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A255679 Least number k such that pi(k*n)^2 = pi(i*n)^2 + pi(j*n)^2 for some 0 < i < j < k.

%I A255679 #14 Jul 11 2015 09:43:14
%S A255679 4,6,4,12,6,17,17,6,11,6,13,14,4,10,31,35,10,8,20,20,70,20,34,23,47,
%T A255679 17,44,25,5,46,4,26,53,13,29,33,35,45,65,10,18,64,117,21,77,19,71,101,
%U A255679 44,80,28,93,36,183,100,98,55,4,53,81,31,120,58,40,53,56,34,49,181,218
%N A255679 Least number k such that pi(k*n)^2 = pi(i*n)^2 + pi(j*n)^2 for some 0 < i < j < k.
%C A255679 Conjecture: a(n) exists for any n > 0. In other words, for each n = 1,2,3,... the sequence pi(k*n) (k = 1,2,3,...) contains a Pythagorean triple.
%D A255679 Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.
%H A255679 Zhi-Wei Sun, <a href="/A255679/b255679.txt">Table of n, a(n) for n = 1..240</a>
%H A255679 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.6641">Problems on combinatorial properties of primes</a>, arXiv:1402.6641 [math.NT], 2014.
%e A255679 a(1) = 4 since pi(4*1)^2 = 2^2 = pi(1*1)^2 + pi(3*1)^2.
%e A255679 a(21) = 70 since pi(70*21)^2 = pi(1470)^2 = 232^2 = 160^2 + 168^2 = pi(45*21)^2 + pi(48*21)^2.
%t A255679 Do[k=0;Label[bb];k=k+1;Do[If[PrimePi[k*n]^2==PrimePi[i*n]^2+PrimePi[j*n]^2,Goto[aa],Goto[cc]];Label[cc];Continue,{j,1,k-1},{i,1,j-1}];Goto[bb];
%t A255679 Label[aa];Print[n," ",k];Continue,{n,1,70}]
%o A255679 (PARI) is_ok(n,k)={ r=0; for(i=1,k-1, for(j=i+1, k-1, if(primepi(k*n)^2 == primepi(i*n)^2 + primepi(j*n)^2,r=1;break(2)) )); return(r);}
%o A255679 a(n)={ k=0; while(!is_ok(n,k), k++); return(k);}
%o A255679 main(size)={ v=vector(size); i=1; m=1; while(i<=size, v[i]=a(i);i++); return(v);} /* _Anders Hellström_, Jul 11 2015 */
%Y A255679 Cf. A000290, A000720, A255677.
%K A255679 nonn
%O A255679 1,1
%A A255679 _Zhi-Wei Sun_, Jul 11 2015