A276663 Sum of primes dividing n-th perfect number (with repetition).

5, 11, 39, 139, 8215, 131103, 524323, 2147483707, 2305843009213694071, 618970019642690137449562287, 162259276829213363391578010288339, 170141183460469231731687303715884105979

Offset: 1

Author

Juri-Stepan Gerasimov, Sep 12 2016

Keywords

• nonn

Comments

Numbers that are equal to the sum of the prime factors (A001414) of some perfect number.

The next term is too large to include.

A001222(a(n)) is 1, 1, 2, 1, 3, 4, 2, 2, 4, 6, 7, 1, 11, ...

Examples

			39 is in this sequence because 39 - 2^(5 - 1) = 31 = 2^5 - 1 and 31 is prime.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..18

Crossrefs

Subsequence of A131898. Supersequence of A276511.

Cf. A000043, A000396, A001222, A001414, A007013, A192436, A239546, A276493.

Programs

Mathematica

Table[Total[Times@@@FactorInteger[PerfectNumber[n]]],{n,15}] (* Harvey P. Dale, Sep 22 2019 *)

PARI

\\ Ochem & Rao: no odd perfect numbers below 10^1500
forprime(p=2,2281, if(ispseudoprime(t=2^p-1), print1(2^p+2*p-3", "))) \\ Charles R Greathouse IV, Sep 18 2016

Formula

a(n) = 2^A000043(n) + 2*A000043(n) - 3, assuming that there are no odd perfect numbers.

a(n) = A001414(A000396(n)). - Michel Marcus, Sep 18 2016