

A283156


Number of preimages of even integers under the sumofproperdivisors function.


8



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, 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, 1, 3, 1, 1, 0, 2, 2, 2, 3, 2, 1, 1, 0, 1, 2, 1, 2, 1, 1
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OFFSET

1,3


COMMENTS

Let sigma(n) denote the sum of divisors function, and s(n):=sigma(n)n. The kth 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}).
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.


LINKS



FORMULA



EXAMPLE

a(1)=0, because 2*1=s(m) has no solutions;
a(2)=1, because 2*2=s(9);
a(3)=2, because 2*3=s(6)=s(25);
a(4)=2, because 2*4=s(10)=s(49);
a(5)=1, because 2*5=s(14).


PROG

(PARI) a(n) = sum(k=1, (2*n1)^2, (sigma(k)  k) == 2*n); \\ Michel Marcus, Mar 04 2017


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



