OFFSET
1,1
COMMENTS
Let i, j and k are nonegtive integers, m > n are positive integers. As [(m^2 - n^2)^(3i+1) * (2mn)^(3j + 2) * (m^2 + n^2)^(3k)]^2 + [(m^2 - n^2)^i * (2mn)^(j + 1) * (m^2 + n^2)^k]^6 = [(m^2 - n^2)^(3i) * (2mn)^(3j + 2) * (m^2 + n^2)^(3k+1)]^2, so that number of form (m^2 - n^2)^(3i) * (2mn)^(3j + 2) * (m^2 + n^2)^(3k+1) is a term in sequence.
When (x, y, z) is solution of x^2 + y^4 = z^2 (i.e., z = A271576(n)), (x^(3i+1) * y^(3j + 1) * z^(3k), (x^i * y^(j + 1) * z^k)^6, x^(3i) * y^(3j + 1) * z^(3k+1) is solution of x^2 + y^6 = z^2.
When (x, y, z) is solution of x^2 + y^6 = z^2, (x^(3i+1) * y^(3j) * z^(3k), x^(3i) * y^(j + 1) * z^k, x^(3i) * y^(3j) * z^(3k+1)) is also.
EXAMPLE
6^2 + 2^6 = 10^2, 10 is a term.
15^2 + 2^6 = 17^2, 17 is a term.
MATHEMATICA
z[n_] := Count[n^2 - Range[(n^2 - 1)^(1/6)]^6, _?(IntegerQ[Sqrt[#]] &)] > 0; Select[Range[3200], z]
CROSSREFS
KEYWORD
nonn
AUTHOR
XU Pingya, Oct 14 2017
STATUS
approved