|
%I #11 Mar 24 2017 22:11:19
%S 2,3,4,4,3,5,4,6,6,7,6,7,6,5,8,8,7,8,7,7,8,7,6,7,9,8,8,11,10,11,10,11,
%T 11,10,12,12,11,10,9,9,8,11,10,9,10,9,8,11,13,13,12,11,10,11,14,15,14,
%U 13,12,14,13,12,13,13,14,14,13,12,11,13,12,15,14,13,14,13,13,17,16,17
%N Largest k such that the product (n+1)(n+2)...(n+k) has at least k distinct prime factors.
%C This sequence is based on a slightly weaker, but still unproved, version of Grimm's conjecture: If there is no prime in the interval [n+1, n+k], then the product (n+1)(n+2)...(n+k) has at least k distinct prime divisors. We have a(n) >= A059686(n), with the two sequences first differing at n=70. Computing a(n) is much faster than computing A059686.
%C It seems that Grimm's conjecture could have another (but not a weak) form: let p(1)...p(i) be a subset of prime numbers such that while n is integer, 0 < n < i, for any n, p(n) < p(n+1). Then there exists such sequence c(1)...c(i) where each term is a composite number, c(n+1) = c(n) + 1, and c(n) == 0 (mod p(n)). - _Sergey Pavlov_, Mar 21 2017
%D See A059686
%H T. D. Noe, <a href="/A101083/b101083.txt">Table of n, a(n) for n=1..1000</a>
%H M. Waldschmidt, <a href="https://arxiv.org/abs/math/0312440">Open Diophantine problems</a>, arXiv:math/0312440 [math.NT], 2003-2004, pages 6-7.
%e a(6) = 5 because 7*8*9*10*11 has 5 prime factors and 7*8*9*10*11*12 does not have 6 prime factors.
%t Table[k=2; While[Length[FactorInteger[Times@@Range[n0+1, n0+k]]]>=k, k++ ]; k-1, {n0, 100}]
%Y Cf. A059686 (Grimm numbers).
%K easy,nonn
%O 1,1
%A _T. D. Noe_, Nov 30 2004
|
