A283156 Number of preimages of even integers under the sum-of-proper-divisors function.

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

Offset: 1

Author

Anton Mosunov, Mar 01 2017

Keywords

• nonn

Comments

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}).

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.

Examples

			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).
		

Links

• Anton Mosunov, Table of n, a(n) for n = 1..10000
• R. K. Guy, J. L. Selfridge, What drives an aliquot sequence?, Math. Comp. 29 (129), 1975, 101-107.
• P. Pollack, C. Pomerance, Some problems of Erdos on the sum-of-divisors function, Trans. Amer. Math. Soc., Ser. B, 3 (2016), 1-26.
• C. Pomerance, The first function and its iterates, A Celebration of the Work of R. L. Graham, S. Butler, J. Cooper, and G. Hurlbert, eds., Cambridge U. Press, to appear.
• C. Pomerance, H.-S. Yang, Variant of a theorem of Erdos on the sum-of-proper-divisors function, Math. Comp., 83 (2014), 1903-1913.T. Oliveira e Silva, Goldbach conjecture verification, 2015.

Crossrefs

Cf. A048138, A283152.

Programs

PARI

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

Formula

a(n) = A048138(2*n). - Michel Marcus, Mar 04 2017