A104903 Numbers n such that sigma(n) = 16*phi(n).

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

Cf. A000043, A062699, A068390, A104900, A104901, A104902.

Sequence in context: A342391 A048960 A252331 * A235952 A207794 A368426

Adjacent sequences: A104900 A104901 A104902 * A104904 A104905 A104906

Keyword

easy,nonn

Author

Farideh Firoozbakht, Apr 01 2005

Status

approved