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Odd numbers n such that 2n/sigma(n) - 1 = 1/x for some positive integer x.
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%I #39 Jan 28 2021 10:31:37

%S 1,3,15,135,315,585,819,1155,1365,1485,2295,4455,9009,9405,9945,11115,

%T 13923,13965,16965,33705,34335,40365,43785,43875,63063,63855,84825,

%U 90675,106425,108927,130815,209385,236115,262845,387387,396495,397575,449295,483945

%N Odd numbers n such that 2n/sigma(n) - 1 = 1/x for some positive integer x.

%C Odd terms from A222264, see there for motivation and further links. In particular, the powers of 2 in A222264 which correspond to perfect numbers are excluded, so all n*x from this sequence are spoof perfect numbers, cf. A174292.

%C Here we do not exclude n if gcd(n,x) > 1 (the first such term is 1155), which would arguably be a "reasonable" additional condition to impose.

%C The first term with odd x is n=a(79)=9018009, x=22021, which yields Descarte's n*x=198585576189, see also A033870, A033871 and A222262.

%C No other term with odd x (and thus no other odd freestyle perfect number) is known as of today, to our best knowledge. See the paper by Banks et al. for some restrictions on such numbers.

%C One can note that when x is even, then sigma(n)/n is of the form (2k-2)/k. For instance, for n=15 we have x=4, and sigma(n)/n = 8/5 with k=5. On the other hand, when x is odd, then sigma(n)/n is of the form (2k-1)/k. For instance, for n=9018009 we have x=22021, and sigma(n)/n = 22021/11011 with k=11011. - _Michel Marcus_, Nov 24 2013

%H Laszlo L. Toth, <a href="/A222263/b222263.txt">Table of n, a(n) for n = 1..692</a> (terms 1..143 terms from M. F. Hasler, 144..500 from Donovan Johnson).

%H W. Banks, A. Guloglu, W. Nevans and F. Saidak, <a href="http://www.math.missouri.edu/~bbanks/papers/2008_Descartes_Final.pdf">Descartes Numbers</a>, in: Anatomy of Integers, 167-174, American Mathematical Society, Providence R.I., 2008.

%H László Tóth, <a href="https://arxiv.org/abs/2101.09718">On the Density of Spoof Odd Perfect Numbers</a>, arXiv:2101.09718 [math.NT], 2021.

%e For a(3)=15, we have sigma(15)=sigma(3*5)=4*6=24, thus 2*15/sigma(15) = 30/24 = 1+1/4. Here, x=4, and if 4 was assumed to be prime, then sigma'(4)=1+4=5 (*) and sigma'(15*4) = sigma(15)*sigma'(4) = 24*5 = 120 = 2*(15*4), i.e. 15*4 is a spoof perfect number.

%e (*) Here, sigma' is the "freestyle" sum-of-divisors function which assumes that 4 is prime, i.e., sigma'(n)=sigma(n/4^k)*(4^(k+1)-1)/3 with 4^k || n, or k=valuation(n,4)=floor(A007814(n)/2).

%o (PARI) forstep(n=1,9e9,2, numerator(2*n/sigma(n)-1)==1 & print1(n","))

%Y Cf. A000203 = sigma, the sum-of-divisors function.

%K nonn

%O 1,2

%A _M. F. Hasler_, Feb 20 2013