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%I #9 Mar 14 2015 18:31:09
%S 16,25,27,32,49,121,125,128,169,225,243,289,343,361,512,529,625,729,
%T 841,961,1000,1225,1331,1369,1681,1849,2025,2048,2187,2197,2209,2401,
%U 2809,3025,3125,3375,3481,3721,3969,4225,4489,4913,5041,5329,5625,5929,6241
%N Perfect powers (A001597) of the form 3p + q + 3, p & q are primes.
%C The sequence has repetitions since different p's and q's will give the same perfect power. Remove the andmap in the program if you want the repetitions.
%C Includes all perfect powers, pp, (A001597) congruent +/- 1 (modulo 6). Also if pp-9 or pp-12 is a prime or if (pp -2)/3 or (pp-3)/3 is a prime.
%C The number of perfect powers of the form 3p + q + 3 <= 10^n: 0,5,21,56,157,433,...,. - _Robert G. Wilson v_, Jun 21 2006
%C In the first one million integers there are 1111 perfect powers (A070428) of which only 433 of them are of the form 3p + q + 3.
%F a(n)=3p+q+3 where p and q are primes and a(n) is a perfect power.
%e a(5)=49 since 3*3+37+3=49 = 5*3+31+3 = 3*11+13+3 = 3*13+7+7 = 7^2.
%e 6859 = 19^3 is in the sequence because there are 116 different ways to combine primes of the form 3p + q + 3, beginning with p=5 & q=6841 and ending with p=2281 & q=13.
%p with(numtheory); egcd := proc(n) local L; L:=map(proc(z) z[2] end, ifactors(n)[2]); igcd(op(L)) end: PW:=[]: for z to 1 do for j from 1 to 100 do for k from 1 to 100 do p:=ithprime(j); q:=ithprime(k); x:=3*p+q+3; if egcd(x)>1 and andmap(proc(w) not(w[3]=x) end, PW) then PW:=[op(PW), [p,q,x]] fi od od od; PW; map(proc(z) z[3] end, PW);
%t fQ[n_] := GCD @@ Last /@ FactorInteger@n > 1; lst = {}; Do[p = Prime@j; q = Prime@k; x = 3p + q + 3; If[fQ@x, AppendTo[lst, x]], {j, 340}, {k, PrimePi[6856 - 3Prime@j]}]; Union@lst (* _Robert G. Wilson v_ *)
%Y Cf. A000040, A001597.
%K nonn
%O 1,1
%A _Walter Kehowski_, Oct 05 2005
%E Edited, corrected and extended by _Robert G. Wilson v_, Jun 21 2006
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