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 1-10 of 23 results. Next

A248827 Row sums of A187783 and A089759.

Original entry on oeis.org

1, 2, 3, 5, 15, 137, 4393, 518625, 239478123, 455196345315, 4054931519864889, 157048983466046778713, 33902817175022723879495899, 33209307490198775518644207834495, 202452250996395056747517509528867065999
Offset: 0

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Author

Tilman Piesk, Oct 28 2014

Keywords

Examples

			          1                  1
        1   1                2
      1   1   1              3
    1   2   1   1            5
  1   6   6   1   1         15
1  24  90  20   1   1      137
		

Programs

  • Mathematica
    Table[Sum[((n-k)*k)! / (k!)^(n-k), {k, 0, n}], {n, 0, 15}] (* Vaclav Kotesovec, Nov 24 2023 *)

Formula

log(a(n)) ~ n^2 * (log(n/2)/4 + 1/(16*log(n))). - Vaclav Kotesovec, Nov 24 2023

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

A000680 a(n) = (2n)!/2^n.

Original entry on oeis.org

1, 1, 6, 90, 2520, 113400, 7484400, 681080400, 81729648000, 12504636144000, 2375880867360000, 548828480360160000, 151476660579404160000, 49229914688306352000000, 18608907752179801056000000, 8094874872198213459360000000, 4015057936610313875842560000000
Offset: 0

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Author

Keywords

Comments

Denominators in the expansion of cos(sqrt(2)*x) = 1 - (sqrt(2)*x)^2/2! + (sqrt(2)*x)^4/4! - (sqrt(2)*x)^6/6! + ... = 1 - x^2 + x^4/6 - x^6/90 + ... By Stirling's formula in A000142: a(n) ~ 2^(n+1) * (n/e)^(2n) * sqrt(Pi*n) - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 20 2001
a(n) is also the constant term in the product: Product_{1<=i, j<=n, i!=j} (1 - x_i/x_j)^2. - Sharon Sela (sharonsela(AT)hotmail.com), Feb 12 2002
a(n) is also the number of lattice paths in the n-dimensional lattice [0..2]^n. - T. D. Noe, Jun 06 2002
Representation as the n-th moment of a positive function on the positive half-axis: a(n) = Integral_{x>=0} (x^n*exp(-sqrt(2*x))/sqrt(2*x)), n=0,1,... - Karol A. Penson, Mar 10 2003
Number of permutations of [2n] with no increasing runs of odd length. Example: a(2) = 6 because we have 1234, 13/24, 14/23, 23/14, 24/13 and 34/12 (runs separated by slashes). - Emeric Deutsch, Aug 29 2004
This is also the number of ways of arranging the elements of n distinct pairs, assuming the order of elements is significant and the pairs are distinguishable. When the pairs are not distinguishable, see A001147 and A132101. For example, there are 6 ways of arranging 2 pairs [1,1], [2,2]: {[1122], [1212], [1221], [2211], [2121], [2112]}. - Ross Drewe, Mar 16 2008
n married couples are seated in a row so that every wife is to the left of her husband. The recurrence a(n+1) = a(n)*((2*n + 1) + binomial(2*n+1, 2)) conditions on whether the (n+1)st couple is seated together or separated by at least one other person. - Geoffrey Critzer, Jun 10 2009
a(n) is the number of functions f:[2n]->[n] such that the preimage of {y} has cardinality 2 for every y in [n]. Note that [k] denotes the set {1,2,...,k} and [0] denotes the empty set. - Dennis P. Walsh, Nov 17 2009
a(n) is also the number of n X 2n (0,1)-matrices with row sum 2 and column sum 1. - Shanzhen Gao, Feb 12 2010
Number of ways that 2n people of different heights can be arranged (for a photograph) in two rows of equal length so that every person in the front row is shorter than the person immediately behind them in the back row.
a(n) is the number of functions f:[n]->[n^2] such that, if floor((f(x))^.5) = floor((f(y))^.5), then x = y. For example, with n = 4, the range of f consists of one element from each of the four sets {1,2,3}, {4,5,6,7,8}, {9,10,11,12,13,14,15}, and {16}. Hence there are 1*3*5*7 = 105 ways to choose the range for f, and there are 4! ways to injectively map {1,2,3,4} to the four elements of the range. Thus there are 105*24 = 2520 such functions. Note also that a(n) = n!*(product of the first n odd numbers). - Dennis P. Walsh, Nov 28 2012
a(n) is also the 2*n th difference of n-powers of A000217 (triangular numbers). For example a(2) is the 4th difference of the squares of triangular numbers. - Enric Reverter i Bigas, Jun 24 2013
a(n) is the multinomial coefficient (2*n) over (2, 2, 2, ..., 2) where there are n 2's in the last parenthesis. It is therefore also the number of words of length 2n obtained with n letters, each letter appearing twice. - Robert FERREOL, Jan 14 2018
Number of ways to put socks and shoes on an n-legged animal, if a sock must be put on before a shoe. - Daniel Bishop, Jan 29 2018

Examples

			For n = 2, a(2) = 6 since there are 6 functions f:[4]->[2] with size 2 preimages for both {1} and {2}. In this case, there are binomial(4, 2) = 6 ways to choose the 2 elements of [4] f maps to {1} and the 2 elements of [4] that f maps to {2}. - _Dennis P. Walsh_, Nov 17 2009
		

References

  • G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, 1998.
  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 283.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 112.
  • Shanzhen Gao and Kenneth Matheis, Closed formulas and integer sequences arising from the enumeration of (0,1)-matrices with row sum two and some constant column sums. In Proceedings of the Forty-First Southeastern International Conference on Combinatorics, Graph Theory and Computing. Congr. Numer. 202 (2010), 45-53.
  • 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).
  • C. B. Tompkins, Methods of successive restrictions in computational problems involving discrete variables. 1963, Proc. Sympos. Appl. Math., Vol. XV pp. 95-106; Amer. Math. Soc., Providence, R.I.

Crossrefs

A diagonal of the triangle in A241171.
Main diagonal of A267479, row sums of A267480.
Row n=2 of A089759.
Column n=2 of A187783.
Even bisection of column k=0 of A097591.

Programs

  • Maple
    A000680 := n->(2*n)!/(2^n);
    a[0]:=1:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]*(2*n-1)*n od: seq(a[n], n=0..16); # Zerinvary Lajos, Mar 08 2008
    seq(product(binomial(2*n-2*k,2),k=0..n-1),n=0..16); # Dennis P. Walsh, Nov 17 2009
  • Mathematica
    Table[Product[Binomial[2 i, 2], {i, 1, n}], {n, 0, 16}]
    polygorial[k_, n_] := FullSimplify[ n!/2^n (k -2)^n*Pochhammer[2/(k -2), n]]; Array[ polygorial[6, #] &, 17, 0] (* Robert G. Wilson v, Dec 26 2016 *)
    Table[(2n)!/2^n,{n,0,20}] (* Harvey P. Dale, Sep 21 2020 *)
  • PARI
    a(n) = (2*n)! / 2^n

Formula

E.g.f.: 1/(1 - x^2/2) (with interpolating zeros). - Paul Barry, May 26 2003
a(n) = polygorial(n, 6) = (A000142(n)/A000079(n))*A001813(n) = (n!/2^n)*Product_{i=0..n-1} (4*i + 2) = (n!/2^n)*4^n*Pochhammer(1/2, n) = gamma(2*n+1)/2^n. - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
a(n) = A087127(n,2*n) = Sum_{i=0..2*n} (-1)^(2*n-i)*binomial(2*n, i)*binomial(i+2, 2)^n. Let T(n,k,j) = ((n - k + j)*(2*n - 2*k + 1))^n*binomial(2*n, 2*k-j+1) then a(n) = Sum{k=0..n} (T(n,k,1) - T(n,k,0)). For example a(12) = A087127(12,24) = Sum_{k=0..12} (T(12,k,1) - T(12,k,0)) = 24!/2^12. - André F. Labossière, Mar 29 2004 [Corrected by Jianing Song, Jan 08 2019]
For even n, a(n) = binomial(2n, n)*(a(n/2))^2. For odd n, a(n) = binomial(2n, n+1)*a((n+1)/2)*a((n-1)/2). For positive n, a(n) = binomial(2n, 2)*a(n-1) with a(0) = 1. - Dennis P. Walsh, Nov 17 2009
a(n) = Product_{i=1..n} binomial(2i, 2).
a(n) = a(n-1)*binomial(2n, 2).
From Peter Bala, Feb 21 2011: (Start)
a(n) = Product_{k = 0..n-1} (T(n) - T(k)), where T(n) = n*(n + 1)/2 is the n-th triangular number.
Compare with n! = Product_{k = 0..n-1} (n - k).
Thus we may view a(n) as a generalized factorial function associated with the triangular numbers A000217. Cf. A010050. The corresponding generalized binomial coefficients a(n)/(a(k)*a(n-k)) are triangle A086645. Also cf. A186432.
a(n) = n*(n + n-1)*(n + n-1 + n-2)*...*(n + n-1 + n-2 + ... + 1).
For example, a(5) = 5*(5+4)*(5+4+3)*(5+4+3+2)*(5+4+3+2+1) = 113400. (End).
G.f.: 1/U(0) where U(k)= x*(2*k - 1)*k + 1 - x*(2*k + 1)*(k + 1)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Oct 28 2012
a(n) = n!*(product of the first n odd integers). - Dennis P. Walsh, Nov 28 2012
a(0) = 1, a(n) = a(n-1)*T(2*n-1), where T(n) is the n-th triangular number. For example: a(4) = a(3)*T(7) = 90*28 = 2520. - Enric Reverter i Bigas, Jun 24 2013
E.g.f.: 1/(1 - x/(1 - 2*x/(1 - 3*x/(1 - 4*x/(1 - 5*x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, May 10 2017
From Amiram Eldar, Jun 25 2020: (Start)
Sum_{n>=0} 1/a(n) = cosh(sqrt(2)).
Sum_{n>=0} (-1)^n/a(n) = cos(sqrt(2)). (End)
D-finite with recurrence a(n) -n*(2*n-1)*a(n-1)=0. - R. J. Mathar, Jan 28 2022
a(n) = n *A007019(n-1), n>0. - R. J. Mathar, Jan 28 2022

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

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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

A034841 a(n) = (n^2)! / (n!)^n.

Original entry on oeis.org

1, 1, 6, 1680, 63063000, 623360743125120, 2670177736637149247308800, 7363615666157189603982585462030336000, 18165723931630806756964027928179555634194028454000000, 53130688706387569792052442448845648519471103327391407016237760000000000
Offset: 0

Views

Author

Keywords

Comments

The number of arrangements of 1,2,...,n^2 in an n X n matrix such that each row is increasing. - Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 12 2001
a(n) == 0 (mod n!). In fact (n^2)! == 0 (mod (n!)^n) by elementary combinatorics, a better result is (n^2)! == 0 (mod (n!)^(n+1)). - Amarnath Murthy, Jul 13 2005
a(n) is also the number of lattice paths from {n}^n to {0}^n using steps that decrement one component by 1. a(2) = 6: [(2,2), (1,2), (0,2), (0,1), (0,0)], [(2,2), (1,2), (1,1), (0,1), (0,0)], [(2,2), (1,2), (1,1), (1,0), (0,0)], [(2,2), (2,1), (1,1), (0,1), (0,0)], [(2,2), (2,1), (1,1), (1,0), (0,0)], [(2,2), (2,1), (2,0), (1,0), (0,0)]. - Alois P. Heinz, May 06 2013
Given n^2 distinguishable balls and n distinguishable urns, a(n) = the number of ways to place n balls in the i-th urn for all 1 <= i <= n, where n = n_1 + n_2 + ... + n_n. - Ross La Haye, Dec 28 2013

Crossrefs

Programs

  • Magma
    [Factorial(n^2) / Factorial(n)^n: n in [0..10]]; // Vincenzo Librandi, Oct 29 2014
  • Maple
    a:= n-> (n^2)! / (n!)^n:
    seq(a(n), n=0..10);  # Alois P. Heinz, Jul 24 2012
  • Mathematica
    Prepend[Table[nn = n^2;nn! Coefficient[Series[(x^n/n!)^n, {x, 0, nn}], x^nn], {n, 1, 15}], 1] (* Geoffrey Critzer, Mar 08 2015 *)
  • PARI
    a(n) = (n^2)! / (n!)^n; \\ Michel Marcus, Oct 28 2014
    

Formula

Using a higher order version of Stirling's formula (the "standard" formula appears in A000142) we have the asymptotic expression: a(n) ~ sqrt(2*Pi) * e^(-1/12) * n^(n^2 - n/2 + 1) / (2*Pi)^(n/2). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 13 2001
a(n) = Product_{k=1..n} binomial(k*n, n). - Vaclav Kotesovec, Mar 10 2019

Extensions

a(0)=1 prepended by Tilman Piesk, Oct 28 2014

A008978 a(n) = (5*n)!/(n!)^5.

Original entry on oeis.org

1, 120, 113400, 168168000, 305540235000, 623360743125120, 1370874167589326400, 3177459078523411968000, 7656714453153197981835000, 19010638202652030712978200000, 48334775757901219912115629238400, 125285878026462826569986857692288000
Offset: 0

Views

Author

Keywords

Comments

Number of paths of length 5n in Z^5 from (0,0,0,0,0) to (n,n,n,n,n).

Crossrefs

Programs

Formula

a(n) ~ 5^(5*n+1/2) / (4 * Pi^2 * n^2). - Vaclav Kotesovec, Mar 07 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n]( F(x)^(120*n) ), where F(x) = 1 + x + 353*x^2 + 318986*x^3 + 408941594*x^4 + 633438203535*x^5 + 1105336091531052*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. (End)
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(5*n,n + k)*binomial(n + k,k)^5.
a(n) = Sum_{k = 0..5*n} (-1)^(n+k)*binomial(5*n,k)*binomial(n + k,k)^5. (End)
From Ilya Gutkovskiy, Nov 23 2017: (Start)
O.g.f.: 4F3(1/5,2/5,3/5,4/5; 1,1,1; 3125*x).
E.g.f.: 4F4(1/5,2/5,3/5,4/5; 1,1,1,1; 3125*x). (End)
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*u)^n] (1 + x + y + z + u )^(5*n). (End)
a(n) = 120*A322252(n). - R. J. Mathar, Jun 21 2023
a(n) = a(n-1)*5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/n^4. - Neven Sajko, Jul 21 2023

A269129 Number A(n,k) of sequences with k copies each of 1,2,...,n avoiding the pattern 12...n; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 5, 1, 0, 0, 1, 43, 23, 1, 0, 0, 1, 374, 1879, 119, 1, 0, 0, 1, 3199, 173891, 102011, 719, 1, 0, 0, 1, 26945, 16140983, 117392909, 7235651, 5039, 1, 0, 0, 1, 224296, 1474050783, 142951955371, 117108036719, 674641325, 40319, 1
Offset: 0

Views

Author

Alois P. Heinz, Feb 19 2016

Keywords

Examples

			Square array A(n,k) begins:
  0,   0,      0,         0,            0,               0, ...
  1,   0,      0,         0,            0,               0, ...
  1,   1,      1,         1,            1,               1, ...
  1,   5,     43,       374,         3199,           26945, ...
  1,  23,   1879,    173891,     16140983,      1474050783, ...
  1, 119, 102011, 117392909, 142951955371, 173996758190594, ...
		

Crossrefs

Programs

  • Maple
    g:= proc(l) option remember; (n-> f(l[1..nops(l)-1])*
          binomial(n-1, l[-1]-1)+add(f(sort(subsop(j=l[j]
          -1, l))), j=1..nops(l)-1))(add(i, i=l))
        end:
    f:= l->(n->`if`(n=0, 1, `if`(l[1]=0, 0, `if`(n=1 or l[-1]=1, 1,
        `if`(n=2, binomial(l[1]+l[2], l[1])-1, g(l))))))(nops(l)):
    A:= (n, k)-> (k*n)!/k!^n - f([k$n]):
    seq(seq(A(n, d-n), n=0..d), d=0..12);
    # second Maple program:
    b:= proc(k, p, j, l, t) option remember;
          `if`(k=0, (-1)^t/l!, `if`(p<0, 0, add(b(k-i, p-1,
           j+1, l+i*j, irem(t+i*j, 2))/(i!*p!^i), i=0..k)))
        end:
    A:= (n, k)-> (n*k)!*(1/k!^n-b(n, k-1, 1, 0, irem(n, 2))*n!):
    seq(seq(A(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, Mar 03 2016
  • Mathematica
    b[k_, p_, j_, l_, t_] := b[k, p, j, l, t] = If[k == 0, (-1)^t/l!, If[p < 0, 0, Sum[b[k-i, p-1, j+1, l + i j, Mod[t + i j, 2]]/(i! p!^i), {i, 0, k}]] ];
    A[n_, k_] := (n k)! (1/k!^n - b[n, k-1, 1, 0, Mod[n, 2]] n!); Table[ Table[ A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Apr 07 2016, after Alois P. Heinz *)

Formula

A(n,k) = A089759(k,n) - A047909(k,n) = A187783(n,k) - A047909(k,n).

A089759 Table T(n,k), 0<=k, 0<=n, read by antidiagonals, defined by T(n,k) = (k*n)! / (n!)^k.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 6, 6, 1, 1, 1, 24, 90, 20, 1, 1, 1, 120, 2520, 1680, 70, 1, 1, 1, 720, 113400, 369600, 34650, 252, 1, 1, 1, 5040, 7484400, 168168000, 63063000, 756756, 924, 1, 1, 1, 40320, 681080400, 137225088000, 305540235000, 11732745024, 17153136, 3432, 1, 1
Offset: 0

Views

Author

Philippe Deléham, Jan 08 2004; revised Jun 08 2005

Keywords

Comments

T(n,k) is the number of lattice paths from {n}^k to {0}^k using steps that decrement one component by 1. - Alois P. Heinz, May 06 2013

Examples

			Row n=0: 1, 1,   1,      1,           1,               1, ... A000012
Row n=1: 1, 1,   2,      6,          24,             120, ... A000142
Row n=2: 1, 1,   6,     90,        2520,          113400, ... A000680
Row n=3: 1, 1,  20,   1680,      369600,       168168000, ... A014606
Row n=4: 1, 1,  70,  34650,    63063000,    305540235000, ... A014608
Row n=5: 1, 1, 252, 756756, 11732745024, 623360743125120, ... A014609
		

Crossrefs

Cf. A000680, A014606, A014608, A014609, A000984, A187783 (transposed version).
Main diagonal gives A034841.

Programs

  • Maple
    T:= (n, k)-> (k*n)!/(n!)^k:
    seq(seq(T(n, d-n), n=0..d), d=0..10);  # Alois P. Heinz, Aug 16 2012
  • Mathematica
    T[n_, k_] := (k*n)!/(n!)^k; Table[T[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 19 2015 *)

Extensions

Corrected by Alois P. Heinz, Aug 16 2012

A071549 a(n) = (7n)!/n!^7.

Original entry on oeis.org

1, 5040, 681080400, 182509367040000, 66475579247327250000, 28837919555681211870935040, 14007180988362844601443040716800, 7363615666157189603982585462030336000, 4104167472585675600759440022842715359250000, 2392741010223442438553822446842770682716580000000
Offset: 0

Views

Author

Benoit Cloitre, May 30 2002

Keywords

Comments

Number of closed paths of length 7n whose steps are 7th roots of unity. - Andrew Howroyd, Nov 01 2018

Crossrefs

Row n=7 of A187783, column k=7 of A089759.
Sequences (k*n)!/n!^k: A000984 (k = 2), A006480 (k =3), A008977 (k = 4), A008978 (k = 5), A008979 (k = 6), A071550 (k = 8), A071551 (k = 9), A071552 (k = 10).

Programs

Formula

From Peter Bala, Feb 14 2020: (Start)
a(n) = C(7*n,n)*C(6*n,n)*C(5*n,n)*C(4*n,n)*C(3*n,n)*C(2*n,n).
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^n](F(x)^(5040*n)), where F(x) = 1 + x + 62528*x^2 + 11087269661*x^3 + 3021437267047869*x^4 + 1045823730475703710735*x^5 + ...
appears to have integer coefficients. For similar results see A008979.
a(n) = [(x*y*z*u*v*w)^n] (1 + x + y + z + u + v + w)^(7*n). (End)

Extensions

a(8)-a(9) added by Andrew Howroyd, Nov 01 2018

A060538 Square array read by antidiagonals of number of ways of dividing n*k labeled items into n labeled boxes with k items in each box.

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 20, 90, 24, 1, 70, 1680, 2520, 120, 1, 252, 34650, 369600, 113400, 720, 1, 924, 756756, 63063000, 168168000, 7484400, 5040, 1, 3432, 17153136, 11732745024, 305540235000, 137225088000, 681080400, 40320, 1, 12870
Offset: 1

Views

Author

Henry Bottomley, Apr 02 2001

Keywords

Examples

			       1        1        1        1
       2        6       20       70
       6       90     1680    34650
      24     2520   369600 63063000
		

Crossrefs

Subtable of A187783.
Rows include A000012, A000984, A006480, A008977, A008978 etc.
Columns include A000142, A000680, A014606, A014608, A014609 etc.
Main diagonal is A034841.

Programs

  • PARI
    T(n,k)=(n*k)!/k!^n;
    for(n=1, 6, for(k=1, 6, print1(T(n,k), ", ")); print) \\ Harry J. Smith, Jul 06 2009

Formula

T(n, k) = (nk)!/k!^n = T(n-1, k)*binomial(nk, k) = T(n-1, k)*A060539(n, k) = A060540(n, k)*A000142(k).
Showing 1-10 of 23 results. Next