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A202941 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.

%I A202941 #15 Oct 24 2024 05:40:31
%S A202941 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 A202941 3003,2730,1,1,1,273,143,143,273,1,1
%N A202941 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 A202941 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 A202941 See also comments in A175669 and A202917.
%F A202941 If conjectural formula in A202367 is true, then A007814(C^(2)(n,m)) =A007814(C(n,m)).
%e A202941 Triangle begins
%e A202941 n/m.|..0.....1.....2.....3.....4.....5.....6.....7
%e A202941 ==================================================
%e A202941 .0..|..1
%e A202941 .1..|..1.....1
%e A202941 .2..|..1....10.....1
%e A202941 .3..|..1....21 ...21.....1
%e A202941 .4..|..1....20....42....20.....1
%e A202941 .5..|..1....11....22....22....11.....1
%e A202941 .6..|..1..2730..3003..2860..3003..2730.....1
%e A202941 .7..|..1.....1...273...143...143...273.....1.....1
%e A202941 .8..|
%Y A202941 Cf. A175669, A053657, A202339, A202367, A202368, A202369, A202917
%K A202941 nonn,tabl
%O A202941 0,5
%A A202941 _Vladimir Shevelev_ and _Peter J. C. Moses_, Dec 26 2011