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%I #7 Sep 30 2017 11:47:37
%S 0,0,0,1,0,1,0,0,2,0,2,0,0,0,1,1,0,1,0,0,0,0,0,2,0,0,0,0,0,4,1,0,0,3,
%T 0,0,1,0,0,0,0,1,0,0,0,2,1,0,0,0,0,0,3,1,0,0,0,0,0,0,1,3,0,0,0,3,0,1,
%U 0,0,0,1,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,2,0,0,0,0,1,0,4,0,0,0,0,0,1
%N Consider the consecutive composite numbers between prime(n) and prime(n+1). Letting k=prime(n+1)-prime(n)-1, a(n) is the number of these numbers that have all primes factors less than k.
%C Grimm's conjecture states that for k consecutive composite numbers there are k distinct prime numbers such that each prime divides one of the composite numbers. As pointed out by Grimm, for a composite number c having a prime factor p>=k, we associate p with c. Hence there are only a(n) numbers remaining between prime(n) and prime(n+1) for which we need to associate a distinct prime. Puzzle 430 gives a heuristic algorithm for finding those a(n) primes. According to Puzzle 430, the largest known value of a(n) is 10, which occurs between the primes 31397 and 31469.
%H T. D. Noe, <a href="/A137678/b137678.txt">Table of n, a(n) for n=1..10000</a>
%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.
%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_430.htm">Puzzle 430: Grimm's Conjecture</a>
%e a(4)=1 because of the 3 numbers between 7 and 11 (8=2^3, 9=3^2, 10=2*5), only 8 has all of its prime factors less than 3. Similarly, a(9)=2 because of the 5 numbers between 23 and 29, only 24 and 27 have all prime factors less than 5.
%t Table[p1=Prime[i]; p2=Prime[i+1]; cnt=0; k=p2-p1-1; Do[If[FactorInteger[n][[ -1,1]]<k, cnt++ ], {n,p1+1,p2-1}]; cnt, {i,150}]
%o (PARI) a(n) = my(p = prime(n), q = nextprime(p+1));sum(c=p+1, q-1, vecmax(factor(c)[,1]) < q-p-1); \\ _Michel Marcus_, Sep 30 2017
%Y Cf. A059686, A101083.
%K nonn
%O 1,9
%A _T. D. Noe_, Feb 04 2008
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