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%I #84 Jun 19 2023 08:13:34
%S 3,4,24,32,81,98,108,168,192,228,256,312,375,500,525,588,648,671,784,
%T 847,864,1014,1029,1183,1225,1261,1323,1344,1372,1536,1824,2048,2187,
%U 2496,2646,2888,2916,3000,3993,4000,4200,4225,4536,4563,4644,4704,5184,5324
%N Numbers k such that k^2 is expressible as the sum of two positive cubes in at least one way.
%C Analogous solutions exist for the sum of two identical cubes z^2 = 2*r^3 (e.g., 864^2 = 2*72^3). Values of 'z' are the terms in A033430, values of 'r' are the terms in A001105.
%C First term whose square can be expressed in two ways is 77976; 77976^2 = 228^3 + 1824^3 = 1026^3 + 1710^3. - _Jud McCranie_
%C First term whose square can be expressed in three ways is 3343221000; 3343221000^2 = 279300^3 + 2234400^3 = 790020^3 + 2202480^3 = 1256850^3 + 2094750^3.
%C First term whose square can be expressed in four ways <= 42794271007595289; 42794271007595289^2 = 14385864402^3 + 122279847417^3 = 55172161278^3 + 118485773289^3 = 64117642953^3 + 116169722214^3 = 96704977369^3 + 97504192058^3.
%C First term whose square can be expressed in five ways <= 47155572445935012696000; 47155572445935012696000^2 = 94405759361550^3 + 1305070263601650^3 = 374224408544280^3 + 1294899176535720^3 = 727959282778000^3 + 1224915311765600^3 = 857010857812200^3 + 1168192425418200^3 = 1009237516560000^3 + 1061381454915600^3.
%C After a(1) = 3 this is always composite, because factorization of the polynomial a^3 + b^3 into irreducible components over Z is a^3 + b^3 = (b+a)*(b^2 - ab + b^2). They may be semiprimes, as with 671 = 11 * 61, and 1261 = 13 * 97. The numbers can be powers in various ways, as with 32 = 2^5, 81 = 3^4, 256 = 2^8, 784 = 2^4 * 7^2 , 1225 = 5^2 * 7^2, and 2187 = 3^7. - _Jonathan Vos Post_, Feb 05 2011
%C If n is a term then n*b^3 is also a term for any b, e.g., 3 is a term hence 3*2^3 = 24, 3*3^3 = 81 and also 3*4^3 = 192 are terms. Sequence of primitive terms may be of interest. - _Zak Seidov_, Dec 11 2013
%C First noncubefree primitive term is 168 = 21*2^3 (21 is not a term of the sequence). - _Zak Seidov_, Dec 16 2013
%C From _XU Pingya_, Apr 10 2021: (Start)
%C Every triple (a, b, c) (with a^2 = b^3 + c^3) can produce a nontrivial parametric solution (x, y, z) of the Diophantine equation x^2 + y^3 + z^3 = d^4.
%C For example, to (1183, 65, 104), there is such a solution (d^2 - (26968032*d)*t^3 + 1183*8232^3*t^6, (376*d)*t - 65*8232^2*t^4, (92*d)*t - 104*8232^2*t^4).
%C To (77976, 228, 1824), there is (d^2 - (272916*d)*t^3 + 77976*57^3*t^6, (52*d)*t - 228*57^2*t^4, (74*d)*t - 1824*57^2*t^4).
%C Or to (77976, 1026, 1719), there is (d^2 - (25992*d)*t^3 + 77976*19^3*t^6, (37*d)*t - 1026*19^2*t^4, (11*d)*t - 1710*19^2*t^4). (End)
%D Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124.
%H Chai Wah Wu, <a href="/A050801/b050801.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..612 from T. D. Noe, terms 613..1000 from Harry J. Smith)
%F a(n) = sqrt(A050802(n)). - _Jonathan Sondow_, Oct 28 2013
%e 1183^2 = 65^3 + 104^3.
%t Select[Range[5350], Reduce[0 < x <= y && #^2 == x^3 + y^3, {x,y}, Integers] =!= False &] (* _Jean-François Alcover_, Mar 30 2011 *)
%t Sqrt[#]&/@Union[Select[Total/@(Tuples[Range[500],2]^3),IntegerQ[ Sqrt[ #]]&]] (* _Harvey P. Dale_, Mar 06 2012 *)
%t Select[Range@ 5400, Length@ DeleteCases[PowersRepresentations[#^2, 2, 3], w_ /; Times @@ w == 0] > 0 &] (* _Michael De Vlieger_, May 20 2017 *)
%o (PARI) nstart=1; astart=3; n=nstart; a=astart-1; until (0, a++; a2=a^2; b1=((a2/2)^(1/3))\1; for (b=b1, a, b3=b^3; c1=1; if (a2 > b3, c1=((a2-b3)^(1/3))\1;); for (c=c1, b, d=b3 + c^3; if (d > a2 && c == 1, break(2)); if (d > a2, break); if (a2 == d, print(n, " ", a); write("b050801.txt", n, " ", a); n++; break(2))))) \\ _Harry J. Smith_, Jan 15 2009
%o (PARI) is(n)=my(N=n^2); for(k=sqrtnint(N\2,3),sqrtnint(N-1,3), if(ispower(N-k^3,3), return(n>1))); 0 \\ _Charles R Greathouse IV_, Dec 13 2013
%Y Cf. A050802, A000404, A033430, A001105, A038597, A050803, A106265, A217248.
%K nonn,nice
%O 1,1
%A _Patrick De Geest_, Sep 15 1999
%E More terms from _Michel ten Voorde_ and _Jud McCranie_
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