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A186432 Triangle associated with the set S of squares {0,1,4,9,16,...}.

%I A186432 #24 Jun 03 2018 03:49:25
%S A186432 1,1,1,1,12,1,1,30,30,1,1,56,140,56,1,1,90,420,420,90,1,1,132,990,
%T A186432 1848,990,132,1,1,182,2002,6006,6006,2002,182,1,1,240,3640,16016,
%U A186432 25740,16016,3640,240,1,1,306,6120,37128,87516,87516,37128,6120,306,1,1,380,9690,77520,251940,369512,251940,77520,9690,380,1
%N A186432 Triangle associated with the set S of squares {0,1,4,9,16,...}.
%C A186432 Given a subset S of the integers Z, Bhargava [1] has shown how to associate with S a generalized factorial function, denoted n!_S, sharing many properties of the classical factorial function n! (which corresponds to the choice S = Z). In particular, he shows that the generalized binomial coefficients n!_S/(k!_S*(n-k)!_S) are always integral for any choice of S. Here we take S = {0,1,4,9,16,...}, the set of squares.
%C A186432 The associated generalized factorial function n!_S is given by the formula
%C A186432 n!_S = Product_{k=0..n} (n^2 - k^2), with the convention 0!_S = 1. This should be compared with n! = Product_{k=0..n} (n - k).
%C A186432 For n >= 1, n!_S = (2*n)!/2 = A002674(n).
%C A186432 Compare this triangle with A086645 and also A186430 - the generalized binomial coefficients for the set S of prime numbers {2,3,5,7,11,...}.
%H A186432 M. Bhargava, <a href="http://dx.doi.org/10.2307/2695734">The factorial function and generalizations</a>, Amer. Math. Monthly, 107 (2000), 783-799.
%F A186432 TABLE ENTRIES
%F A186432 T(n,k) = n!_S/(k!_S*(n-k)!_S),
%F A186432 which simplifies to
%F A186432 T(n,k) = 2*binomial(2*n,2*k) for 1 <= k < n,
%F A186432 with boundary conditions T(n,0) = 1 and T(n,n) = 1 for n >= 0.
%F A186432 RELATIONS WITH OTHER SEQUENCES
%F A186432 Denote this triangle by T. The first column of the inverse T^-1 (see A186433) begins [1, -1, 11, -301, 15371, ...] and, apart from the initial 1, is a signed version of the Glaisher's H' numbers A002114.
%F A186432 The first column of (1/2)*T^2 begins [1/2, 1, 7, 31, 127, ...] and, apart from the initial term, equals A000225(2*n-1), counting the preferential arrangements on (2*n - 1) labeled elements having less than or equal to two ranks.
%F A186432 The first column of (1/3)*T^3 begins [1/3, 1, 13, 181, 1933, ...] and, apart from the initial term, is A101052(2*n-1), which gives the number of preferential arrangements on (2*n-1) labeled elements having less than or equal to three ranks.
%e A186432 Triangle begins
%e A186432 n/k.|..0.....1.....2.....3.....4.....5.....6.....7
%e A186432 ==================================================
%e A186432 .0..|..1
%e A186432 .1..|..1.....1
%e A186432 .2..|..1....12.....1
%e A186432 .3..|..1....30....30.....1
%e A186432 .4..|..1....56...140....56.....1
%e A186432 .5..|..1....90...420...420....90.....1
%e A186432 .6..|..1...132...990..1848...990...132.....1
%e A186432 .7..|..1...182..2002..6006..6006..2002...182.....1
%e A186432 ...
%t A186432 Table[2 Binomial[2 n, 2 k] - Boole[Or[k == 0, k == n]], {n, 0, 10}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, May 23 2017 *)
%Y A186432 Cf. A002114, A086645, A186430, A186433 (inverse).
%K A186432 nonn,easy,tabl
%O A186432 0,5
%A A186432 _Peter Bala_, Feb 22 2011