|
|
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.
|
|
3
|
|
|
1, 1, 1, 1, 42, 1, 1, 5, 5, 1, 1, 1092, 130, 1092, 1, 1, 1, 26, 26, 1, 1, 1, 11970, 285, 62244, 285, 11970, 1, 1, 11, 3135, 627, 627, 3135, 11, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,5
|
|
COMMENTS
|
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.
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
Triangle begins
n/m.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1......1
.2..|..1.....42.....1
.3..|..1......5 ....5......1
.4..|..1...1092...130...1092.....1
.5..|..1......1....26.....26.....1......1
.6..|..1..11970...285..62244...285..11970....1
.7..|..1.....11..3135....627...627...3135...11.....1
.8..|
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|