A348332 a(n) = (2^n-1) * 2^(2^n-n).

2, 12, 224, 61440, 4160749568, 18158513697557839872, 337623910929368631717566993311207522304, 115339776388732929035197660848497720713218148788040405586178452820382218977280

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

Table of n, a(n) for n=1..8.

Art of Problem Solving, Kazakhstan National Olympiad 2015, Day 1, problem 2.

Diophante, A499 - Bon souvenir d'Astana (in French).

Index to sequences related to Olympiads.

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

Cf. A058891, A319511.

Sequence in context: A009525 A009683 A132879 * A101712 A156484 A333447

Adjacent sequences: A348329 A348330 A348331 * A348333 A348334 A348335

Keyword

nonn,easy

Author

Bernard Schott, Oct 13 2021

Status

approved