A289398 Least integer m > n such that (n^2 + m^2)/2 is a square.

7, 14, 21, 28, 35, 42, 17, 56, 63, 70, 77, 84, 91, 34, 105, 112, 31, 126, 133, 140, 51, 154, 47, 168, 175, 182, 189, 68, 203, 210, 49, 224, 231, 62, 85, 252, 259, 266, 273, 280, 113, 102, 301, 308, 315, 94, 79, 336, 71, 350, 93, 364, 371, 378, 385, 136, 399, 406, 413, 420

Offset

1,1

Comments

From first 100 terms, in 65 cases a(n) = 7*n. In general, a(n) <= 7*n.

From Robert Israel, Jul 07 2017: (Start)

For any p in A042999, a(n) == 0 (mod p) if and only if n == 0 (mod p), with a(p*k) = p*a(k).

Thus if n = m*r where all prime factors of m are in A042999, a(n) = m*a(r).

In particular, if all prime factors of n are in A042999, then a(n) = 7*n.

Conjecture: this is "if and only if".

(End)

Alternatively: A306236(n) is the smallest integer m > n with integer j > m that makes n^2, m^2 and j^2 an arithmetic progression. This is the sequence of j. - Jinyuan Wang, Feb 09 2019.

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

a(1)=7: (1^2 + 7^2)/2 = 5^2;
a(7)=17: (7^2 + 17^2)/2 = 5^2.

Maple

f:= proc(n) local m; for m from n+2 by 2 do if issqr((n^2+m^2)/2) then return m fi od end proc:
map(f, [$1..100]); # Robert Israel, Jul 07 2017

Mathematica

n=0; Table[n++; m=n+1; While[!IntegerQ[Sqrt[(n^2+m^2)/2]], m++]; m, {100}]

Prog

(PARI) a(n) = my(m=n+1); while(!issquare((n^2+m^2)/2), m++); m; \\ Michel Marcus, Jul 07 2017

Crossrefs

Cf. A042999, A075892, A306236.

Sequence in context: A282147 A283444 A109048 * A044892 A004959 A020646

Adjacent sequences: A289395 A289396 A289397 * A289399 A289400 A289401

Keyword

nonn,look

Author

Zak Seidov, Jul 05 2017

Status

approved