2,3

It appears that this is the sequence of k's for A110357. - Michel Marcus, Aug 16 2019

If n-m = s, then n = s+m and n-m | n^2+m^2 is equivalent to s | (s^2 + 2*s*m + 2*m^2). So n-m | n^2+m^2 is equivalent to n-m | 2*m^2. If n-k = s, then n = s+k and n-k | n*(n+k) is equivalent to s | (s^2 + 3*s*k + 2*k^2). So n-k | n*(n+k) is equivalent to n-k | 2*k^2. Therefore n-m | n^2+m^2 is equivalent to n-k | n*(n+k) and the k's from A110357 and the m's from this sequence are the same. - Bob Andriesse, Dec 26 2022

Let n-m = s; then m = n-s and n-m | n^2 + m^2 is equivalent to s | n^2 + (n-s)^2 or s | 2*n^2. If n is an odd prime, s must be 2. So if n is an odd prime, a(n) = m = n-2. Examples: a(7) = 5, a(11) = 9. - Bob Andriesse, Jul 13 2023

Clark Kimberling, Table of n, a(n) for n = 2..1000

a(n) = H(n, A110357(n)) - n where H is the harmonic mean. - Bob Andriesse, Jan 03 2023

Write x#y if x|y is false; then 7#65, 6#68, 5#73, 4|80, so a(8) = 4.

For n = 11, A110357(11) = 110 and a(11) = H(11, 110) - 11 = 20 - 11 = 9. - Bob Andriesse, Jan 03 2023

Table[m = 1; While[! Divisible[n^2+m^2, n-m], m++]; m, {n, 2, 100}]

(PARI) a(n) = my(m=1); while(denominator((n^2+m^2)/(n-m)) != 1, m++); m; \\ Michel Marcus, Aug 16 2019

(Python)

from sympy.abc import x, y

from sympy.solvers.diophantine.diophantine import diop_quadratic

def A214750(n): return min(int(x) for x, y in diop_quadratic(n*(n-y)+x*(y+x)) if x>0) # Chai Wah Wu, Oct 06 2023

Cf. A110357, A214749.

Sequence in context: A263216 A141663 A011153 * A317585 A132226 A197702

Adjacent sequences: A214747 A214748 A214749 * A214751 A214752 A214753

nonn,easy

Clark Kimberling, Jul 29 2012

approved