%I #21 Jun 03 2018 02:04:06
%S 0,1,2,2,1,1,2,2,1,2,2,1,1,1,1,2,1,1,1,3,2,2,2,1,2,0,2,1,1,1,2,3,1,1,
%T 1,1,3,3,1,1,1,1,1,0,2,2,1,0,1,2,1,2,4,2,2,1,2,1,1,0,1,0,1,1,2,1,3,2,
%U 1,3,1,1,0,2,2,2,3,2,1,1,0,1,2,1,2,1,1
%N Number of preimages of even integers under the sum-of-proper-divisors function.
%C Let sigma(n) denote the sum of divisors function, and s(n):=sigma(n)-n. The k-th element a(k) corresponds to the number of solutions to 2k=s(m) in positive integers, where m is a variable. In 2016, C. Pomerance proved that, for every e > 0, the number of solutions is O_e((2k)^{2/3+e}).
%C Note that for odd numbers n the problem of solving n=s(m) is quite different from the case when n is even. According to a slightly stronger version of Goldbach's conjecture, for every odd number n there exist primes p and q such that n = s(pq) = p + q + 1. This conjecture was verified computationally by Oliveira e Silva to 10^18. Thus the problem is (almost) equivalent to counting the solutions to n=p+q+1 in primes.
%H Anton Mosunov, <a href="/A283156/b283156.txt">Table of n, a(n) for n = 1..10000</a>
%H R. K. Guy, J. L. Selfridge, <a href="https://doi.org/10.1090/S0025-5718-1975-0384669-X">What drives an aliquot sequence?</a>, Math. Comp. 29 (129), 1975, 101-107.
%H P. Pollack, C. Pomerance, <a href="https://doi.org/10.1090/btran/10">Some problems of Erdos on the sum-of-divisors function</a>, Trans. Amer. Math. Soc., Ser. B, 3 (2016), 1-26.
%H C. Pomerance, <a href="https://math.dartmouth.edu/~carlp/aliquot.pdf">The first function and its iterates</a>, A Celebration of the Work of R. L. Graham, S. Butler, J. Cooper, and G. Hurlbert, eds., Cambridge U. Press, to appear.
%H C. Pomerance, H.-S. Yang, <a href="https://doi.org/10.1090/S0025-5718-2013-02775-5">Variant of a theorem of Erdos on the sum-of-proper-divisors function</a>, Math. Comp., 83 (2014), 1903-1913.T. Oliveira e Silva, <a href="http://sweet.ua.pt/tos/goldbach.html">Goldbach conjecture verification</a>, 2015.
%F a(n) = A048138(2*n). - _Michel Marcus_, Mar 04 2017
%e a(1)=0, because 2*1=s(m) has no solutions;
%e a(2)=1, because 2*2=s(9);
%e a(3)=2, because 2*3=s(6)=s(25);
%e a(4)=2, because 2*4=s(10)=s(49);
%e a(5)=1, because 2*5=s(14).
%o (PARI) a(n) = sum(k=1, (2*n-1)^2, (sigma(k) - k) == 2*n); \\ _Michel Marcus_, Mar 04 2017
%Y Cf. A048138, A283152.
%K nonn
%O 1,3
%A _Anton Mosunov_, Mar 01 2017