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Length of longest run of consecutive even integers having exactly n divisors.
1

%I #16 Apr 06 2019 04:11:01

%S 0,1,1,3,1,2,1,3,1,1,1,7,1,1,1,3,1,2,1,3,1,1,1

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

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

%C 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.

%C 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.

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

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

%Y Cf. A325117.

%K nonn,more,hard

%O 1,4

%A _David Wasserman_, Mar 27 2019