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A138588
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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).
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1
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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
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OFFSET
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1,1
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COMMENTS
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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).
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LINKS
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EXAMPLE
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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.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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