Diophantine equation x+y^2+z^3=n^4

[zak seidov]

Dear seqfans,

I've just submitted:

solutions to equation x+y^2+z^3=n^4 (see below).
 
1) Can anyone check/extend entries?

2) What about case of all distinct x,y,z:
   (x-y)(x-z)(y-z)=/=0?

2a) n=2 3 (of total 5) solns:
    {x,y,z}={6,3,1},{7,1,2},{11,2,1};

2b) n=2 23 (of total 27) solns:
{x,y,z}={5,7,3},{8,3,4},{9,8,2},{13,2,4},{16,1,4},{16,8,1},{18,6,3},{24,7,2},{29,5,3},{31,7,1},{37,6,2},{38,4,3},{44,6,1},{48,5,2},{50,2,3},{53,1,3},{55,5,1},{57,4,2},{64,3,2},{64,4,1},{71,3,1},{72,1,2},{76,2,1}

2c) n=1..10: 0,3,23,69,155,293,508,799,1205,1732 solns
 

thanks, Zak


%I A000001
%S A000001 0, 5, 27, 75, 163, 303, 521, 815, 1223,
1753
%N A000001 a(n) = number of solutions to the
Diophantine equation 
x+y^2+z^3=n^4 with positive x,y,z.
%e A000001 a(2)=5 because there are five solutions to
equation 
x+y^2+z^3=2^4 with
{x,y,z}={4,2,2},{6,3,1},{7,1,2},{11,2,1},{14,1,1};
a(3)=27 because there are 27 solutions to equation
x+y^2+z^3=3^4 with
{x,y,z}={1,4,4},{5,7,3},{8,3,4},{9,8,2},{13,2,4},{16,1,4},{16,8,1},{18,6,3},{24,7,2},{29,5,3},{31,7,1},{37,6,2},{38,4,3},{44,6,1},{45,3,3},{48,5,2},{50,2,3},{53,1,3},{55,5,1},{57,4,2},{64,3,2},{64,4,1},{69,2,2},{71,3,1},{72,1,2},{76,2,1},{79,1,1}
%O A000001 1
%K A000001 ,more,nice,nonn,
%A A000001 Zak Seidov (zakseidov at yahoo.com), Sep 09
2006



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