A306236 a(n) is the smallest integer m > n with integer j > m makes n^2, m^2 and j^2 an arithmetic progression.

5, 10, 15, 20, 25, 30, 13, 40, 45, 50, 55, 60, 65, 26, 75, 80, 25, 90, 95, 100, 39, 110, 37, 120, 125, 130, 135, 52, 145, 150, 41, 160, 165, 50, 65, 180, 185, 190, 195, 200, 85, 78, 215, 220, 225, 74, 65, 240, 61, 250, 75, 260, 265, 270, 275, 104, 285, 290

Offset: 1

Author

Jinyuan Wang, Feb 08 2019

Keywords

• nonn

Comments

a(n) and n have the same parity.

If k is a term in A058529, gcd(k, a(k)) does not necessarily equal 1. For example, k = 217, 289, 343, 497, 529, 553, 679, 889, 961, 1127, ...

Conjecture: if gcd(k, a(k)) = 1, then k is a term in A058529.

Proof: if k is not in A058529, then k either is even or has a prime factor p == 3, 5 (mod 8). If k is even, then a(k) is also even, so 2 divides gcd(k, a(k)). If k has a prime factor p == 3, 5 (mod 8), then 2*m^2 == j^2 (mod p), 2^((p-1)/2)*m^(p-1) == -m^(p-1) == j^(p-1) (mod p), so m and j must both be multiples of p. As a result, p divides gcd(k, a(k)). - Jianing Song, Feb 09 2019

Examples

			a(1) = 5 because 1^2, 5^2 and 7^2 are an arithmetic progression.
		

Crossrefs

Cf. A003629, A058529, A289398 (integer j).

Programs

Mathematica

Array[Block[{m = # + 2}, While[! IntegerQ@ Sqrt[2 m^2 - #^2], m += 2]; m] &, 58] (* Michael De Vlieger, Feb 15 2019 *)

PARI

a(n) = {m=n+2; while(issquare(2*m^2-n^2)==0, m=m+2); m;}

Formula

a(n) = sqrt((n^2 + A289398(n)^2)/2).

For positive integer k, a(2*k^2 - 1) = 2*k^2 + 2*k + 1.

a(A003629(k)) = 5*A003629(k).

a(n) <= 5*n.

a(k*n) = k*a(n) for all k not in A058529. - Jianing Song, Feb 15 2019