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a(n) is the smallest integer m > n with integer j > m makes n^2, m^2 and j^2 an arithmetic progression.
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%I #43 Apr 10 2019 08:24:38

%S 5,10,15,20,25,30,13,40,45,50,55,60,65,26,75,80,25,90,95,100,39,110,

%T 37,120,125,130,135,52,145,150,41,160,165,50,65,180,185,190,195,200,

%U 85,78,215,220,225,74,65,240,61,250,75,260,265,270,275,104,285,290

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

%C a(n) and n have the same parity.

%C 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, ...

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

%C 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

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

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

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

%F a(n) <= 5*n.

%F a(k*n) = k*a(n) for all k not in A058529. - _Jianing Song_, Feb 15 2019

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

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

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

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

%K nonn

%O 1,1

%A _Jinyuan Wang_, Feb 08 2019