OFFSET
1,2
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
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.
EXAMPLE
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.
MATHEMATICA
Select[Range[1000000], Transpose[FactorInteger[#]][[1]] == Transpose[FactorInteger[DivisorSigma[1, #]]][[1]] &] (* T. D. Noe, Dec 08 2012 *)
PROG
(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
(GAP) Filtered([1..1000000], n->Set(Factors(n))=Set(Factors(Sigma(n)))); # Muniru A Asiru, Feb 21 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Edited by N. J. A. Sloane, Jul 12 2008 at the suggestion of R. J. Mathar
STATUS
approved