A186432 - OEIS

%I #24 Jun 03 2018 03:49:25

%S 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 1848,990,132,1,1,182,2002,6006,6006,2002,182,1,1,240,3640,16016,

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

%C 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 The associated generalized factorial function n!_S is given by the formula

%C 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 For n >= 1, n!_S = (2*n)!/2 = A002674(n).

%C 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 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 TABLE ENTRIES

%F T(n,k) = n!_S/(k!_S*(n-k)!_S),

%F which simplifies to

%F T(n,k) = 2*binomial(2*n,2*k) for 1 <= k < n,

%F with boundary conditions T(n,0) = 1 and T(n,n) = 1 for n >= 0.

%F RELATIONS WITH OTHER SEQUENCES

%F 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 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 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 Triangle begins

%e n/k.|..0.....1.....2.....3.....4.....5.....6.....7

%e ==================================================

%e .0..|..1

%e .1..|..1.....1

%e .2..|..1....12.....1

%e .3..|..1....30....30.....1

%e .4..|..1....56...140....56.....1

%e .5..|..1....90...420...420....90.....1

%e .6..|..1...132...990..1848...990...132.....1

%e .7..|..1...182..2002..6006..6006..2002...182.....1

%e ...

%t 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 Cf. A002114, A086645, A186430, A186433 (inverse).

%K nonn,easy,tabl

%O 0,5

%A _Peter Bala_, Feb 22 2011