A186432 Triangle associated with the set S of squares {0,1,4,9,16,...}.
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
Offset: 0
Examples
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 ...
Links
- M. Bhargava, The factorial function and generalizations, Amer. Math. Monthly, 107 (2000), 783-799.
Programs
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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 *)
Formula
TABLE ENTRIES
T(n,k) = n!_S/(k!_S*(n-k)!_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*n-1), counting the preferential arrangements on (2*n - 1) labeled 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*n-1), which gives the number of preferential arrangements on (2*n-1) labeled elements having less than or equal to three ranks.
Comments