A322154 - OEIS

%I #30 Oct 19 2021 21:33:06

%S 693,1386,1463,1881,2379,2926,4389,4758,8778,9516,11895,13167,16653,

%T 18018,19032,23790,24180,25641,26169,26334,33306,37271,40443,43890,

%U 45201,52668,54717,57057,61380,65835,73150,78507,105336,109725,111813,114114,131670,157014,166530,169959

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

%C 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 Amiram Eldar, <a href="/A322154/b322154.txt">Table of n, a(n) for n = 1..1000</a>

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

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

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

%o (GP, Sage Cell Server)

%o for (x=1, 1000000,if(gcd(x^2,sigma(x^2))>sigma(x),print(x)))

%o (PARI) isok(n) = gcd(n^2,sigma(n^2)) > sigma(n); \\ _Michel Marcus_, Nov 29 2018

%o (GAP) Filtered([1..170000],n->Gcd(n^2,Sigma(n^2))>Sigma(n)); # _Muniru A Asiru_, Dec 06 2018

%o (Python)

%o from sympy import divisor_sigma, gcd

%o for n in range(1,170000):

%o     if gcd(n**2,divisor_sigma(n**2))>divisor_sigma(n):

%o         print(n) # _Stefano Spezia_, Dec 07 2018

%Y Cf. A000203 (sigma), A065764.

%K nonn

%O 1,1

%A _Jose Arnaldo Bebita Dris_, Nov 29 2018

%E More terms from _Michel Marcus_, Nov 29 2018