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%I #20 Mar 16 2023 08:50:59
%S 1,4,7,9,13,16,19,21,25,28,31,36,37,39,43,49,49,52,57,61,63,64,67,73,
%T 76,79,81,84,91,91,93,97,100,103,109,111,112,117,121,124,127,129,133,
%U 133,139,144,147,148,151,156,157,163,169,169,171,172,175,181,183
%N Consider the Diophantine equations x+y=a, x^2+y^2=b, x^3+y^3=c. There are three values of a for each pair b,c with b>0 and c>0. Sequence gives values of b.
%C From _Andrew Howroyd_, Mar 14 2023: (Start)
%C The following information was extracted from the Derive script and notes.
%C The values of x and y are not required to be integers or even real numbers.
%C Eliminating x and y from the equations gives:
%C a^3 - 3*a*b + 2*c = 0.
%C This is the Diophantine equation that is being solved. For a given b and c there will be three possibly complex roots. It is required to find those values of b and c where all three roots are integers.
%C The values for b are the integers of the form (d^2 + d*e + e^2)/3 where d and e are positive integers and d <= e. This sequence lists these values in order with repetition.
%C The corresponding values for c are d*e*(d+e)/2 given in A108940, and the three values for a are d, e and -(d+e).
%C (End)
%H Andrew Howroyd, <a href="/A108287/b108287.txt">Table of n, a(n) for n = 1..1000</a>
%H James R FitzSimons, <a href="/A108287/a108287.txt">Derive program</a>
%o (Derive) See Links section
%o (PARI) upto(blim)={my(L=List()); for(e=1, sqrtint(3*blim), for(d=1, e, my(b=(d^2+d*e+e^2)/3); if(b<=blim && !frac(b), listput(L, b)))); listsort(L); Vec(L)} \\ _Andrew Howroyd_, Mar 13 2023
%Y For values of c see A108940.
%K nonn
%O 1,2
%A James R FitzSimons (cherry(AT)getnet.net), Jun 22 2007
%E Terms a(42) and beyond from _Andrew Howroyd_, Mar 14 2023
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