A059751 - OEIS

%I #21 Nov 20 2017 03:30:47

%S 7,6,5,10,9,14,13,12,15,22,21,20,19,20,23,22,21,20,19,18,27,26,25,24,

%T 29,30,29,28,27,26,25,24,31,34,41,40,39,46,47,46,45,44,43,42,41,44,43,

%U 42,43,42,43,42,41,40,55,54,53,60,59,58,57,58,57,56,57,56,55,54,59,58,57

%N 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.

%C 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

%D D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section XII.15, p. 438.

%H T. D. Noe, <a href="/A059751/b059751.txt">Table of n, a(n) for n = 1..1000</a>

%H Paul Erdős and Carl Pomerance, <a href="https://math.dartmouth.edu/~carlp/PDF/38.pdf">An analogue of Grimm's problem of finding distinct prime factors of consecutive integers</a>, Util. Math. 24 (1983), 45-65. [From _T. D. Noe_, Feb 17 2009]

%H C. A. Grimm, <a href="http://www.jstor.org/stable/2317188">A conjecture on consecutive composite numbers</a>, Amer. Math. Monthly, 76 (1969), 1126-1128.

%e 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.

%Y Cf. A059686, A059752.

%K nonn,easy,nice

%O 1,1

%A _N. J. A. Sloane_, Feb 11 2001

%E More terms from _Naohiro Nomoto_, Mar 03 2001

%E Extended by _T. D. Noe_, Feb 17 2009