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
Keywords
Examples
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.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..7914; terms < 2^33 (first 1000 terms from Amiram Eldar)
- 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.
Programs
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GAP
Filtered([1..170000],n->Gcd(n^2,Sigma(n^2))>Sigma(n)); # Muniru A Asiru, Dec 06 2018
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Maple
with(numtheory): select(n->gcd(n^2,sigma(n^2))>sigma(n),[$1..170000]); # Muniru A Asiru, Dec 06 2018
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Mathematica
Select[Range[10^6], GCD[#^2, DivisorSigma[1, #^2]] > DivisorSigma[1, #] &]
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PARI
isok(n) = gcd(n^2,sigma(n^2)) > sigma(n); \\ Michel Marcus, Nov 29 2018
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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
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.
Extensions
More terms from Michel Marcus, Nov 29 2018
Comments