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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

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%I A348332 #41 Aug 22 2025 14:58:42
%S A348332 2,12,224,61440,4160749568,18158513697557839872,
%T A348332 337623910929368631717566993311207522304,
%U A348332 115339776388732929035197660848497720713218148788040405586178452820382218977280
%N A348332 a(n) = (2^n-1) * 2^(2^n-n).
%C A348332 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:
%C A348332 -> x and y are both even.
%C A348332 -> This equation has solution iff x = y.
%C A348332 -> The equation becomes x^(2x) = (2x)^z.
%C A348332 -> Corresponding triples (x,y,z) that are solutions satisfy:
%C A348332      x = y = 2^(2^n-1) = A058891(n+1) for n >= 1, and,
%C A348332      z = a(n) = (2^n-1) * 2^(2^n-n) for n >= 1.
%C A348332 -> First triples solutions are (x,y,z) = (2,2,2), (8,8,12), (128,128,224), (32768,32768,61440), ...
%H A348332 Art of Problem Solving, <a href="https://artofproblemsolving.com/community/c4048_2015_kazakhstan_national_olympiad">Kazakhstan National Olympiad 2015, Day 1, problem 2</a>.
%H A348332 Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a4-equations-diophantiennes/3721-a499-bon-souvenir-d-astana">A499 - Bon souvenir d'Astana</a> (in French).
%H A348332 <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.
%F A348332 a(n) = 2 * A319511(n,n-1). - _Kevin Ryde_, Oct 13 2021
%e A348332 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.
%p A348332 Sequence = seq((2^n-1)*2^(2^n-n), n=1..8);
%t A348332 Table[(2^n - 1)*2^(2^n - n), {n, 1, 8}] (* _Amiram Eldar_, Oct 15 2021 *)
%o A348332 (PARI) a(n) = (2^n-1)<<(2^n-n) \\ _Charles R Greathouse IV_, Oct 20 2021
%o A348332 (Python)
%o A348332 def A348332(n): return ((1<<n)-1)<<(1<<n)-n # _Chai Wah Wu_, Dec 12 2022
%Y A348332 Cf. A058891, A319511.
%K A348332 nonn,easy,changed
%O A348332 1,1
%A A348332 _Bernard Schott_, Oct 13 2021