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A202917
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For n >= 0, let n!^(1) = A053657(n+1) and, for 0 <= m <= n, C^(1)(n,m) = n!^(1)/(m!^(1)*(n-m)!^(1)). The sequence gives a triangle of numbers C^(1)(n,m) with rows of length n+1.
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10
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1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 60, 10, 60, 1, 1, 1, 10, 10, 1, 1, 1, 126, 21, 1260, 21, 126, 1, 1, 1, 21, 21, 21, 21, 1, 1
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OFFSET
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0,5
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COMMENTS
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1) Note that A053657(n+1) is the LCM of the denominators of the coefficients of the polynomials Q^(1)_n(x) which, for integer x=k, are defined by the recursion Q^(1)_0(x)=1, for n>=1, Q^(1)_n(x) = Sum_{i=1..k} i*Q^(1)_(n-1)(i). Also note that Q^(1)_n(k) = S(k+n,k), where the numbers S(l,m) are Stirling numbers of the second kind. The sequence of polynomials {Q^(1)_n(x)} includes the family of sequences of polynomials {{Q^(r)_n}}_(r>=0) described in a comment at A175669. In particular, the LCM of the denominators of the coefficients of Q^(0)_n(x) is n!.
2) This triangle differs from triangle A186430 which is defined according to the theory of factorials over sets by Bhargava. Unfortunately, this theory does not have a conversion theorem. Therefore it is not known if there is a set A such that n!^(1) = n!_A in the Bhargava sense.
3) If p is an odd prime, then the (p-1)-th row contains two 1's and p-2 numbers that are multiples of p. For a conjectural generalization, see comment in A175669.
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LINKS
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FORMULA
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EXAMPLE
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Triangle begins
n/m.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1.....6.....1
.3..|..1.....1 ... 1 .....1
.4..|..1....60....10......60.....1
.5..|..1.....1....10......10.....1.....1
.6..|..1...126....21....1260....21...126.....1
.7..|..1.....1....21......21....21....21.....1.....1
.8..|
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MATHEMATICA
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A053657[n_] := Product[p^Sum[Floor[(n-1)/((p-1) p^k)], {k, 0, n}], {p, Prime[Range[n]]}]; f1[n_] := A053657[n+1]; C1[n_, m_] := f1[n]/(f1[m] * f1[n-m]); Table[C1[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2016 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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