|
|
A113926
|
|
Triangle where a(1,1)=1. a(n,m) = GCD of m and sum of terms in row (n-1).
|
|
0
|
|
|
1, 1, 1, 1, 2, 1, 1, 2, 1, 4, 1, 2, 1, 4, 1, 1, 1, 3, 1, 1, 3, 1, 2, 1, 2, 5, 2, 1, 1, 2, 1, 2, 1, 2, 7, 2, 1, 2, 3, 2, 1, 6, 1, 2, 9, 1, 1, 3, 1, 1, 3, 1, 1, 9, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 11, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 3, 2, 1, 6, 1, 2, 9, 2, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,5
|
|
COMMENTS
|
The sum of the terms in the n-th row is A100675(n+1).
|
|
LINKS
|
|
|
EXAMPLE
|
Row 6 is[1,1,3,1,1,3], which has a sum of 10.
So row 7 is [GCD(1,10),GCD(2,10),GCD(3,10),GCD(4,10),GCD(5,10),GCD(6,10),GCD(7,10)]
= [1,2,1,2,5,2,1].
|
|
MAPLE
|
T[1, 1]:=1: for n from 1 to 15 do for k from 1 to n do T[n, k]:=gcd(k, add(T[n-1, j], j=1..n-1)) od: od: for n from 1 to 15 do seq(T[n, k], k=1..n) od; # yields sequence in triangular form - Emeric Deutsch, Feb 13 2006
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|