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A262408
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Positive integers m such that pi(m^2) = pi(j^2) + pi(k^2) for no 0 < j <= k < m.
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12
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1, 2, 5, 10, 21, 23, 46, 103, 105, 193, 222, 232, 285, 309, 345, 392, 404, 476, 587, 670, 779, 912, 1086, 1162, 1249, 2508, 2592, 2852, 2964, 3362, 3673, 3895, 4218, 4732, 5452, 6417, 7667, 7759, 8430, 8796, 9606, 11096, 11953, 12014, 12125, 13956, 14474, 15018, 17854, 18861, 18879, 19307, 22843, 28106, 29423, 31576, 37182
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OFFSET
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1,2
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COMMENTS
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Conjecture: (i) There are infinitely many positive integers m such that pi(m^2) = pi(x^2) + pi(y^2) for some 0 < x <= y < m. Also, the current sequence has infinitely many terms.
(ii) For every n = 4,5,... the equation pi(x^n) + pi(y^n) = pi(z^n) has no integral solution with 0 < x <= y < z.
It is interesting to compare this conjecture with Fermat's Last Theorem. See also A262409 for the equation pi(x^3) + pi(y^3) = pi(z^3).
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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(3) = 5 since pi(5^2) = 9 cannot be written as pi(j^2) + pi(k^2) with 0 < j <= k < 5. Note that pi(1^2) = 0, pi(2^2) = 2, pi(3^2) = 4 and pi(4^2) = 6 are all even.
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MATHEMATICA
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f[n_]:=PrimePi[n^2]
T[n_]:=Table[f[k], {k, 1, n}]
n=0; Do[Do[If[MemberQ[T[m-1], f[m]-f[k]], Goto[aa]], {k, 1, m-1}]; n=n+1; Print[n, " ", m]; Label[aa]; Continue, {m, 1, 32000}]
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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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