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%I #30 Jun 24 2023 01:29:21
%S 1,2,4,7,8,9,10,11,14,16,18,21,22,23,25,26,28,32,33,34,35,36,37,38,40,
%T 44,46,49,50,56,57,63,64,65,70,72,78,81,84,86,88,90,91,92,95,98,99,
%U 100,104,105,110,112,114,121,122,126,128,129,130,132,136,140,144,148,152,154,158,160,162,169,170,175,176,177,183,184,189,190,193,196,198,200
%N Positive integers x such that there exist positive integers y and z satisfying x^3 + y^3 = z^2.
%C A001105 is a subset (excluding 0), since (x, y, z) = (A001105(k), A001105(k), A033430(k)) satisfies x^3 + y^3 = z^2. - _R. J. Mathar_, Sep 11 2006
%C A parametric solution: {x,y,z} = {g*(4*e + g)*(4*e^2 + 8*e*g + g^2), 2*g*(4*e + g)*(-2*e^2 +2*e*g + g^2), 3*g^2*(4*e + g)^2*(4*e^2 + 2*e*g + g^2)}, provided (-2*e^2 +2*e*g + g^2) > 0. - _James Mc Laughlin_, Jan 27 2007
%C Allowing y = 0 would give the same sequence, since x^3 = z^2 implies x is a square, and all squares are terms since (t^2)^3 + (2*t^2)^3 = (3*t^3)^2. On the other hand, allowing y to be negative would introduce new terms: 71, 74, and 155 would be terms since 71^3 + (-23)^3 = 588^2, 74^3 + (-47)^3 = 549^2, and 155^3 + (-31)^3 = 1922^2. See A356720. - _Jianing Song_, Aug 24 2022
%H Fritz Beukers, <a href="https://projecteuclid.org/euclid.dmj/1077231890">The Diophantine equation Ax^p+By^q=Cz^r</a>, Duke Math. J. 91 (1998), 61-88.
%e x=7, y=21, 7^3 + 21^3 = 98^2. 7 is the 4th term in the list.
%e Other solutions are (x, y, z)=(1, 2, 3), (4, 8, 24), (7, 21, 98), (9, 18, 81), (10, 65, 525), (11, 37, 228), (14, 70, 588), (16, 32, 192), (21, 7, 98), (22, 26, 168), (23, 1177, 40380), ...
%o (Magma) [ k : k in [1..200] | exists{P : P in IntegralPoints(EllipticCurve([0,k^3])) | P[1] gt 0 and P[2] ne 0 } ]; // Geoff Bailey, Jan 28 2007
%Y See A103255 for another version.
%K nonn
%O 1,2
%A _Cino Hilliard_, Mar 20 2005
%E Recomputed and extended to 48 terms by Geoff Bailey (geoff(AT)maths.usyd.edu.au) using Magma, Jan 28 2007
%E Terms 104..200 added by _Joerg Arndt_, Sep 29 2012
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