A073825 Numbers n such that Sum_{k=1..n} k^k is prime.

2, 5, 6, 10, 30

Offset: 1

Author

Rick L. Shepherd, Aug 13 2002

Keywords

• nonn

Comments

Any additional terms are greater than 1320 with the next prime having more than 4120 digits.

No terms out to 3000. The next term would yield a prime with over 10000 digits. - John Sillcox (johnsillcox(AT)hotmail.com), Aug 05 2003

For every n, a(n) must be equal to 1 or 2 (mod 4) because Sum[k^k,{k,a(n)}] must be odd. Any additional terms are greater than 5368 with the next prime having more than 20025 digits. - Farideh Firoozbakht, Aug 09 2003

Soundararajan finds an asymptotic upper bound of log k / log log k prime numbers of the form 1^1 + 2^2 + ... + n^n less than k; that is, n << log a(n) / log log a(n). - Charles R Greathouse IV, Aug 27 2008

According to Andersen, the next term is larger than 28000, see Rivera link. - M. F. Hasler, Mar 01 2009

Conjecture: This sequence is infinite. - Daniel Hoying, Jul 20 2020

Links

• Carlos Rivera, Puzzle 404. Sigma(x^x), for x=1 to n, The Prime Puzzles & Problems Connection.
• K. Soundararajan, Primes in a Sparse Sequence, Journal of Number Theory 43:2 (1993), pp. 220-227.

Crossrefs

Cf. A073826 (corresponding primes), A001923 (Sum k^k, k=1..n).

Programs

Mathematica

v={}; Do[If[(Mod[n, 4]==1||Mod[n, 4]==2)&&PrimeQ[Sum[k^k, {k, n}]], v=Insert[v, n, -1]; Print[v]], {n, 5368}]

PARI

s=0; for(k=1,1320, s=s+k^k; if(isprime(s), print1(k,", ")))

Formula

log a(n) >> n log^2 n. - Charles R Greathouse IV, May 17 2016

Extensions

Edited by Charles R Greathouse IV, Oct 27 2010