A111231 Numbers which are perfect powers m^k equal to the sum of m distinct primes.

0, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, 343, 512, 625, 729, 1000, 1024, 1296, 1331, 1728, 2048, 2187, 2197, 2401, 2744, 3125, 3375, 4096, 4913, 5832, 6561, 6859, 7776, 8000, 8192, 9261, 10000, 10648, 12167, 13824, 14641, 15625, 16384, 16807

Offset: 1

Author

Giovanni Teofilatto, Oct 28 2005

Keywords

• nonn

Comments

Perfect powers m^k with k >= 3, m = 0 or m > 1.

Is a(n) = A076467(n) for all n > 1? - R. J. Mathar, May 22 2009

A sum of m distinct primes is >= A007504(m) ~ m^2(log m)/2 > m^2, also for small m, therefore the second condition excludes squares m^2. On the other hand, considering results related to Goldbach's conjecture (e.g., every even number >= 4 is the sum of at most 4 primes), it is increasingly improbable that some m^k with k >= 3 is not the sum of m primes. This explains the first comment - but can it be rigorously proved? - M. F. Hasler, May 25 2018

Examples

			a(1) = 0 because 0 = 0^2 = 0^3 is the sum of 0 primes;
a(2) = 8 because 8 = 2^3 = 3 + 5, sum of 2 primes;
a(3) = 16 because 16 = 2^4 = 3 + 13, sum of 2 primes.
a(4) = 27 because 27 = 3^3 = 3 + 11 + 13, sum of 3 primes.
		

Crossrefs

Cf. A007504, A076467.

Programs

PARI

is(n,d)={if(d=ispower(n), fordiv(d,e,e>1&&forvec(v=vector(d=sqrtnint(n,e)-1,i,[1,primepi((n-1)\2-d+3)]),prime(v[#v])<(d=n-vecsum(apply(i->prime(i),v)))&&isprime(d)&&return(1),2)), !n)} \\ M. F. Hasler, May 25 2018

Extensions

Offset corrected by R. J. Mathar, May 25 2009

Edited by M. F. Hasler, May 25 2018