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 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 (list; table; graph; refs; listen; history; text; internal format)
 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*(n-k)!_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^2 - k^2), with the convention 0!_S = 1. This should be compared with n! = Product_{k=0..n} (n - k). 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 M. Bhargava, The factorial function and generalizations, Amer. Math. Monthly, 107 (2000), 783-799. 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. 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

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Last modified September 25 22:55 EDT 2020. Contains 337346 sequences. (Running on oeis4.)