A138588 a(n) = the least integer > n such that r(1)|a(n), r(2)|(a(n)+1), r(3)|(a(n)+2),... and r(n)|(a(n)+n-1), where (r(1),r(2),r(3),...,r(n)) is some permutation of (1,2,3,...,n).

2, 3, 4, 6, 6, 20, 24, 48, 48, 110, 110, 110, 243, 403, 402, 2504, 2352, 12219, 25200, 60458, 14256, 95760, 120120, 582090, 582096, 186120, 3299404, 11060250, 28648620, 376576202, 9469950, 832431604, 832431603, 962161203, 1403352722

Offset: 1

Author

Leroy Quet, May 13 2008

Keywords

• nonn

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).

Examples

			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

Cf. A071373, A140257, A140275.

Extensions

More terms from Don Reble, May 15 2008