A178473 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.

1, 1, 1, 1, 2, 1, 1, 273, 273, 1, 1, 68, 9282, 68, 1, 1, 55, 1870, 1870, 55, 1, 1, 546, 15015, 3740, 15015, 546, 1, 1, 29, 7917, 1595, 1595, 7917, 29, 1

Offset: 0

Author

Vladimir Shevelev and Peter J. C. Moses, Jan 02 2012

Keywords

• nonn
• tabl

Comments

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.

Examples

			Triangle begins
n/m.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1......1
.2..|..1......2......1
.3..|..1....273 ...273......1
.4..|..1.....68...9282.....68......1
.5..|..1.....55...1870...1870.....55......1
.6..|..1....546..15015...3740..15015....546....1
.7..|..1.....29...7917...1595...1595...7917...29.....1
.8..|
		

Crossrefs

Cf. A175669, A053657, A202339, A202367, A202368, A202369, A202917, A202941, A203484.

Formula

Conjecture. A007814(C^(4)(n,m)) = A007814(C(n,m)).