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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.
1
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
1,1
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
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
EXAMPLE
a(1) = 5 because 1^2, 5^2 and 7^2 are an arithmetic progression.
MATHEMATICA
Array[Block[{m = # + 2}, While[! IntegerQ@ Sqrt[2 m^2 - #^2], m += 2]; m] &, 58] (* Michael De Vlieger, Feb 15 2019 *)
PROG
(PARI) a(n) = {m=n+2; while(issquare(2*m^2-n^2)==0, m=m+2); m; }
CROSSREFS
Cf. A003629, A058529, A289398 (integer j).
Sequence in context: A313730 A313731 A313732 * A297305 A182340 A313733
KEYWORD
nonn
AUTHOR
Jinyuan Wang, Feb 08 2019
STATUS
approved