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
KEYWORD
nonn
AUTHOR
Jose Arnaldo Bebita Dris, Nov 29 2018
EXTENSIONS
More terms from Michel Marcus, Nov 29 2018
STATUS
approved