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A203484 For n>=0, let n!^(3) = A202368(n+1) and, for 0<=m<=n, C^(3)(n,m) = n!^(3)/(m!^(3)*(n-m)!^(3)). The sequence gives triangle of numbers C^(3)(n,m) with rows of length n+1.

%I A203484 #6 May 01 2014 02:44:24
%S A203484 1,1,1,1,42,1,1,5,5,1,1,1092,130,1092,1,1,1,26,26,1,1,1,11970,285,
%T A203484 62244,285,11970,1,1,11,3135,627,627,3135,11,1
%N A203484 For n>=0, let n!^(3) = A202368(n+1) and, for 0<=m<=n, C^(3)(n,m) = n!^(3)/(m!^(3)*(n-m)!^(3)). The sequence gives triangle of numbers C^(3)(n,m) with rows of length n+1.
%C A203484 Conjecture. If p is prime of the form 3*k+1, then the k-th row contains two 1's and k-1 numbers multiple of p; if p is prime of the form 3*k+2, then the (2*k+1)-th row contains two 1's and 2*k numbers multiple of p.
%F A203484 Conjecture. A007814(C^(3)(n,m)) = A007814(C(n,m)).
%e A203484 Triangle begins
%e A203484 n/m.|..0.....1.....2.....3.....4.....5.....6.....7
%e A203484 ==================================================
%e A203484 .0..|..1
%e A203484 .1..|..1......1
%e A203484 .2..|..1.....42.....1
%e A203484 .3..|..1......5 ....5......1
%e A203484 .4..|..1...1092...130...1092.....1
%e A203484 .5..|..1......1....26.....26.....1......1
%e A203484 .6..|..1..11970...285..62244...285..11970....1
%e A203484 .7..|..1.....11..3135....627...627...3135...11.....1
%e A203484 .8..|
%Y A203484 Cf. A175669, A053657, A202339, A202367, A202368, A202369, A202917, A202941.
%K A203484 nonn,tabl
%O A203484 0,5
%A A203484 _Vladimir Shevelev_ and _Peter J. C. Moses_, Jan 02 2012