A322154 Numbers k satisfying gcd(k^2, sigma(k^2)) > sigma(k), where sigma is the sum-of-divisors function.

693, 1386, 1463, 1881, 2379, 2926, 4389, 4758, 8778, 9516, 11895, 13167, 16653, 18018, 19032, 23790, 24180, 25641, 26169, 26334, 33306, 37271, 40443, 43890, 45201, 52668, 54717, 57057, 61380, 65835, 73150, 78507, 105336, 109725, 111813, 114114, 131670, 157014, 166530, 169959

Offset

1,1

Comments

Let N = q^k*n^2 be an odd perfect number with special prime q. If k = 1, it follows that sigma(q^k) < n. Since 2n^2/sigma(q^k) = gcd(n^2, sigma(n^2)), if k = 1 then we have gcd(n^2, sigma(n^2)) > 2n > sigma(n) (since n is deficient, because q^k n^2 is perfect). [See (Dris, 2017)]

Links

Amiram Eldar, Table of n, a(n) for n = 1..1000

Feng-Juan Chen and Yong-Gao Chen, On the index of an odd perfect number, Colloquium Mathematicum, Vol. 136, No. 1 (2014), pp. 41-49.

Jose Arnaldo Bebita Dris, On a curious biconditional involving the divisors of odd perfect numbers, Notes on Number Theory and Discrete Mathematics, Vol. 23, No. 4 (2017), pp. 1-13.

Pascal Ochem and Michaël Rao, Odd perfect numbers are greater than {10}^{1500}, Mathematics of Computation, Vol. 81, No. 279 (2012), pp. 1869-1877.

Formula

If N = q^k*n^2 is an odd perfect number with special prime q, then it is easy to show that sigma(n^2)/q^k = 2n^2/sigma(q^k) = gcd(n^2,sigma(n^2)). From the last equation, it is easy to prove that D(n^2)/s(q^k) = 2s(n^2)/D(q^k) = gcd(n^2,sigma(n^2)), where D(x)=2x-sigma(x) is the deficiency of x and s(x)=sigma(x)-x is the sum of the aliquot divisors of x.

Note that, if k = 1, then sigma(q^k) < n, from which it would follow that q^k < n. [See (Dris, 2017).] Therefore, if k = 1, we have that N = q^k n^2 < n^3. Using Ochem and Rao's lower bound for an odd perfect number, we get n^3 > N > 10^1500, from which we obtain n > 10^500. [See (Ochem and Rao, 2012).]

Thus, if k = 1, we have the lower bound sigma(n^2)/q^k = 2n^2/sigma(q^k) = gcd(n^2, sigma(n^2)) > 2n > n > 10^500 which significantly improves on the corresponding result in (Chen and Chen, 2014).

The assertion k = 1 is known as the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers.

Example

a(1) = 693 is in the sequence because gcd((693)^2, sigma((693)^2)) = gcd(480249, sigma(480249)) > sigma(693), where sigma(480249) = 917301 = 3*7*11^2*19^2, and 480249 = 3^4*7^2*11^2, therefore gcd(480249, sigma(480249)) = 3*7*11^2 = 2541 but sigma(693) = 1248.

Maple

with(numtheory): select(n->gcd(n^2, sigma(n^2))>sigma(n), [$1..170000]); # Muniru A Asiru, Dec 06 2018

Mathematica

Select[Range[10^6], GCD[#^2, DivisorSigma[1, #^2]] > DivisorSigma[1, #] &]

Prog

(GP, Sage Cell Server)
for (x=1, 1000000, if(gcd(x^2, sigma(x^2))>sigma(x), print(x)))
(PARI) isok(n) = gcd(n^2, sigma(n^2)) > sigma(n); \\ Michel Marcus, Nov 29 2018
(GAP) Filtered([1..170000], n->Gcd(n^2, Sigma(n^2))>Sigma(n)); # Muniru A Asiru, Dec 06 2018
(Python)
from sympy import divisor_sigma, gcd
for n in range(1, 170000):
    if gcd(n**2, divisor_sigma(n**2))>divisor_sigma(n):
        print(n) # Stefano Spezia, Dec 07 2018

Crossrefs

Cf. A000203 (sigma), A065764.

Sequence in context: A292833 A004240 A004241 * A129913 A031946 A049356

Adjacent sequences: A322151 A322152 A322153 * A322155 A322156 A322157

Keyword

nonn

Author

Jose Arnaldo Bebita Dris, Nov 29 2018

Extensions

More terms from Michel Marcus, Nov 29 2018

Status

approved