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

%I A322154 #39 Jun 22 2025 09:18:35
%S A322154 693,1386,1463,1881,2379,2926,4389,4758,8778,9516,11895,13167,16653,
%T A322154 18018,19032,23790,24180,25641,26169,26334,33306,37271,40443,43890,
%U A322154 45201,52668,54717,57057,61380,65835,73150,78507,105336,109725,111813,114114,131670,157014,166530,169959
%N A322154 Numbers k satisfying gcd(k^2, sigma(k^2)) > sigma(k), where sigma is the sum-of-divisors function.
%C A322154 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)]
%H A322154 Antti Karttunen, <a href="/A322154/b322154.txt">Table of n, a(n) for n = 1..7914; terms < 2^33</a> (first 1000 terms from Amiram Eldar)
%H A322154 Feng-Juan Chen and Yong-Gao Chen, <a href="http://www.doi.org/10.4064/cm136-1-4">On the index of an odd perfect number</a>, Colloquium Mathematicum, Vol. 136, No. 1 (2014), pp. 41-49.
%H A322154 Jose Arnaldo Bebita Dris, <a href="http://nntdm.net/volume-23-2017/number-4/01-13/">On a curious biconditional involving the divisors of odd perfect numbers</a>, Notes on Number Theory and Discrete Mathematics, Vol. 23, No. 4 (2017), pp. 1-13.
%H A322154 Pascal Ochem and Michaël Rao, <a href="https://doi.org/10.1090/S0025-5718-2012-02563-4">Odd perfect numbers are greater than {10}^{1500}</a>, Mathematics of Computation, Vol. 81, No. 279 (2012), pp. 1869-1877.
%F A322154 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.
%F A322154 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).]
%F A322154 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).
%F A322154 The assertion k = 1 is known as the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers.
%e A322154 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.
%p A322154 with(numtheory): select(n->gcd(n^2,sigma(n^2))>sigma(n),[$1..170000]); # _Muniru A Asiru_, Dec 06 2018
%t A322154 Select[Range[10^6], GCD[#^2, DivisorSigma[1, #^2]] > DivisorSigma[1, #] &]
%o A322154 (GP, Sage Cell Server)
%o A322154 for (x=1, 1000000,if(gcd(x^2,sigma(x^2))>sigma(x),print(x)))
%o A322154 (PARI) isok(n) = gcd(n^2,sigma(n^2)) > sigma(n); \\ _Michel Marcus_, Nov 29 2018
%o A322154 (GAP) Filtered([1..170000],n->Gcd(n^2,Sigma(n^2))>Sigma(n)); # _Muniru A Asiru_, Dec 06 2018
%o A322154 (Python)
%o A322154 from sympy import divisor_sigma, gcd
%o A322154 for n in range(1,170000):
%o A322154     if gcd(n**2,divisor_sigma(n**2))>divisor_sigma(n):
%o A322154         print(n) # _Stefano Spezia_, Dec 07 2018
%Y A322154 Cf. A000203 (sigma), A065764.
%K A322154 nonn
%O A322154 1,1
%A A322154 _Jose Arnaldo Bebita Dris_, Nov 29 2018
%E A322154 More terms from _Michel Marcus_, Nov 29 2018