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Counterbalanced numbers: Composite numbers k such that phi(k)/(sigma(k)-k) is an integer.
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%I #21 Nov 16 2021 11:07:17

%S 133,403,583,713,817,2077,2623,2923,4453,4717,5311,5773,7093,7747,

%T 9313,11023,11581,11653,12877,14353,15553,19303,20803,21409,21733,

%U 21971,24307,31169,35033,39283,39337,43873,46297,46357,50573,50879,53863

%N Counterbalanced numbers: Composite numbers k such that phi(k)/(sigma(k)-k) is an integer.

%C Banks and Luca (2007) showed that the number of terms <= x, N(x) <= x * exp(-((1/3)*(log(8))^(1/3) + o(1))*(log(x))^(1/3)*(log(log(x)))^(1/3)) as x -> infinity, and that under Dickson's conjecture this sequence is infinite, since for each positive integer m, if p = 5m + 1 and q = 20m + 13 are primes, then p*q is a term. - _Amiram Eldar_, Apr 13 2020

%H Donovan Johnson, <a href="/A055940/b055940.txt">Table of n, a(n) for n = 1..10000</a>

%H William D. Banks and Florian Luca, <a href="https://doi.org/10.1017/S0013091505001161">When the sum of aliquots divides the totient</a>, Proceedings of the Edinburgh Mathematical Society, Vol. 50, No. 3 (2007), pp. 563-569.

%H Douglas E. Iannucci, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL20/Iannucci/ian5.html">On the Equation sigma(n) = n + phi(n)</a>, Journal of Integer Sequences, Vol. 20 (2017), Article 17.6.2.

%e k = 133 = 7*19: phi(133)=108, sigma(133)-133 = 1+7+19 = 27, q = 4.

%t Do[s=EulerPhi[n]/(DivisorSigma[1, n]-n); If[ !PrimeQ[n]&&IntegerQ[s], Print[n]], {n, 2, 1000000}]

%t Select[Range[54000],CompositeQ[#]&&IntegerQ[EulerPhi[#]/(DivisorSigma[ 1,#]-#)]&] (* _Harvey P. Dale_, Nov 16 2021 *)

%o (PARI) is(n)=!isprime(n) && n>1 && eulerphi(n)%(sigma(n)-n)==0 \\ _Charles R Greathouse IV_, Jan 02 2014

%Y Cf. A000010, A001065, A068418, A020492, A070037.

%K nonn

%O 1,1

%A _Robert G. Wilson v_, Jul 22 2000