OFFSET
1,1
COMMENTS
It is easy to see that every term of this sequence exists, because the stretch of n terms, n!-n to n!-1, is such that n|(n!-n), (n-1)|(n!-n+1),...,2|(n!-2), 1|(n!-1).
EXAMPLE
Example, n = 7:
For all stretches of 7 consecutive integers, with the least integer m in each stretch such that m >=8 and m <= 19, there are at least 2 primes (each > 7) in the stretch. Now both primes cannot be divided by any positive integer <= 7 except 1. But there is only one 1 in the permutation (r(1),r(2),...,r(7)). So a(7) is > 19.
If the least integer in the stretch of 7 consecutive integers is 20, 21, or 22, then there is only one prime in the stretch, but there are two integers, 22 and 26, that aren't divisible by any integer <= 7 except 1 and 2. (And there is already a prime, 23, that needs to be divided by 1.)
So a(7) is > 22. If the least integer in the stretch of 7 consecutive integers is 23, then there are 2 primes in the stretch. But if the smallest integer of the stretch is 24, then we have 4|24, 5|25, 2|26, 3|27, 7|28, 1|29 and 6|30. And the sequence of 7 divisors (4,5,2,3,7,1,6) is a permutation of (1,2,3,4,5,6,7). So a(7) = 24.
CROSSREFS
KEYWORD
nonn
AUTHOR
Leroy Quet, May 13 2008
EXTENSIONS
More terms from Don Reble, May 15 2008
STATUS
approved