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%I A104903 #21 Dec 04 2019 18:19:30
%S A104903 20790,26040,43890,268380,368280,377580,415380,426720,547470,566580,
%T A104903 777480,906780,996030,1659000,1744470,2102730,2179320,2454270,2699970,
%U A104903 3682770,4373880,5053860,5340060,5791170,5874660,5894070,5936280,6035040,7067340,8013060
%N A104903 Numbers n such that sigma(n) = 16*phi(n).
%C A104903 If p>3 and 2^p-1 is prime (a Mersenne prime) then 105*2^(p-2)*(2^p-1) is in the sequence. So 105*2^(A000043-2)*(2^A000043-1) is a subsequence of this sequence. It seems that 10 divides all terms of this sequence.
%H A104903 Amiram Eldar, Table of n, a(n) for n = 1..10000 (calculated using data from Jud McCranie, terms 1..1000 from Donovan Johnson)
%H A104903 Kevin A. Broughan and Daniel Delbourgo, On the Ratio of the Sum of Divisors and Euler’s Totient Function I, Journal of Integer Sequences, Vol. 16 (2013), Article 13.8.8.
%H A104903 Kevin A. Broughan and Qizhi Zhou, On the Ratio of the Sum of Divisors and Euler's Totient Function II, Journal of Integer Sequences, Vol. 17 (2014), Article 14.9.2.
%e A104903 p>2, q=2^p-1(q is prime); m=105*2^(p-2)*q so sigma(m)=192*(2^(p-1)-1)*2^p=16*(48*2^(p-3)*(2^p-2))=16*phi(m) hence m is in the sequence.
%e A104903 sigma(1659000)=5990400=16*374400=16*phi(1659000) so 1659000 is in the sequence but 1659000 is not of the form 105*2^(p-2)*(2^p-1).
%t A104903 Do[If[DivisorSigma[1, m] == 16*EulerPhi[m], Print[m]], {m, 10000000}]
%o A104903 (PARI) is(n)=sigma(n)==16*eulerphi(n) \\ _Charles R Greathouse IV_, May 09 2013
%Y A104903 Cf. A000043, A062699, A068390, A104900, A104901, A104902.
%K A104903 easy,nonn
%O A104903 1,1
%A A104903 _Farideh Firoozbakht_, Apr 01 2005
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