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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.
3

%I #6 May 01 2014 02:44:24

%S 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 62244,285,11970,1,1,11,3135,627,627,3135,11,1

%N 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 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 Conjecture. A007814(C^(3)(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.....42.....1

%e .3..|..1......5 ....5......1

%e .4..|..1...1092...130...1092.....1

%e .5..|..1......1....26.....26.....1......1

%e .6..|..1..11970...285..62244...285..11970....1

%e .7..|..1.....11..3135....627...627...3135...11.....1

%e .8..|

%Y Cf. A175669, A053657, A202339, A202367, A202368, A202369, A202917, A202941.

%K nonn,tabl

%O 0,5

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Jan 02 2012