A387127 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.

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

Offset

1,1

Links

Table of n, a(n) for n=1..16.

Mrexcel, Efficient algorithm for solving Diophantine equation x^2+y^3+z^5=w^7 with gcd(x,y,z)=1.

Chinese BBS, An efficient algorithm of x^2 + y^3 + z^5 = w^7, GCD(x, y, z) = 1.

Example

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.

Mathematica

f[w_] := (sol = {}; w7 = w^7;
  Do[yy = w7 - z^5; Do[xx = yy - y^3; x = Sqrt@xx;
    If[IntegerQ@x, If[GCD[x, y, z] == 1, AppendTo[sol, {x, y, z}]; ]],
{y, Floor[yy^(1/3)]}], {z, Floor[w7^(1/5)]}]; {Length@sol, w, sol});
Do[w=1; Until[f[w][[1]]==n, w++]; Print[f[w]], {n, 5}]

Prog

(Python)
from itertools import count
from math import isqrt, gcd
from sympy.ntheory.primetest import is_square
def A387127(n):
    for w in count(1):
        c = 0
        w7 = w**7
        for z in count(1):
            z5 = w7-z**5
            if z5<2:
                break
            for y in count(1):
                y3 = z5-y**3
                if y3<1:
                    break
                if is_square(y3):
                    x = isqrt(y3)
                    if gcd(x, y, z)==1:
                        c += 1
                        if c>n:
                            break
        if c == n:
            return w # Chai Wah Wu, Nov 09 2025

Crossrefs

Cf. A386373, A386988.

Sequence in context: A072397 A077208 A321534 * A212483 A084313 A140268

Adjacent sequences: A387124 A387125 A387126 * A387128 A387129 A387130

Keyword

nonn,more

Author

Zhining Yang, Aug 17 2025

Extensions

a(12), a(14)-a(16) from Zhao Hui Du, Sep 22 2025

Status

approved