%I M4287 N1793
%S 1,1,6,90,2520,113400,7484400,681080400,81729648000,12504636144000,
%T 2375880867360000,548828480360160000,151476660579404160000,
%U 49229914688306352000000,18608907752179801056000000,8094874872198213459360000000,4015057936610313875842560000000
%N a(n) = (2n)!/2^n.
%C 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)mydeja.com), Apr 20 2001
%C a(n) is also the constant term in the product : product 1 <= i,j <= n, i different from j (1  x_i/x_j)^2.  Sharon Sela (sharonsela(AT)hotmail.com), Feb 12 2002
%C a(n) is also the number of lattice paths in the ndimensional lattice [0..2]^n.  _T. D. Noe_, Jun 06 2002
%C Representation as the nth moment of a positive function on the positive halfaxis: a(n) = Integral_{x>=0} (x^n*exp(sqrt(2*x))/sqrt(2*x)), n=0,1,...  _Karol A. Penson_, Mar 10 2003
%C Sum of consecutive combinatorial differences whose result gives (2*n)! for its numerator and 2^n for its denominator and which is the last coefficient for the lines presented in the table of sequence A087127. That is, a(n) = Sum_{i=1..n} [ C(2*n2,2*i2)*C(2*n2*i+2,2*n2*i)^(n1) C(2*n2,2*i1)*C(2*n2*i+1,2*n2*i1)^(n1) ]. E.g. a(13)= Sum_{i=1..13} [C(24,2*i2)*C(282*i,262*i)^12 C(24,2*i1)*C(272*i,252*i)^12 ] = 24!/2^12 = 4!!/2^12 = 151476660579404160000.  _AndrÃ© F. LabossiÃ¨re_, Mar 29 2004
%C 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
%C 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
%C 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]*((2n+1) + Binomial[2n+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
%C 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
%C 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
%C 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.
%C 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
%C a(n) is also the 2*n th difference of npowers 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
%C Also a(0) = 1 and a(n) = a(n1) * T(2n  1) (where T(n) is the nth triangular number). For example: a(4) = a(3) * T7 (that is, 2520 = 90 * 28).  _Enric Reverter i Bigas_, Jun 24 2013
%C 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
%D G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, 1998.
%D 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.
%D 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 AddisonWesley, Reading, MA, 1962, Vol. 1, p. 112.
%D Gao, Shanzhen, and Matheis, Kenneth, 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 FortyFirst Southeastern International Conference on Combinatorics, Graph Theory and Computing. Congr. Numer. 202 (2010), 4553.
%D S. A. Joffe, Calculation of the first thirtytwo Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 47 (1914), 103126.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D C. B. Tompkins, Methods of successive restrictions in computational problems involving discrete variables. 1963, Proc. Sympos. Appl. Math., Vol. XV pp. 95106; Amer. Math. Soc., Providence, R.I.
%H T. D. Noe, <a href="/A000680/b000680.txt">Table of n, a(n) for n = 0..100</a>
%H Daniel Dockery, <a href="http://danieldockery.com/res/math/polygorials.pdf"> Polygorials, Special "Factorials" of Polygonal Numbers.</a>
%H M. Ghebleh, <a href="https://doi.org/10.1016/j.laa.2014.06.021">Antichains of (0, 1)matrices through inversions</a>, Linear Algebra and its Applications, Volume 458, Oct 01 2014, Pages 503511.
%H J.C. Novelli, J.Y. Thibon, <a href="http://arxiv.org/abs/1403.5962">Hopf Algebras of mpermutations,(m+1)ary trees, and mparking functions</a>, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
%H Robert A. Proctor, <a href="http://arxiv.org/abs/math/0606404">Let's Expand Rota's Twelvefold Way For Counting Partitions!</a>, arXiv:math/0606404 [math.CO], 20062007.
%H D. Walsh, <a href="http://www.mtsu.edu/~dwalsh/PREIMAGE.pdf">Counting integer functions with size2 preimage constraints</a>, (preprint).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LatticePath.html">Lattice Path</a>
%H <a href="/index/Di#divseq">Index to divisibility sequences</a>
%H <a href="/index/Par#partN">Index entries for related partitioncounting sequences</a>
%F E.g.f.: 1/(1x^2/2) (with interpolating zeros).  _Paul Barry_, May 26 2003
%F A000680(n) = Polygorial(n, 6) = A000142(n)/A000079(n)*A001813(n) = n!/2^n*product(4*i+2, i=0..n1) = n!/2^n*4^n*pochhammer(1/2, n) = GAMMA(2*n+1)/2^n.  Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
%F 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((n1)/2). For positive n, a(n) = binomial(2n,2)*a(n1) with a(0)=1.  _Dennis P. Walsh_, Nov 17 2009
%F a(n) = Product_{i=1..n} binomial(2i,2).
%F a(n) = a(n1)* binomial(2n,2).
%F From _Peter Bala_, Feb 21 2011: (Start)
%F a(n) = Product_{k = 0..n1} (T(n)T(k)), where T(n) = n*(n+1)/2 is the nth triangular number.
%F Compare with n! = Product_{k = 0..n1} (nk).
%F 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(nk)) are triangle A086645. Also cf. A186432.
%F a(n) = n*(n + n1)*(n + n1 + n2)*...*(n + n1 + n2 + ... + 1).
%F For example, a(5) = 5*(5+4)*(5+4+3)*(5+4+3+2)*(5+4+3+2+1) = 113400. (End).
%F G.f.: 1/U(0) where U(k)= x*(2*k1)*k + 1  x*(2*k+1)*(k+1)/U(k+1); (continued fraction, Euler's 1st kind, 1step).  _Sergei N. Gladkovskii_, Oct 28 2012
%F a(n) = n!*(product of the first n odd integers).  _Dennis P. Walsh_, Nov 28 2012
%F 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
%e 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
%p A000680 := n>(2*n)!/(2^n);
%p a[0]:=1:a[1]:=1:for n from 2 to 50 do a[n]:=a[n1]*(2*n1)*n od: seq(a[n], n=0..16); # _Zerinvary Lajos_, Mar 08 2008
%p seq(product(binomial(2*n2*k,2),k=0..n1),n=0..16); # _Dennis P. Walsh_, Nov 17 2009
%t Table[Product[Binomial[2 i, 2], {i, 1, n}], {n, 0, 16}]
%t 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 *)
%o (PARI) a(n) = (2*n)! / 2^n
%Y Cf. A084939, A084940, A084941, A084942, A084943, A084944, A087127, A001147, A132101.
%Y A diagonal of the triangle in A241171.
%Y Main diagonal of A267479, row sums of A267480.
%Y Row n=2 of A089759.
%Y Column n=2 of A187783.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_
