|
| |
|
|
A374959
|
|
a(n) is the least k such that binomial(A349958(n), k) is a multiple of n.
|
|
4
|
|
|
|
0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 3, 2, 4, 1, 3, 1, 2, 1, 2, 1, 3, 1, 1, 3, 2, 3, 2, 1, 4, 2, 3, 1, 3, 1, 3, 2, 8, 1, 5, 1, 2, 2, 6, 1, 4, 2, 3, 2, 2, 1, 3, 1, 2, 4, 1, 4, 4, 1, 2, 6, 4, 1, 5, 1, 2, 2, 4, 5, 2, 1, 3, 1, 2, 1, 3, 3, 4, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,6
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
For n = 12: the first multiple of 12 in Pascal's triangle appears in row 9; this row contains: 1, 9, 36, 84, 126, 126, 84, 36, 9, 1; the first multiple of 12 (36), appears at (0-based) index 2; so a(12) = 2.
|
|
|
PROG
|
(PARI) a(n) = { my (r = [1 % n], j); for (i = 0, oo, if (vecmin(r, &j)==0, return (j-1), r = (concat(0, r) + concat(r, 0)) % n; ); ); }
(Python)
from math import comb
def A374959(n): return next(k for j in range(n+1) for k in range(j+1) if not (comb(j, k) % n)) # Chai Wah Wu, Jul 30 2024
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
