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A111231
Numbers which are perfect powers m^k equal to the sum of m distinct primes.
5
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
1,2
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
EXAMPLE
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.
PROG
(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
CROSSREFS
Sequence in context: A377847 A036966 A076467 * A111307 A246549 A076405
KEYWORD
nonn
AUTHOR
Giovanni Teofilatto, Oct 28 2005
EXTENSIONS
Offset corrected by R. J. Mathar, May 25 2009
Edited by M. F. Hasler, May 25 2018
STATUS
approved