

A262722


Positive integers m such that pi(k^3+m^3) is a cube for some k = 1..m, where pi(x) denotes the number of primes not exceeding x.


1



1, 41, 56, 74, 103, 157, 384, 491, 537, 868, 1490, 1710, 4322, 4523, 4877, 4942, 5147, 5407, 7564, 17576, 67722, 131455, 220641, 438895, 443475, 553878, 571473, 625611
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OFFSET

1,2


COMMENTS

Conjecture: (i) There are infinitely many distinct primes p,q,r such that pi(p^2+q^2) = r^2.
(ii) The Diophantine equation pi(x^3+y^3) = z^3 with 0 < x <= y and z > 0 only has the following 13 solutions: (x,y,z) = (1,1,1), (5,41,19), (47,56,29), (28,74,33), (2,103,44), (3,103,44), (6,157,65), (235,384,160), (266,491,198), (91,537,206), (359,868,331), (783,1490,565), (1192,1710,677).
(iii) The Diophantine equation pi(x^n+y^n) = z^n with n > 3 and x,y,z > 0 has no solution.
Part (ii) of the conjecture implies that the current sequence only has 12 terms as shown here.
Conjecture (ii) is false as there are more terms beyond 1710. It is likely the sequence has an infinite number of terms. (x,y,z) for 1710 < y <= 7564: (1429, 4322, 1514), (1974, 4523, 1604), (3361, 4877, 1840), (3992, 4942, 1949), (3253, 5147, 1902), (971, 5407, 1859), (935, 7564, 2563).  Chai Wah Wu, Apr 12 2021
Solutions (x,y,z) for 7564 < y <= 67722: (3484, 17576, 5783), (25184, 67722, 21604).  Chai Wah Wu, Apr 17 2021
Solutions (x,y,z) for 67722 < y <= 625611: (61021, 131455, 41715), (93577, 220641, 68507), (394510, 438895, 155930), (3086, 443475, 131933), (338485, 553878, 175133), (239982, 571473, 172855), (610794, 625611, 228409).  Chai Wah Wu, Apr 26 2021


REFERENCES

ZhiWei 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 ChinaJapan Seminar (Fukuoka, Oct. 28  Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169187.


LINKS



EXAMPLE

a(2) = 41 since pi(5^3+41^3) = pi(125+68921) = pi(69046) = 6859 = 19^3.


MATHEMATICA

f[x_, y_]:=PrimePi[x^3+y^3]
CQ[n_]:=IntegerQ[n^(1/3)]
n=0; Do[Do[If[CQ[f[x, y]], n=n+1; Print[n, " ", y]; Goto[aa]], {x, 1, y}]; Label[aa]; Continue, {y, 1, 1800}]


CROSSREFS

Cf. A000040, A000290, A000578, A000720, A262408, A262409, A262443, A262462, A262536, A262698, A262700, A262707.


KEYWORD

nonn,more,hard


AUTHOR



EXTENSIONS



STATUS

approved



