

A322154


Numbers n satisfying gcd(n^2, sigma(n^2)) > sigma(n), where sigma is the sumofdivisors function.


0



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
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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

Table of n, a(n) for n=1..40.
F. J. Chen and Y.G. Chen, On the index of an odd perfect number, Colloquium Mathematicum, 136(1) (2014), 4149.
Jose Arnaldo Bebita Dris, On a curious biconditional involving the divisors of odd perfect numbers, Notes on Number Theory and Discrete Mathematics, 23(4) (2017), 113.
P. Ochem and M. Rao, Odd perfect numbers are greater than {10}^{1500}, Mathematics of Computation, 81(279) (2012), 18691877.


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)=2xsigma(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 DescartesFrenicleSorli 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).
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



