|
|
A164798
|
|
a(n) = the smallest integer >= n such that a(n)!/(n-1)! is divisible by every prime from 2 to the largest prime divisor of a(n)!/(n-1)!. (a(1)=1.)
|
|
3
|
|
|
1, 2, 4, 4, 6, 6, 10, 8, 10, 14, 15, 12, 22, 15, 16, 16, 26, 18, 34, 21, 38, 38, 38, 24, 46, 46, 46, 46, 46, 30, 58, 32, 62, 62, 36, 36, 62, 74, 74, 74, 74, 82, 82, 86, 86, 86, 86, 48, 94, 94, 94, 94, 94, 54, 106, 106, 106, 106, 106, 60, 118, 122, 66, 64, 122, 122, 122, 134
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
|
|
LINKS
|
|
|
EXAMPLE
|
Consider the products of consecutive integers, m!/9!, m >= 10. First, 10 is divisible by 2 and 5, but there is a prime gap since 3 is missing from the factorization. 10*11 is divisible by 2, 5, and 11, but 3 and 7 are missing. 10*11*12 is divisible by 2, 3, 5, and 11, but 7 is missing. 10*11*12*13 is divisible by all primes up to 13, except 7. But 10*11*12*13*14 is indeed divisible by every prime from 2 to 13. So a(10) = 14.
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|