Displaying 1-10 of 23 results found.
1, 2, 3, 5, 15, 137, 4393, 518625, 239478123, 455196345315, 4054931519864889, 157048983466046778713, 33902817175022723879495899, 33209307490198775518644207834495, 202452250996395056747517509528867065999
EXAMPLE
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
MATHEMATICA
Table[Sum[((n-k)*k)! / (k!)^(n-k), {k, 0, n}], {n, 0, 15}] (* Vaclav Kotesovec, Nov 24 2023 *)
De Bruijn's S(3,n): (3n)!/(n!)^3.
(Formerly M4284)
+10
97
1, 6, 90, 1680, 34650, 756756, 17153136, 399072960, 9465511770, 227873431500, 5550996791340, 136526995463040, 3384731762521200, 84478098072866400, 2120572665910728000, 53494979785374631680, 1355345464406015082330, 34469858696831179429500, 879619727485803060256500, 22514366432046593564460000
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 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 (0<x<4/27), generate this sequence via the period-energy function T(x) = 2*Pi*2F1(1/3,2/3; 1; (27/4)*x). - Bradley 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
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
Marko Petkovsek, Herbert Wilf and Doron Zeilberger, A=B, A K Peters, 1996, p. 22.
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 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 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 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 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)
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 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)
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 a(n) = (-1)^n * [x^(2*n)] ( (1 - x)^(4*n) * Legendre_P(2*n, (1 + x)/(1 - x)) ).
EXAMPLE
G.f.: 1 + 6*x + 90*x^2 + 1680*x^3 + 34650*x^4 + 756756*x^5 + 17153136*x^6 + ...
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 *) CoefficientList[Series[Hypergeometric2F1[1/3, 2/3, 1, 27*x], {x, 0, 5}], x] (* Bradley Klee, Feb 28 2018 *)
PROG
(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 */ (Magma) [Factorial(3*n)/(Factorial(n))^3: n in [0..20] ]; // Vincenzo Librandi, Aug 20 2011 (Maxima) makelist(multinomial_coeff(n, n, n), n, 0, 24); /* Emanuele Munarini, Oct 25 2016 */ (GAP) List([0..20], n->Factorial(3*n)/Factorial(n)^3); # Muniru A Asiru, Mar 31 2018 (Python)
from math import factorial
EXTENSIONS
Terms a(17) and beyond from T. D. Noe, Jun 29 2008
a(n) = (2n)!/2^n.
(Formerly M4287 N1793)
+10
81
1, 1, 6, 90, 2520, 113400, 7484400, 681080400, 81729648000, 12504636144000, 2375880867360000, 548828480360160000, 151476660579404160000, 49229914688306352000000, 18608907752179801056000000, 8094874872198213459360000000, 4015057936610313875842560000000
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
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.
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).
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 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
EXAMPLE
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
MAPLE
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 *)
PROG
(PARI) a(n) = (2*n)! / 2^n
CROSSREFS
A diagonal of the triangle in A241171. Even bisection of column k=0 of A097591.
1, 24, 2520, 369600, 63063000, 11732745024, 2308743493056, 472518347558400, 99561092450391000, 21452752266265320000, 4705360871073570227520, 1047071828879079131681280, 235809301462142612780721600, 53644737765488792839237440000, 12309355935372581458927646400000
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
LINKS
N. Narumiya and H. Shiga, The mirror map for a family of K3 surfaces induced from the simplest 3-dimensional reflexive polytope, Proceedings on Moonshine and related topics (Montréal, QC, 1999), 139-161, CRM Proc. Lecture Notes, 30, Amer. Math. Soc., Providence, RI, 2001. MR1877764 (2002m:14030).
FORMULA
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 G.f.: hypergeom([1/4, 2/4, 3/4], [1, 1], 256*x). - Michael Somos, Aug 12 2014 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 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)
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
EXAMPLE
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 + ...
MATHEMATICA
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 *)
PROG
(Magma) [Factorial(4*n)/Factorial(n)^4: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014 (Python)
from math import factorial
CROSSREFS
Cf. A000984, A006480, A008978, A178529, A000897, A002894, A002897, A006480, A008979, A186420, A188662.
1, 1, 6, 1680, 63063000, 623360743125120, 2670177736637149247308800, 7363615666157189603982585462030336000, 18165723931630806756964027928179555634194028454000000, 53130688706387569792052442448845648519471103327391407016237760000000000
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
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
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 *)
PROG
(Magma) [Factorial(n^2) / Factorial(n)^n: n in [0..10]]; // Vincenzo Librandi, Oct 29 2014
1, 120, 113400, 168168000, 305540235000, 623360743125120, 1370874167589326400, 3177459078523411968000, 7656714453153197981835000, 19010638202652030712978200000, 48334775757901219912115629238400, 125285878026462826569986857692288000
COMMENTS
Number of paths of length 5n in Z^5 from (0,0,0,0,0) to (n,n,n,n,n).
FORMULA
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) 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)
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)
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) = a(n-1)*5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/n^4. - Neven Sajko, Jul 21 2023
PROG
(Magma) [Factorial(5*n)/Factorial(n)^5: n in [0..10]]; // Vincenzo Librandi, Mar 08 2014
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.
+10
21
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
LINKS
J. D. Horton and A. Kurn, Counting sequences with complete increasing subsequences, Congressus Numerantium, 33 (1981), 75-80. MR 681905
EXAMPLE
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, ...
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!):
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 *)
CROSSREFS
Columns k=0-10 give: A057427, A033312, A267532, A269113, A269114, A269115, A269116, A269117, A269118, A269119, A269120. Rows n=0-10 give: A000004, A000007, A000012, A269121, A269122, A269123, A269124, A269125, A269126, A269127, A269128.
Table T(n,k), 0<=k, 0<=n, read by antidiagonals, defined by T(n,k) = (k*n)! / (n!)^k.
+10
18
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
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
EXAMPLE
Row n=0: 1, 1, 1, 1, 1, 1, ... A000012Row n=1: 1, 1, 2, 6, 24, 120, ... A000142Row n=2: 1, 1, 6, 90, 2520, 113400, ... A000680Row n=3: 1, 1, 20, 1680, 369600, 168168000, ... A014606Row n=4: 1, 1, 70, 34650, 63063000, 305540235000, ... A014608Row n=5: 1, 1, 252, 756756, 11732745024, 623360743125120, ... A014609
MAPLE
T:= (n, k)-> (k*n)!/(n!)^k:
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 *)
1, 5040, 681080400, 182509367040000, 66475579247327250000, 28837919555681211870935040, 14007180988362844601443040716800, 7363615666157189603982585462030336000, 4104167472585675600759440022842715359250000, 2392741010223442438553822446842770682716580000000
COMMENTS
Number of closed paths of length 7n whose steps are 7th roots of unity. - Andrew Howroyd, Nov 01 2018
FORMULA
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)
PROG
(Magma) [Factorial(7*n)/Factorial(n)^7: n in [0..20]]; // Vincenzo Librandi, Aug 13 2014
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.
+10
7
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
EXAMPLE
1 1 1 1
2 6 20 70
6 90 1680 34650
24 2520 369600 63063000
PROG
(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
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