cp's OEIS Frontend

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).

%I A138588 #9 Apr 08 2019 01:32:01
%S A138588 2,3,4,6,6,20,24,48,48,110,110,110,243,403,402,2504,2352,12219,25200,
%T A138588 60458,14256,95760,120120,582090,582096,186120,3299404,11060250,
%U A138588 28648620,376576202,9469950,832431604,832431603,962161203,1403352722
%N 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).
%C A138588 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).
%e A138588 Example, n = 7:
%e A138588 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.
%e A138588 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.)
%e A138588 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.
%Y A138588 Cf. A071373, A140257, A140275.
%K A138588 nonn
%O A138588 1,1
%A A138588 _Leroy Quet_, May 13 2008
%E A138588 More terms from _Don Reble_, May 15 2008