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A059751
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Grimm numbers (2): a(n) = largest k so that for each composite m in {n+1, n+2, ..., n+k} there corresponds a different divisor d_m with 1 < d_m < m.
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3
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7, 6, 5, 10, 9, 14, 13, 12, 15, 22, 21, 20, 19, 20, 23, 22, 21, 20, 19, 18, 27, 26, 25, 24, 29, 30, 29, 28, 27, 26, 25, 24, 31, 34, 41, 40, 39, 46, 47, 46, 45, 44, 43, 42, 41, 44, 43, 42, 43, 42, 43, 42, 41, 40, 55, 54, 53, 60, 59, 58, 57, 58, 57, 56, 57, 56, 55, 54, 59, 58, 57
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| Comment from T. D. Noe, Feb 18 2009: Erdos and Pomerance conjectured that the number n+a(n)+1, which "blocks" a(n) from becoming larger, is always an odd semiprime. They verified this conjecture up to n=492 and proved it for large n. The numbers n at which n+a(n)+1 increases also appear to be semiprimes.
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REFERENCES
| C. A. Grimm, A conjecture on consecutive composite numbers, Amer. Math. Monthly, 76 (1969), 1126-1128.
D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section XII.15, p. 438.
Paul Erdos and Carl Pomerance, An analogue of Grimm's problem of finding distinct prime factors of consecutive integers, Util. Math. 24 (1983), 45-65. [From T. D. Noe (noe(AT)sspectra.com), Feb 17 2009]
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..1000
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EXAMPLE
| For n=4 we look at the sequence {5, 6, 7, ... } and we must choose distinct proper divisors for as many composites as we can. We can choose 2 for 6, 4 for 8, 3 for 9, 5 for 10, 6 for 12 and 7 for 14, but now all the proper divisors of 15 have appeared, so we stop and a(4) = 14-4 = 10.
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CROSSREFS
| Cf. A059686, A059752.
Sequence in context: A092874 A198109 A015791 * A019859 A188736 A102769
Adjacent sequences: A059748 A059749 A059750 * A059752 A059753 A059754
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KEYWORD
| nonn,easy,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Feb 11 2001
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EXTENSIONS
| More terms from Naohiro Nomoto (6284968128(AT)geocities.co.jp), Mar 03 2001
Extended by T. D. Noe (noe(AT)sspectra.com), Feb 17 2009
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