A289398 -id:A289398 - cp's OEIS frontend

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

Jinyuan Wang, Feb 08 2019

nonn

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

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

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

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

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

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

A379596 a(n) is the least positive integer k for which k^2 + (k + n)^2 is a square.

Original entry on oeis.org

3, 6, 9, 12, 15, 18, 5, 24, 27, 30, 33, 36, 39, 10, 45, 48, 7, 54, 57, 60, 15, 66, 12, 72, 75, 78, 81, 20, 87, 90, 9, 96, 99, 14, 25, 108, 111, 114, 117, 120, 36, 30, 129, 132, 135, 24, 16, 144, 11, 150, 21, 156, 159, 162, 165, 40, 171, 174, 177, 180, 183, 18, 45

Offset: 1

Felix Huber, Feb 15 2025

a(n) is also the smallest short leg of a Pythagorean triangle where the difference between the two legs is n.

A289398(n) is the least integer m > n for which (n^2 + m^2)/2 is a square. This is equivalent to the least positive integer k for which (n^2 + (n + 2*k)^2)/2 = k^2 + (n + k)^2 is a square. From m = n + 2*k follows a(n) = (A289398(n) - n)/2.

			a(1) = 3 because 3^2 + (3 + 1)^2 = 5^2 and there is no smaller positive integer k than 3 with that property.
a(28) = 20 because 20^2 + (20 + 28)^2 = 52^2 and there is no smaller positive integer k than 20 with that property.

Felix Huber, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Pythagorean Triple

Cf. A089015, A009004, A120682, A132404, A289398, A379830.

A379596:=proc(n)
    local k;
    for k do
        if issqr(k^2+(k+n)^2) then
            return k
        fi
    od
end proc;
seq(A379596(n),n=1..63);

s={};Do[k=0;Until[IntegerQ[Sqrt[k^2+(k+n)^2]],k++];AppendTo[s,k],{n,63}];s (* James C. McMahon, Mar 02 2025 *)

a(n) = my(k=1); while (!issquare(k^2 + (k + n)^2), k++); k; \\ Michel Marcus, Feb 15 2025

from itertools import count
from sympy.ntheory.primetest import is_square
def A379596(n): return next(k for k in count(1) if is_square(k**2+(k+n)**2)) # Chai Wah Wu, Mar 02 2025

a(n) = (A289398(n) - n)/2.

cp's OEIS Frontend

A306236 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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