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A050801
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Numbers k such that k^2 is expressible as the sum of two positive cubes in at least one way.
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22
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3, 4, 24, 32, 81, 98, 108, 168, 192, 228, 256, 312, 375, 500, 525, 588, 648, 671, 784, 847, 864, 1014, 1029, 1183, 1225, 1261, 1323, 1344, 1372, 1536, 1824, 2048, 2187, 2496, 2646, 2888, 2916, 3000, 3993, 4000, 4200, 4225, 4536, 4563, 4644, 4704, 5184, 5324
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OFFSET
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1,1
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COMMENTS
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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.
First term whose square can be expressed in two ways is 77976; 77976^2 = 228^3 + 1824^3 = 1026^3 + 1710^3. - Jud McCranie
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.
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.
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.
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
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
First noncubefree primitive term is 168 = 21*2^3 (21 is not a term of the sequence). - Zak Seidov, Dec 16 2013
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.
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).
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).
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)
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REFERENCES
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Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124.
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LINKS
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FORMULA
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EXAMPLE
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1183^2 = 65^3 + 104^3.
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MATHEMATICA
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Select[Range[5350], Reduce[0 < x <= y && #^2 == x^3 + y^3, {x, y}, Integers] =!= False &] (* Jean-François Alcover, Mar 30 2011 *)
Sqrt[#]&/@Union[Select[Total/@(Tuples[Range[500], 2]^3), IntegerQ[ Sqrt[ #]]&]] (* Harvey P. Dale, Mar 06 2012 *)
Select[Range@ 5400, Length@ DeleteCases[PowersRepresentations[#^2, 2, 3], w_ /; Times @@ w == 0] > 0 &] (* Michael De Vlieger, May 20 2017 *)
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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