A383484 Integers k such that there exists an integer 0

3, 33, 69, 77, 133, 161, 235, 267, 287, 321, 385, 699, 715, 1235, 1379, 1437, 1529, 1595, 1653, 1719, 2047, 2233, 2241, 2569, 2727, 2829, 3237, 3269, 3999, 4585, 4683, 4911, 5075, 5163, 5215, 5497, 5667, 5691, 7085, 7089, 7587, 7761, 7797, 7945, 8259, 9159, 9659, 10653

Offset: 1

Author

S. I. Dimitrov, Apr 28 2025

Keywords

• nonn

Comments

From David A. Corneth, May 04 2025: (Start)

If (t, u) is a divisor pair of sigma(k)^2 then m = (t + u - 2*k)/2, sigma(m) = m + k - t.

Proof:

Since sigma(m)^2 + sigma(k)^2 = (m+k)^2 we have sigma(k)^2 = (m+k)^2 - sigma(m)^2 = (m + k - sigma(m)) * (m + k + sigma(m)) = t * u where t, u | sigma(k)^2.

This gives the system (m + k - sigma(m)) = t and (m + k + sigma(m)) = u. Solving gives

m = (t + u - 2*k)/2, sigma(m) = m + k - t. For every pair (t, u) of divisors of sigma(k)^2 we can test if the given values of m and sigma(m) hold. If at least one of them holds then k is in the sequence. Q. E. D.

Are there any even terms? There are none in the first 1006 terms. (End)

Examples

			(2, 3) is such a pair because sigma^2(2)+sigma^2(3) = 3^2+4^2 = (2+3)^2.
33 is in the sequence. As sigma(33)^2 = 2304 and for the divisor pair (32, 72) we have m = (32 + 72 - 2*33)/2 = 19 and sigma(m) = m + k - 32 = 19 + 33 - 32 = 20 and indeed sigma(19) = 20. - _David A. Corneth_, May 04 2025
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..2003
• S. I. Dimitrov, Generalizations of amicable numbers, arXiv:2408.07387 [math.NT], 2024.

Crossrefs

Cf. A063990, A259180, A383239.

Programs

PARI

isok(k) = for (m=1, k, if (sigma(m)^2+sigma(k)^2==(m+k)^2, return(1))); \\ Michel Marcus, Apr 28 2025

PARI

is(n) = {my(sn = sigma(n)^2, d = divisors(sn)); for(i = 1, #d / 2, k = (d[i] + d[#d + 1 - i] - 2*n) / 2; if(denominator(k) == 1, sk = n + k - d[i]; if(k < n && sigma(k) == sk, return(1)))); 0} \\ David A. Corneth, May 04 2025

Extensions

More terms from Michel Marcus, Apr 28 2025