OFFSET
1,1
COMMENTS
Solutions z of the Diophantine equation x^y * y^x = (x+y)^z with 1 <= x <= y. This Diophantine equation was problem 2 of Day 1 for the 2015 Kazakhstan National Olympiad, see Art of Problem Solving and Diophante links. Some results:
-> x and y are both even.
-> This equation has solution iff x = y.
-> The equation becomes x^(2x) = (2x)^z.
-> Corresponding triples (x,y,z) that are solutions satisfy:
x = y = 2^(2^n-1) = A058891(n+1) for n >= 1, and,
z = a(n) = (2^n-1) * 2^(2^n-n) for n >= 1.
-> First triples solutions are (x,y,z) = (2,2,2), (8,8,12), (128,128,224), (32768,32768,61440), ...
LINKS
Art of Problem Solving, Kazakhstan National Olympiad 2015, Day 1, problem 2.
Diophante, A499 - Bon souvenir d'Astana (in French).
FORMULA
a(n) = 2 * A319511(n,n-1). - Kevin Ryde, Oct 13 2021
EXAMPLE
For n = 2, a(2) = (2^2-1) * 2^(2^2-2) = 3 * 4 = 12 = z; x = y = 2^(2^2-1) = 8 and 8^8 * 8^8 = (8+8)^12 = 281474976710656.
MAPLE
Sequence = seq((2^n-1)*2^(2^n-n), n=1..8);
MATHEMATICA
Table[(2^n - 1)*2^(2^n - n), {n, 1, 8}] (* Amiram Eldar, Oct 15 2021 *)
PROG
(PARI) a(n) = (2^n-1)<<(2^n-n) \\ Charles R Greathouse IV, Oct 20 2021
(Python)
def A348332(n): return ((1<<n)-1)<<(1<<n)-n # Chai Wah Wu, Dec 12 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bernard Schott, Oct 13 2021
STATUS
approved