(1-t))^(2*x+1) = 1/2 + (2*x+1)*t + (2*x+1)^2*t^2 + (2*x+1)*(8*x^2+8*x+3)/3*t^3 + ... . Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(2*x+1;b,c) at b = 0, c = -1. Compare with A142979.
The polynomial p_n(x) is the unique polynomial solution to the difference equation (2*x+1)*{f(x+1/2) - f(x-1/2)} = 2*n*f(x), normalized so that f(0) = 1. The function p_n(x) is also the unique polynomial solution to the difference equation (2*x+1)*{(x+1)*f(x+1) + x*f(x-1)} = ((2*x+1)^2 + 2*n^2)*f(x), normalized so that f(0) = 1.
The zeros of p_n(x) lie on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials p_n(x-1), n = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p.4 of [BUMP et al.]).
For n > 0, the entries in row n of the array occur in series acceleration formulas for log(2): 2*log(2) = 1 + (1/2 - 1/6 +...+(-1)^n/(n*(n-1))) + (-1)^(n+1)*Sum_{k >= 1} 1/(k*T(n,k-1)*T(n,k)). For example, the fourth row of the table (n = 3) gives 2*log(2) = 4/3 + 1/(1*1*19) + 1/(2*19*85) + 1/(3*85*231) + ... .
The corresponding result for column k is 2*log(2) = 1 + (1/(1*3) + 1/(2*3*5) +...+ 1/(k*(2*k-1)*(2k+1)) + (2*k+1)*Sum_{n >= 1} (-1)^(n+1)/(n*(n+1)*T(n,k)* T(n+1,k)).
For example, the third column of the table (k = 2) gives 2*log(2) = 41/30 + 5*(1/(1*2*5*25) - 1/(2*3*25*85) + 1/(3*4*85*225) - ... ).
For the main diagonal calculation suggests the result: 2*log(2) = 4/3 + Sum_{n >= 1} (-1)^(n+1)*(5*n+3)/(n*(n+1)*T(n,n)*T(n+1,n+1)).
Similar series acceleration formulas for log(2) come from the row, column and diagonal entries of the square array of Delannoy numbers, A008288 (which may viewed as the array of crystal ball sequences for the product lattices A_1 x...x A_1). For corresponding results for the constants zeta(2) and zeta(3) see A108625 and A143007 respectively.
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