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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 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 LCM of denominators of coefficients of 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 S(l,m) are Stirling numbers of the second kind.The sequence of polynomials {Q^(1)_n(x)} includes to the family of sequences of polynomials {{Q^(r)_n}}_(r>=0) described in comment to A175669. In particular, LCM of denominators of coefficients of Q^(0)_n(x) is n!
2) This triangle differs from triangle A186430 which defined according to theory of factorials over sets by Bhargava. Unfortunately, this theory has not 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 odd prime, then the (p-1)-th row contains two 1's and p-2 numbers multiple of p. For a conjectural generalization, see comment in A175669.
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LINKS
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Table of n, a(n) for n=0..35.
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FORMULA
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A007814(C^(1)(n,m))=A007814(C(n,m)).
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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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CROSSREFS
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Cf. A175669, A053657, A202339, A202367, A202368, A202369
Sequence in context: A087253 A197686 A143532 * A167155 A080219 A040037
Adjacent sequences: A202914 A202915 A202916 * A202918 A202919 A202920
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KEYWORD
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nonn,tabl
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AUTHOR
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Vladimir Shevelev and Peter Moses, Dec 26 2011
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STATUS
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approved
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