OFFSET
1,1
COMMENTS
We prove that the solutions of 4^x + 4^y + 4^z = k^2 are (x,y,2y-x-1), for any arbitrary integer x,y. We calculate z. 4^x + 4^y + 4^z is square if positive integers m and odd integer t are such as : 1 + 4^(y-x) + 4^(z-x) = (1 + t*2^m)^2, that's why : (1 + 4^(z-y)*( 4^(y-x)) = t(1 + t*2^(m+1)) t.2^(m+1), and then m = 2y - 2x - 1. If we report this value in the precedent equation, we obtain : t-1 = (2^(z-2y+x+1) + t)(2^(z-2y+x+1) - t) * 4^(y-x-1). Because t is odd, z = 2y - x - 1. Finally, this values gives the square (2^x + 2^(2y-x-1))^2 = k^2.
From Frederik P.J. Vandecasteele, Jun 06 2025: (Start)
For a given n, the exponents are x = A384688(n-1), y = A138099(n), z = A000267(n-1) so that a(n) = 2^A384688(n-1) + 2^A000267(n-1).
Terms are all and only those k whose binary expansion is two 1 bits an odd distance apart. (End)
REFERENCES
T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.
J. M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres. Ellipses, 2004.
H. N. Shapiro, Introduction to the Theory of Numbers, John Wiley & Sons, 1983.
LINKS
David A. Corneth, Table of n, a(n) for n = 1..9999 (terms <= 10^60)
FORMULA
k = 2^x + 2^(2y-x-1), and z = 2y - x - 1.
Conjecture: a(n) = 3*A263132(n). - George Beck, May 05 2021
EXAMPLE
x = 0, y = 1 then z = 1, and k = 3.
x = 1, y = 2 then z = 2, and k = 6.
x = 0, y = 2 then z = 3, and k = 9.
MAPLE
for x from 0 to 1000 do :for y from x to 1000 do: n := evalf(2^x + 2^(2*y-x-1)): print (n) ; od :od :
MATHEMATICA
Take[Union[Select[Sqrt[Flatten[Table[(2^x + 2^(2*y - x - 1))^2, {x, 0, 13}, {y, 0, 13}]]], IntegerQ]], 49] (* Jean-François Alcover, Sep 13 2011 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Michel Lagneau, Feb 12 2010
STATUS
approved