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%I A202917 #23 Mar 24 2017 00:33:24
%S A202917 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,
%T A202917 1,1,1,21,21,21,21,1,1
%N A202917 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.
%C A202917 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!.
%C A202917 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.
%C A202917 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.
%F A202917 A007814(C^(1)(n,m)) = A007814(C(n,m)).
%e A202917 Triangle begins
%e A202917 n/m.|..0.....1.....2.....3.....4.....5.....6.....7
%e A202917 ==================================================
%e A202917 .0..|..1
%e A202917 .1..|..1.....1
%e A202917 .2..|..1.....6.....1
%e A202917 .3..|..1.....1 ... 1  .....1
%e A202917 .4..|..1....60....10......60.....1
%e A202917 .5..|..1.....1....10......10.....1.....1
%e A202917 .6..|..1...126....21....1260....21...126.....1
%e A202917 .7..|..1.....1....21......21....21....21.....1.....1
%e A202917 .8..|
%t A202917 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 *)
%Y A202917 Cf. A175669, A053657, A202339, A202367, A202368, A202369
%K A202917 nonn,tabl
%O A202917 0,5
%A A202917 _Vladimir Shevelev_ and _Peter J. C. Moses_, Dec 26 2011

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