A276711 Number of primes p <= n such that n - p is a perfect power (including 0 and 1).
0, 1, 2, 1, 1, 2, 2, 1, 1, 1, 4, 2, 2, 2, 2, 1, 2, 2, 3, 2, 3, 1, 3, 1, 1, 1, 4, 2, 3, 3, 2, 4, 2, 2, 3, 1, 3, 5, 4, 2, 3, 2, 3, 3, 4, 2, 4, 2, 3, 2, 4, 2, 3, 3, 3, 4, 2, 1, 3, 2, 3, 4, 3, 1, 2, 3, 4, 5, 4, 2, 3, 3, 3, 2, 5, 1, 4, 2, 4, 4
Offset: 1
Keywords
Examples
a(2) = 1 since 2 = 2 + 0^2 with 2 prime. a(3) = 2 since 3 = 2 + 1^2 = 3 + 0^2 with 2 and 3 prime. a(4) = 1 since 4 = 3 + 1^2 with 3 prime. a(64) = 1 since 64 = 37 + 3^3 with 37 prime. a(328) = 1 since 328 = 103 + 15^2 with 103 prime. a(370) = 1 since 370 = 127 + 3^5 with 127 prime. a(841) = 1 since 841 = 809 + 2^5 with 809 prime. a(1204) = 1 since 1204 = 1123 + 9^2 with 1123 prime. a(1243) = 1 since 1243 = 919 + 18^2 with 919 prime. a(1549) = 1 since 1549 = 1549 + 0^2 with 1549 prime. a(1681) = 1 since 1681 = 1553 + 2^7 with 1553 prime. a(1849) = 1 since 1849 = 1721 + 2^7 with 1721 prime. a(2146) = 1 since 2146 = 2137 + 3^2 with 2137 prime. a(2986) = 1 since 2986 = 2861 + 5^3 with 2861 prime. a(10404) = 1 since 10404 = 10061 + 7^3 with 10061 prime. a(46656) = 1 since 46656 = 431 + 215^2 with 431 prime. a(52900) = 1 since 52900 = 16963 + 33^3 with 16963 prime. a(112896) = 1 since 112896 = 112771 + 5^3 with 112771 prime.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Programs
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Maple
N:= 1000: # to get a(1) .. a(N) Primes:= select(isprime, [2,seq(i,i=3..N,2)]): Powers:= {0,1,seq(seq(b^k,k=2..floor(log[b](N))),b=2..floor(sqrt(N)))}: G:= expand(add(x^p,p=Primes)*add(x^r,r=Powers)): seq(coeff(G,x,i),i=1..N); # Robert Israel, Sep 27 2016 -
Mathematica
Do[r=0;Do[Do[If[IntegerQ[(n-Prime[j])^(1/k)],r=r+1;Goto[aa]],{k,2,If[n-Prime[j]>1,Log[2,n-Prime[j]],2]}];Label[aa];Continue,{j,1,PrimePi[n]}];Print[n," ",r];Continue,{n,1,80}]
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