A385008 Integers k such that there exists an integer 0

4, 284, 1210, 2924, 4892, 5564, 6368, 9962, 10425, 10856, 13130, 14595, 18416, 28130, 29631, 35584, 53296, 53912, 64617, 66992, 67268, 71145, 76084, 86812, 87633, 88730, 100695, 102364, 104805, 122390, 123152, 124155, 139815, 147610, 153176, 165596, 168730, 176336, 180848

Offset: 1

Author

S. I. Dimitrov, Jun 15 2025

Keywords

• nonn

Comments

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)

Examples

			(1, 4) is such a pair because sigma(1)^2+sigma(4)^2 = 1^2+7^2 = 2*(1+4)^2.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..154 (terms <= 2500000)
• David A. Corneth, PARI program
• S. I. Dimitrov, Generalizations of amicable numbers, arXiv:2408.07387 [math.NT], 2024.

Crossrefs

Cf. A000203, A002046, A063990, A383484.

Programs

PARI

isok(k) = for (m=1, k-1, if (sigma(m)^2 + sigma(k)^2 == 2*(m+k)^2, return(m))); \\ Michel Marcus, Jun 15 2025

Extensions

a(17)-a(26) from Michel Marcus, Jun 15 2025

a(27)-a(39) from Michael S. Branicky, Jun 26 2025