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

%I A306236 #43 Apr 10 2019 08:24:38
%S A306236 5,10,15,20,25,30,13,40,45,50,55,60,65,26,75,80,25,90,95,100,39,110,
%T A306236 37,120,125,130,135,52,145,150,41,160,165,50,65,180,185,190,195,200,
%U A306236 85,78,215,220,225,74,65,240,61,250,75,260,265,270,275,104,285,290
%N A306236 a(n) is the smallest integer m > n with integer j > m makes n^2, m^2 and j^2 an arithmetic progression.
%C A306236 a(n) and n have the same parity.
%C A306236 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 A306236 Conjecture: if gcd(k, a(k)) = 1, then k is a term in A058529.
%C A306236 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 A306236 a(n) = sqrt((n^2 + A289398(n)^2)/2).
%F A306236 For positive integer k, a(2*k^2 - 1) = 2*k^2 + 2*k + 1.
%F A306236 a(A003629(k)) = 5*A003629(k).
%F A306236 a(n) <= 5*n.
%F A306236 a(k*n) = k*a(n) for all k not in A058529. - _Jianing Song_, Feb 15 2019
%e A306236 a(1) = 5 because 1^2, 5^2 and 7^2 are an arithmetic progression.
%t A306236 Array[Block[{m = # + 2}, While[! IntegerQ@ Sqrt[2 m^2 - #^2], m += 2]; m] &, 58] (* _Michael De Vlieger_, Feb 15 2019 *)
%o A306236 (PARI) a(n) = {m=n+2; while(issquare(2*m^2-n^2)==0, m=m+2); m;}
%Y A306236 Cf. A003629, A058529, A289398 (integer j).
%K A306236 nonn
%O A306236 1,1
%A A306236 _Jinyuan Wang_, Feb 08 2019