cp's OEIS Frontend

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

%I A111231 #21 Jun 01 2018 01:50:52
%S A111231 0,8,16,27,32,64,81,125,128,216,243,256,343,512,625,729,1000,1024,
%T A111231 1296,1331,1728,2048,2187,2197,2401,2744,3125,3375,4096,4913,5832,
%U A111231 6561,6859,7776,8000,8192,9261,10000,10648,12167,13824,14641,15625,16384,16807
%N A111231 Numbers which are perfect powers m^k equal to the sum of m distinct primes.
%C A111231 Perfect powers m^k with k >= 3, m = 0 or m > 1.
%C A111231 Is a(n) = A076467(n) for all n > 1? - _R. J. Mathar_, May 22 2009
%C A111231 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
%e A111231 a(1) = 0 because 0 = 0^2 = 0^3 is the sum of 0 primes;
%e A111231 a(2) = 8 because 8 = 2^3 = 3 + 5, sum of 2 primes;
%e A111231 a(3) = 16 because 16 = 2^4 = 3 + 13, sum of 2 primes.
%e A111231 a(4) = 27 because 27 = 3^3 = 3 + 11 + 13, sum of 3 primes.
%o A111231 (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
%Y A111231 Cf. A007504, A076467.
%K A111231 nonn
%O A111231 1,2
%A A111231 _Giovanni Teofilatto_, Oct 28 2005
%E A111231 Offset corrected by _R. J. Mathar_, May 25 2009
%E A111231 Edited by _M. F. Hasler_, May 25 2018