|
%I #12 May 01 2014 02:44:23
%S 1,1,1,1,2,1,1,273,273,1,1,68,9282,68,1,1,55,1870,1870,55,1,1,546,
%T 15015,3740,15015,546,1,1,29,7917,1595,1595,7917,29,1
%N For n>=0, let n!^(4) = A202369(n+1) and, for 0<=m<=n, C^(4)(n,m) = n!^(4)/(m!^(4)*(n-m)!^(4)). The sequence gives triangle of numbers C^(4)(n,m) with rows of length n+1.
%C Conjecture. If p is prime of the form 4*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 4*k+3, then the (2*k+1)-th row contains two 1's and 2*k numbers multiple of p.
%F Conjecture. A007814(C^(4)(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......2......1
%e .3..|..1....273 ...273......1
%e .4..|..1.....68...9282.....68......1
%e .5..|..1.....55...1870...1870.....55......1
%e .6..|..1....546..15015...3740..15015....546....1
%e .7..|..1.....29...7917...1595...1595...7917...29.....1
%e .8..|
%Y Cf. A175669, A053657, A202339, A202367, A202368, A202369, A202917, A202941, A203484.
%K nonn,tabl
%O 0,5
%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Jan 02 2012
|
