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A253238
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Number of ways to write n as a sum of a perfect power (>1) and a prime.
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1
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0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 0, 1, 1, 4, 2, 2, 2, 1, 3, 2, 2, 3, 1, 2, 4, 4, 2, 2, 1, 2, 2, 4, 2, 3, 1, 3, 2, 4, 2, 2, 2, 3, 4, 2, 1, 2, 1, 2, 3, 3, 1, 2, 3, 3, 4, 4, 2, 2, 2, 2, 1, 5, 1, 4, 2, 3, 3, 2, 1, 5, 2, 1, 4, 4, 3, 2, 1, 2, 4, 3, 2, 3, 2, 2, 4, 2, 2, 2, 2, 3, 2, 6, 2, 4, 2, 2, 4, 5, 2, 3, 1, 3, 3, 5, 2, 3, 1, 2, 4, 4, 3, 3, 2, 1, 6
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OFFSET
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1,11
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COMMENTS
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In this sequence, "perfect power" does not include 0 or 1, "prime" does not include 1. Both "perfect power" and "prime" must be positive.
In the past, I conjectured that a(n) > 0 for all n>24, but this is not true. My PARI program found that a(1549) = 0.
I also asked which a(n) are 1. For example, 331 is a de Polignac number (A006285), so it cannot be written as 2^n+p with p prime, and 331-6^n must divisible by 5, 331-10^n must divisible by 3, ..., 331-18^2 = 331-324 = 7 is prime (and it is the only prime of the form 331-m^n, with m, n natural numbers, m>1, n>1), so a(331) = 1. Similarly, a(3319) = 1. Conjecture: a(n) > 1 for all n > 3319.
This conjecture is not true: a(1771561) = 0. (See A119748)
Another conjecture: For every number m>=0, there is a number k such that a(n)>=m for all n>=k.
Another conjecture: Except for k=2, first occurrence of k must be earlier then first occurrence of k+1.
For n such that a(n) = 0, see A119748.
For n such that a(n) = 1, see the following a-file of this sequence.
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LINKS
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MATHEMATICA
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nn = 128; pwrs = Flatten[Table[Range[2, Floor[nn^(1/ex)]]^ex, {ex, 2, Floor[Log[2, nn]]}]]; pp = Prime[Range[PrimePi[nn]]]; t = Table[0, {nn}]; Do[ t[[i[[1]]]] = i[[2]], {i, Tally[Sort[Select[Flatten[Outer[Plus, pwrs, pp]], # <= nn &]]]}]; t
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PROG
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(PARI) a(n) = sum(k=1, n-1, ispower(k) && isprime(n-k))
(PARI) a(n)=sum(e=2, log(n)\log(2), sum(b=2, sqrtnint(n, e), isprime(n-b^e)&&!ispower(b))) \\ Charles R Greathouse IV, May 28 2015
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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