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A255679
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Least number k such that pi(k*n)^2 = pi(i*n)^2 + pi(j*n)^2 for some 0 < i < j < k.
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3
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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, 17, 44, 25, 5, 46, 4, 26, 53, 13, 29, 33, 35, 45, 65, 10, 18, 64, 117, 21, 77, 19, 71, 101, 44, 80, 28, 93, 36, 183, 100, 98, 55, 4, 53, 81, 31, 120, 58, 40, 53, 56, 34, 49, 181, 218
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OFFSET
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1,1
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COMMENTS
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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.
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REFERENCES
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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.
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LINKS
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EXAMPLE
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a(1) = 4 since pi(4*1)^2 = 2^2 = pi(1*1)^2 + pi(3*1)^2.
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.
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MATHEMATICA
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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];
Label[aa]; Print[n, " ", k]; Continue, {n, 1, 70}]
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PROG
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(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); }
a(n)={ k=0; while(!is_ok(n, k), k++); return(k); }
main(size)={ v=vector(size); i=1; m=1; while(i<=size, v[i]=a(i); i++); return(v); } /* Anders Hellström, Jul 11 2015 */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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