|
|
A283156
|
|
Number of preimages of even integers under the sum-of-proper-divisors 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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
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.
|
|
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*n-1)^2, (sigma(k) - k) == 2*n); \\ Michel Marcus, Mar 04 2017
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|