A262409 Positive integers m such that pi(m^3) = pi(j^3) + pi(k^3) for some 0 < j <= k < m.

4, 89, 97, 101, 110, 196, 237, 372, 410, 1457, 2522, 3327, 4244, 4437, 5684, 5777, 7647, 8827, 9608, 9680, 9807, 10744, 17563, 19146, 21208, 23188, 27153, 28286, 34086, 35443, 40057, 49338, 49613, 54425, 55360, 56906, 61304, 69147, 69515, 73694, 84508, 95674

Offset: 1

Author

Zhi-Wei Sun, Sep 22 2015

Keywords

• nonn

Comments

Conjecture: The Diophantine equation pi(x^3) + pi(y^3) = pi(z^3) with 0 < x <= y < z has infinitely many solutions.

The 25 terms we have found yield the following 25 solutions to the equation: (x,y,z) = (3,3,4), (54,80,89), (63,85,97), (27,100,101), (47,106,110), (80,190,196), (122,223,237), (229,335,372), (151,401,410), (263,1453,1457), (1302,2382,2522), (879,3301,3327), (2190,4011,4244), (498,4434,4437), (3792,4991,5684), (4496,4584,5777), (3113,7442,7647), (5239,8090,8827), (6904,8116,9608), (5659,8910,9680), (5323,9187,9807), (5527,10168,10744), (7395,17050,17563), (11637,17438,19146), (4486,21125,21208).

See also the conjecture in A262408 involving the n-th powers with n = 2,4,5,....

Solution triples (x,y,z) corresponding to a(n) for n = 26..42: (16440, 19774, 23188), (4775, 27091, 27153), (10708, 27687, 28286), (25272, 28248, 34086), (6302, 35360, 35443), (3941, 40040, 40057), (16336, 48639, 49338), (33631, 43365, 49613), (6206, 54390, 54425), (6741, 55317, 55360), (28160, 54247, 56906), (25339, 59637, 61304), (41473, 63300, 69147), (27684, 67825, 69515), (29690, 71841, 73694), (65989, 67172, 84508), (55781, 88294, 95674) - Chai Wah Wu, May 24 2018

Examples

			a(1) = 4 since pi(4^3) = pi(64) = 18 = 9 + 9 = pi(27) + pi(27) = pi(3^3) + pi(3^3).
a(2) = 89 since pi(89^3) = 56924 = 14479 + 42445 = pi(157464) + pi(512000) = pi(54^3) + pi(80^3).
a(22) = 10744 since pi(10744^3) = pi(1240217910784) = 46266787130 = 6805722064 + 39461065066 = pi(168837298183) + pi(1051251461632) = pi(5527^3) + pi(10168^3).
a(23) = 17563 since pi(17563^3) = pi(5417464872547) = 191548794617 = 15745791385 + 175803003232 = pi(404403154875) + pi(4956477625000) = pi(7395^3) + pi(17050^3).
a(24) = 19146 since pi(19146^3) = pi(7018336124136) = 245897610272 = 58267274193 + 187630336079 = pi(1575879851853) + pi(5302614071672) = pi(11637^3) + pi(17438^3).
a(25) = 21208 since pi(21208^3) = pi(9538918630912) = 330649999352 = 3733416265 + 326916583087 = pi(90277143256) + pi(9427361328125) = pi(4486^3) + pi(21125^3).
		

References

• 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.

Links

• Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.

Crossrefs

Cf. A000720, A000578, A262403, A262408, A262443.

Programs

Mathematica

f[n_]:=PrimePi[n^3]
T[1]:={0}
T[n_]:=Union[T[n-1],{f[n]}]
Do[n=0;Do[If[MemberQ[T[m-1],f[m]-f[k]],n=n+1;Print[n," ",m];Goto[aa]],{k,1,m-1}];Label[aa];Continue,{m,1,21350}]

Extensions

a(26)-a(42) from Chai Wah Wu, May 24 2018