|
| |
|
|
A058007
|
|
Freestyle perfect numbers n = Product_{i=1,..,k} f_i^e_i where 1 < f_1 < ... < f_k, e_i > 0, such that 2n = Product_{i=1,..,k} (f_i^(e_i+1)-1)/(f_i-1).
|
|
3
|
|
|
|
6, 28, 60, 84, 90, 120, 336, 496, 840, 924, 1008, 1080, 1260, 1320, 1440, 1680, 1980, 2016, 2160, 2184, 2520, 2772, 3024, 3420, 3600, 3780, 4680, 5040, 5940, 6048, 6552, 7440, 7560, 7800, 8128, 8190, 8280, 9240, 9828, 9900, 10080, 10530, 11088, 11400, 13680
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Only one odd freestyle perfect number is known: 198585576189, found by Descartes.
This sequence consists of perfect numbers A000396 and those which aren't, called spoof-perfect numbers A174292. Roughly said, a spoof-perfect number is a number that would be perfect if one or more of its composite factors were wrongly assumed to be prime, i.e., taken as a "spoof prime". - Daniel Forgues, Nov 15 2009 (slightly rephrased)
The right hand side of the second equation in the definition, 2n = ..., equals the sum of divisors sigma(n), if all of the f_i are distinct primes. If they aren't, there arise some ambiguities: See A174292 for further discussion. - M. F. Hasler, Jan 13 2013
|
|
|
REFERENCES
|
R. K. Guy, Unsolved Problems in Number Theory, B1.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
n = 60 = (3^1)*(4^1)*(5^1), s = 120 = [(3^2-1)*(4^2-1)*(5^2-1)]/[(3-1)*(4-1)*(5-1)]. s-n = 120-60 = n. So 60 is in the sequence.
|
|
|
MATHEMATICA
|
r[s_, n_, f_] := Catch[If[n == 1, s == 1, Block[{p, e}, Do[e = 1; While[ Mod[n, p^e] == 0, r[s*(p^(e+1) - 1)/(p-1), n/p^e, p] && Throw@True; e++], {p, Select[Divisors@n, f < # &]}]]; False]];
spoofQ[n_] := r[1/2/n, n, 1] && DivisorSigma[-1, n] != 2;
perfectQ[n_] := DivisorSigma[1, n] == 2*n;
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,nice
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
