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A104903
Numbers n such that sigma(n) = 16*phi(n).
7
20790, 26040, 43890, 268380, 368280, 377580, 415380, 426720, 547470, 566580, 777480, 906780, 996030, 1659000, 1744470, 2102730, 2179320, 2454270, 2699970, 3682770, 4373880, 5053860, 5340060, 5791170, 5874660, 5894070, 5936280, 6035040, 7067340, 8013060
OFFSET
1,1
COMMENTS
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.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (calculated using data from Jud McCranie, terms 1..1000 from Donovan Johnson)
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.
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.
EXAMPLE
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.
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).
MATHEMATICA
Do[If[DivisorSigma[1, m] == 16*EulerPhi[m], Print[m]], {m, 10000000}]
PROG
(PARI) is(n)=sigma(n)==16*eulerphi(n) \\ Charles R Greathouse IV, May 09 2013
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Farideh Firoozbakht, Apr 01 2005
STATUS
approved