A027598 Numbers k such that the set of prime divisors of k is equal to the set of prime divisors of sigma(k).

1, 6, 28, 120, 270, 496, 672, 1080, 1638, 1782, 3780, 8128, 18600, 20580, 24948, 26208, 30240, 32640, 32760, 35640, 41850, 44226, 55860, 66960, 164640, 167400, 185220, 199584, 273000, 293760, 401310, 441936, 446880, 502740, 523776, 614250, 707616, 802620, 819000

Offset: 1

Author

Jud McCranie

Keywords

• nonn

Comments

Multiplicities are ignored.

All even perfect numbers are in the sequence. It seems that 1 is the only odd term of the sequence. - Farideh Firoozbakht, Jul 01 2008

sigma() is the multiplicative sum-of-divisors function. - Walter Nissen, Dec 16 2009

Pollack and Pomerance call these "prime-perfect numbers" and show that there are << x^(1/3+e) of these up to x for any e > 0. - Charles R Greathouse IV, May 09 2013

Except for unity for the obvious reason, the primitive terms are the perfect numbers (A000396). - Robert G. Wilson v, Feb 19 2019

If an odd term > 1 exists, it is larger than 5*10^23. - Giovanni Resta, Jun 02 2020

Examples

			273000 = 2^3*3*5^3*7*13 and sigma(273000) = 1048320 = 2^8*3^2*5*7*13 so 273000 is in the sequence.
		

References

• R. K. Guy, Unsolved Problems in Number Theory, B19.

Links

• Donovan Johnson, Table of n, a(n) for n = 1..500 (first 100 terms from T. D. Noe)
• Paul Pollack and Carl Pomerance, Prime-Perfect Numbers, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 12a, Paper A14, 2012.

Crossrefs

Intersection of A105402 and A175200. - Amiram Eldar, Jun 02 2020

Cf. A110751, A110819, A055744, A081377, A000203.

Programs

GAP

Filtered([1..1000000],n->Set(Factors(n))=Set(Factors(Sigma(n)))); # Muniru A Asiru, Feb 21 2019

Mathematica

Select[Range[1000000], Transpose[FactorInteger[#]][[1]] == Transpose[FactorInteger[DivisorSigma[1, #]]][[1]] &] (* T. D. Noe, Dec 08 2012 *)

PARI

a(n) = {for (i=1, n, fn = factor(i); fs = factor(sigma(i)); if (fn[,1] == fs[,1], print1(i, ", ")););} \\ Michel Marcus, Nov 18 2012

PARI

is(n)=my(f=factor(n),fs=[],t);for(i=1,#f[,1], t=factor((f[i,1]^(f[i,2]+1)-1)/(f[i,1]-1))[,1]; fs=vecsort(concat(fs,t~),,8); if(#setminus(fs,f[,1]~), return(0))); fs==f[,1]~ \\ Charles R Greathouse IV, May 09 2013

Extensions

Edited by N. J. A. Sloane, Jul 12 2008 at the suggestion of R. J. Mathar