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A059751
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.
3
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
OFFSET
1,1
COMMENTS
Erdős 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. - T. D. Noe, Feb 18 2009
REFERENCES
D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section XII.15, p. 438.
LINKS
Paul Erdős 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, Feb 17 2009]
C. A. Grimm, A conjecture on consecutive composite numbers, Amer. Math. Monthly, 76 (1969), 1126-1128.
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.
CROSSREFS
Sequence in context: A092874 A371322 A198109 * A019859 A188736 A265304
KEYWORD
nonn,easy,nice
AUTHOR
N. J. A. Sloane, Feb 11 2001
EXTENSIONS
More terms from Naohiro Nomoto, Mar 03 2001
Extended by T. D. Noe, Feb 17 2009
STATUS
approved