cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Views

Author

N. J. A. Sloane

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).

Links

Crossrefs

Cf. A000984, A001459, A008977, A050983, A050984, A091683, A161581, A181545.
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

A023983 Sum of exponents in prime-power factorization of multinomial coefficient M(5n;2n,2n,n).

Original entry on oeis.org

0, 3, 6, 8, 10, 12, 14, 19, 17, 18, 21, 24, 24, 29, 27, 29, 27, 31, 29, 33, 35, 37, 39, 41, 40, 40, 45, 45, 43, 48, 48, 54, 46, 48, 51, 52, 51, 56, 57, 59, 58, 57, 60, 61, 62, 65, 64, 69, 63, 68, 66, 69, 74, 76, 74, 78, 75, 76, 81, 82, 81, 85, 87, 83, 77, 81, 82, 84, 81, 87, 86, 93, 87, 91, 92, 93, 96
Offset: 0

Views

Author

Clark Kimberling

Keywords

Links

Crossrefs

Cf. A001222, A001459, A022559.
Cf. A023978, A023979, A023980, A023981, A023982, A023984, A023985, A023986, A023987, A023990, A023991, A023992, A023993.

Programs

  • Mathematica
    a[n_] := PrimeOmega[Multinomial[2*n, 2*n, n]]; Array[a, 100, 0] (* Amiram Eldar, Jun 11 2025 *)
  • PARI
    a(n) = bigomega((5*n)! / ((2*n)!^2*n!)); \\ Amiram Eldar, Jun 11 2025

Formula

From Amiram Eldar, Jun 11 2025: (Start)
a(n) = A001222(A001459(n)).
a(n) = A022559(5*n) - 2*A022559(2*n) - A022559(n). (End)

Extensions

Offset changed to 0 and a(0) prepended by Amiram Eldar, Jun 11 2025

A001450 a(n) = binomial(5*n,2*n).

Original entry on oeis.org

1, 10, 210, 5005, 125970, 3268760, 86493225, 2319959400, 62852101650, 1715884494940, 47129212243960, 1300853625660225, 36052387482172425, 1002596421878664480, 27963143931814663880, 781879430625942976880, 21910242651571684460050, 615167304833936727234180
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A001451, A001459, A001460, A000108, A060941, A066802, A259550.

Programs

  • Magma
    [Binomial(5*n, 2*n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014
  • Maple
    f := n->(5*n)!/((3*n)!*(2*n)!);
  • Mathematica
    Table[Hypergeometric2F1[-3n,-2n,1,1],{n,0,60}] (* John M. Campbell, Jul 15 2011 *)
Table[Binomial[5n,2n],{n,0,20}] (* Harvey P. Dale, Nov 09 2011 *)
  • PARI
    a(n) = binomial(5*n,2*n) \\ Altug Alkan, Oct 06 2015

Formula

a(n) = (5*n)!/((3*n)!*(2*n)!).
a(n) = 2F1[-3n,-2n,1,1] (see Mathematica code below). - John M. Campbell, Jul 15 2011
G.f.: hypergeom([1/5, 2/5, 3/5, 4/5], [1/3, 1/2, 2/3], (3125/108)*x). - Robert Israel, Aug 07 2014
From Peter Bala, Oct 05 2015: (Start)
a(n) = [x^n] ( (1 + x)*C(x) )^(5*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
a(n) = 5*A259550(n) for n >= 1.
exp( (1/5) * Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 23*x^2 + 377*x^3 + ... is the o.g.f. for the sequence of Duchon numbers A060941. (End)
a(n) = [x^(2*n)] 1/(1 - x)^(3*n+1). - Ilya Gutkovskiy, Oct 10 2017
D-finite with recurrence 6*n*(3*n-1)*(2*n-1)*(3*n-2)*a(n) -5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
a(n) = Sum_{k = 0..2*n} binomial(3*n+k-1, k). Cf. A066802. - Peter Bala, Jun 04 2024
Right-hand side of the identity Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)* binomial(4*n-k, 2*n-k) = binomial(5*n, 2*n). Compare with the identity Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(4*n-k, 2*n-k) = binomial(3*n, n). - Peter Bala, Jun 05 2024
From Karol A. Penson, May 07 2025: (Start)
G.f. denoted by h(x) satisfies the following algebraic equation of order 10:
8 - 3125*x + 20*(-13 + 3125*x)*h(x) - 45*(-74 + 9375*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^3 + 25*(1809 + 53125*x)*h(x)^4 + (34375*x - 738)*(3125*x - 108)*h(x)^5 + 15*(3125*x + 297)*(3125*x - 108)*h(x)^6 + 5*(3125*x - 108)^2*h(x)^7 + 135*(3125*x - 108)^2*h(x)^8 + (3125*x - 108)^3*h(x)^10=0.
a(n) = Integral_{x=0..3125/108} x^n*W(x)*dx, n>=0, where W(x) = W1(x)+W2(x)+W3(x)+W4(x) can be expressed with four generalized hypergeometric functions of type 4F3:
W1(x) = sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([1/5, 8/15, 7/10, 13/15], [2/5, 3/5, 4/5], (108*x)/3125)/(10*Pi*x^(4/5)),
W2(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([2/5, 11/15, 9/10, 16/15], [3/5, 4/5, 6/5], (108*x)/3125)/(50*Pi*x^(3/5)),
W3(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([3/5, 14/15, 11/10, 19/15], [4/5, 6/5, 7/5], (108*x)/3125)/(125*Pi*x^(2/5)), and
W4(x) = (7*sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([4/5, 17/15, 13/10, 22/15], [6/5, 7/5, 8/5], (108*x)/3125))/(1250*Pi*x^(1/5)).
Using the formula for a(n) only, W(x) can be shown to be a positive function. It is singular at x=0 and at x=3125/108. This integral representation is unique since W(x) is the solution of the Hausdorff moment problem. (End)
From Peter Bala, Jun 21 2025: (Start)
a(n) = [x^(3*n)] 1/(1 - x)^(2*n+1).
a(n) = Sum_{k = 0..3*n} binomial(2*n+k-1, k). (End)
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.