

A186432


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


4



1, 1, 1, 1, 12, 1, 1, 30, 30, 1, 1, 56, 140, 56, 1, 1, 90, 420, 420, 90, 1, 1, 132, 990, 1848, 990, 132, 1, 1, 182, 2002, 6006, 6006, 2002, 182, 1, 1, 240, 3640, 16016, 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
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OFFSET

0,5


COMMENTS

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*(nk)!_S) are always integral for any choice of S. Here we take S = {0,1,4,9,16,...}, the set of squares.
The associated generalized factorial function n!_S is given by the formula
n!_S = product {k = 0..n } (n^2k^2), with the convention 0!_S = 1. This should be compared with n! = product {k = 0..n } (nk).
For n >= 1, n!_S = (2*n)!/2 = A002674(n).
Compare this triangle with A086645 and also A186430  the generalized binomial coefficients for the set S of prime numbers {2,3,5,7,11,...}.


LINKS

Table of n, a(n) for n=0..65.
M. Bhargava, The factorial function and generalizations, Amer. Math. Monthly, 107 (2000), 783799.


FORMULA

TABLE ENTRIES
T(n,k) = n!_S/(k!_S*(nk)!_S),
which simplifies to
T(n,k) = 2*binomial(2*n,2*k) for 1 <= k < n,
with boundary conditions T(n,0) = 1 and T(n,n) = 1 for n >= 0.
RELATIONS WITH OTHER SEQUENCES
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.
The first column of 1/2*T^2 begins [1/2,1,7,31,127,...] and, apart from the initial term, equals A000225(2*n1), counting the number of preferential arrangements on (2*n  1) labelled elements having less than or equal to two ranks.
The first column of 1/3*T^3 begins [1/3,1,13,181,1933,...] and, apart from the initial term, is A101052(2*n1), which gives the number of preferential arrangements on (2*n1) labelled elements having less than or equal to three ranks.


EXAMPLE

Triangle begins
n/k...0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0....1
.1....1.....1
.2....1....12.....1
.3....1....30....30.....1
.4....1....56...140....56.....1
.5....1....90...420...420....90.....1
.6....1...132...990..1848...990...132.....1
.7....1...182..2002..6006..6006..2002...182.....1
...


MATHEMATICA

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 *)


CROSSREFS

Cf. A002114, A086645, A186430, A186433 (inverse).
Sequence in context: A051457 A174450 A166343 * A176489 A174039 A174148
Adjacent sequences: A186429 A186430 A186431 * A186433 A186434 A186435


KEYWORD

nonn,easy,tabl


AUTHOR

Peter Bala, Feb 22 2011


STATUS

approved



