A289290 Numbers n such that the sum of the divisors of n is of the form m^2+1.

1, 657, 3681, 10809, 15777, 17937, 24201, 28521, 54657, 81441, 122697, 154881, 230481, 265257, 336321, 346041, 455337, 473481, 547137, 613017, 718857, 833121, 898137, 1161009, 1226457, 1274841, 1305081, 1510281, 1584801, 1618497, 1695609, 1752417, 1846161, 1965609

Offset: 1

Author

Michel Lagneau, Jul 02 2017

Keywords

• nonn

Comments

The corresponding values of m are 0, 31, 73, 125, 151, 161, 187, 203, 281, ...

For n > 1, it appears that a(n) is of the form 9p where p is prime congruent to 1 (mod 24). The corresponding primes p are 73, 409, 1201, 1753, 1993, 2689, 3169, 6073, 9049, 13633, 17209, ... (a subsequence of A107008?).

For n > 1, the divisors of a(n) are of the form {1, 3, 9, p, 3p, 9p} and sigma(a(n)) = 13(1 + p) = 1 + m^2.

Proof:

We use the formula (a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2 with:

a = 2, b = 3, c = (3m + 2)/13 and d = (2m - 3)/13. We obtain:

a^2 + b^2 = 13, c^2 + d^2 = 1 + p = (m^2 + 1)/13 => p = (m^2 - 12)/13,

(ac - bd)^2 = 1 and (ad + bc)^2 = m^2.

The first terms not of the form 9p are 2493961, 7106353, 8325721, 10708297, and 14120281. Note that solving the Diophantine equation m^2 + 1 = sigma(k)*(1+p), for various numbers k, it is possible to identify other potentially infinite subsequences similar to 9p. For example, 47^2*p, 7^4*p, 3^6*p, 3^2*11^2*p, and so on. - Giovanni Resta, Jul 02 2017

Examples

			657 is in the sequence because sigma(657) = 962 = 31^2 + 1.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000203, A002522, A107008.

Programs

Maple

with(numtheory):nn:=10^6:
for n from 1 to nn do:
    y:=sqrt(sigma(n)-1):
     if y=floor(y) then printf(`%d, `,n):
       else
      fi:
od:

Mathematica

Select[Range[10^6], IntegerQ[Sqrt[DivisorSigma[1, #] - 1]] &] (* Giovanni Resta, Jul 02 2017 *)

PARI

isok(n) = issquare(sigma(n) - 1); \\ Michel Marcus, Jul 02 2017

Extensions

Name edited by Robert Israel, Jul 03 2017