A325116 Length of longest run of consecutive even integers having exactly n divisors.

0, 1, 1, 3, 1, 2, 1, 3, 1, 1, 1, 7, 1, 1, 1, 3, 1, 2, 1, 3, 1, 1, 1

Offset

1,4

Comments

The start of the first run of exactly k consecutive even integers having exactly n divisors is A325117(n,k).

If m does not divide n, then a(n) < 2^(m-1). Proof: 2^(m-1) consecutive even integers must include one congruent to 2^(m-1) mod 2^m, and the number of divisors of this one is a multiple of m.

For m > 2, if 2*(m-1) does not divide n, then a(n) < 2^(m-1). Proof: 2^(m-1) consecutive even integers must include two congruent to 2^(m-2) mod 2^(m-1). These are 2^(m-2)*x and 2^(m-2)*(x+2) for some odd x, and x and x+2 each have n/(m-1) divisors. If 2*(m-1) does not divide n, then n/(m-1) is odd. Any number with an odd number of divisors is a square, so x and x+2 are squares, but squares cannot differ by 2.

14 <= a(24) <= 15. Dickson's conjecture implies a(24)=15.

Links

Table of n, a(n) for n=1..23.

Crossrefs

Cf. A119479 (with consecutive integers), A319045 (with consecutive odd integers).

Cf. A325117.

Sequence in context: A332929 A361019 A157520 * A227339 A030777 A353375

Adjacent sequences: A325113 A325114 A325115 * A325117 A325118 A325119

Keyword

nonn,more,hard

Author

David Wasserman, Mar 27 2019

Status

approved