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For n>=0, let n!^(2)=A202367(n+1) and, for 0<=m<=n, C^(2)(n,m)=n!^(2)/(m!^(2)*(n-m)!^(2)). The sequence gives triangle of numbers C^(2)(n,m) with rows of length n+1.
9

%I #15 Oct 24 2024 05:40:31

%S 1,1,1,1,10,1,1,21,21,1,1,20,42,20,1,1,11,22,22,11,1,1,2730,3003,2860,

%T 3003,2730,1,1,1,273,143,143,273,1,1

%N For n>=0, let n!^(2)=A202367(n+1) and, for 0<=m<=n, C^(2)(n,m)=n!^(2)/(m!^(2)*(n-m)!^(2)). The sequence gives triangle of numbers C^(2)(n,m) with rows of length n+1.

%C Conjecture. If p is an odd prime, then the ((p-1)/2)-th row contains two 1's and (p-3)/2 numbers multiple of p.

%C See also comments in A175669 and A202917.

%F If conjectural formula in A202367 is true, then A007814(C^(2)(n,m)) =A007814(C(n,m)).

%e Triangle begins

%e n/m.|..0.....1.....2.....3.....4.....5.....6.....7

%e ==================================================

%e .0..|..1

%e .1..|..1.....1

%e .2..|..1....10.....1

%e .3..|..1....21 ...21.....1

%e .4..|..1....20....42....20.....1

%e .5..|..1....11....22....22....11.....1

%e .6..|..1..2730..3003..2860..3003..2730.....1

%e .7..|..1.....1...273...143...143...273.....1.....1

%e .8..|

%Y Cf. A175669, A053657, A202339, A202367, A202368, A202369, A202917

%K nonn,tabl

%O 0,5

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Dec 26 2011