cp's OEIS Frontend

The numbers m and k form a GM-amicable pair. See Dimitrov link.

The numbers m and k form a HM(2,1)-amicable pair (HM = harmonic mean). See Dimitrov link.

The numbers m and k form a WPM(2)-amicable pair (WPM = weighted power mean). See Dimitrov link.

The numbers m and k form a PM(2,2)-amicable pair (PM = Power Mean). See Dimitrov link.

An amicable pair forms a PM(2,2)-amicable pair, so the larger member of an amicable pair A002046 is a term of this sequence.

From David A. Corneth, Jun 29 2025: (Start)

Terms <= 2500000 are 2, 3 or 4 (mod 6). Are there any terms from a different residue class?

m > ceiling(sqrt(2*k^2 + sigma(k)^2) - 2*k).

Proof: m + sigma(k)^2 < sigma(m)^2 + sigma(k)^2 = 2*(m+k)^2.

Solving m + sigma(k)^2 < 2*(m+k)^2 gives the desired result.

Also 8*k^2 > sigma(k)^2.

Proof: sigma(k)^2 < sigma(m)^2 + sigma(k)^2 = 2*(m+k)^2 < 2*(k+k)^2 = 8*k^2.

Combining the two we have 2*k^2 < sigma(k)^2 < 8*k^2.

In a search it helps to choose an odd prime p and then classify numbers m to points (m mod p, sigma(m) mod p).

Then if sigma(m)^2 + sigma(k)^2 = 2*(m+k)^2 then sigma(m)^2 == (2*(m+k)^2 - sigma(k)^2) (mod p).

for 0 <= m <= p-1 assume the equivalence class of m (mod p) which would give an equivalence class of sigma(m) mod p and reduce the numbers to be checked. (End)

The most interesting question that arises here is whether there exist such pairs for which sigma(m) = sigma(k), which would imply that sigma(m)^2 = sigma(k)^2 = m^2+k^2. None have been found for m < k <= 10^8.