A101083 Largest k such that the product (n+1)(n+2)...(n+k) has at least k distinct prime factors.

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, 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, 13, 12, 14, 13, 12, 13, 13, 14, 14, 13, 12, 11, 13, 12, 15, 14, 13, 14, 13, 13, 17, 16, 17

Offset

1,1

Comments

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.

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 an 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

References

See A059686

Links

T. D. Noe, Table of n, a(n) for n=1..1000

M. Waldschmidt, Open Diophantine problems, arXiv:math/0312440 [math.NT], 2003-2004, pages 6-7.

Example

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.

Mathematica

Table[k=2; While[Length[FactorInteger[Times@@Range[n0+1, n0+k]]]>=k, k++ ]; k-1, {n0, 100}]

Crossrefs

Cf. A059686 (Grimm numbers).

Sequence in context: A329526 A160371 A059686 * A373557 A097935 A188940

Adjacent sequences: A101080 A101081 A101082 * A101084 A101085 A101086

Keyword

easy,nonn

Author

T. D. Noe, Nov 30 2004

Status

approved