A387127 - OEIS

%I #31 Nov 09 2025 12:32:09

%S 2,21,13,41,8,73,145,113,201,393,377,1489,793,1097,1657,2161

%N a(n) is the smallest integer w such that the Diophantine equation x^2 + y^3 + z^5 = w^7 where GCD(x,y,z)=1 has exactly n positive integer solutions.

%H Mrexcel, <a href="https://math.stackexchange.com/questions/5097413/efficient-algorithm-for-solving-diophantine-equation-x-2y-3z-5-w-7-w">Efficient algorithm for solving Diophantine equation x^2+y^3+z^5=w^7 with gcd(x,y,z)=1</a>.

%H Chinese BBS, <a href="https://bbs.emath.ac.cn/thread-50064-1-1.html">An efficient algorithm of x^2 + y^3 + z^5 = w^7, GCD(x, y, z) = 1</a>.

%e a(3) = 13 because 13^7 = 7783^2 + 91^3 + 17^5 = 5921^2 + 277^3 + 23^5 = 6484^2 + 58^3 + 29^5 and for no integer smaller than 13 we have 3 solutions.

%t f[w_] := (sol = {}; w7 = w^7;

%t   Do[yy = w7 - z^5; Do[xx = yy - y^3; x = Sqrt@xx;

%t     If[IntegerQ@x, If[GCD[x, y, z] == 1, AppendTo[sol, {x, y, z}];]],

%t {y, Floor[yy^(1/3)]}], {z, Floor[w7^(1/5)]}]; {Length@sol, w, sol});

%t Do[w=1;Until[f[w][[1]]==n,w++];Print[f[w]],{n,5}]

%o (Python)

%o from itertools import count

%o from math import isqrt, gcd

%o from sympy.ntheory.primetest import is_square

%o def A387127(n):

%o     for w in count(1):

%o         c = 0

%o         w7 = w**7

%o         for z in count(1):

%o             z5 = w7-z**5

%o             if z5<2:

%o                 break

%o             for y in count(1):

%o                 y3 = z5-y**3

%o                 if y3<1:

%o                     break

%o                 if is_square(y3):

%o                     x = isqrt(y3)

%o                     if gcd(x,y,z)==1:

%o                         c += 1

%o                         if c>n:

%o                             break

%o         if c == n:

%o             return w # _Chai Wah Wu_, Nov 09 2025

%Y Cf. A386373, A386988.

%K nonn,more

%O 1,1

%A _Zhining Yang_, Aug 17 2025

%E a(12), a(14)-a(16) from _Zhao Hui Du_, Sep 22 2025