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A256179 Sequence of power towers in ascending order, using all possible permutations of 2s and 3s. 5

%I #17 Jan 02 2023 12:30:51

%S 2,3,4,8,9,16,27,81,256,512,6561,19683,65536,43046721,134217728,

%T 7625597484987,2417851639229258349412352,

%U 443426488243037769948249630619149892803,115792089237316195423570985008687907853269984665640564039457584007913129639936

%N Sequence of power towers in ascending order, using all possible permutations of 2s and 3s.

%C a(n) is found by treating the digits of A248907(n) as power towers, so the sequence starts 2, 3, 2^2=4, 2^3=8, 3^2=9, 2^(2^2)=16, 3^3=27, 3^(2^2)=81, 2^(2^3)=256...

%H Vladimir Reshetnikov, <a href="http://list.seqfan.eu/oldermail/seqfan/2015-March/014595.html">2-3 sequence puzzle</a>, SeqFan list, Mar 18 2015.

%F A256179 = A256229 o A248907 = A256229 o A032810 o A185969, i.e., a(n) = A256229(A248907(n)) = A256229(A032810(A185969(n))).

%F Recurrence: a(1)=2, a(2)=3, a(3)=2^2, a(4)=2^3, a(5)=3^2, a(6)=2^(2^2), a(7)=3^3, a(8)=3^(2^2), a(9)=2^(2^3), a(10)=2^(3^2), a(11)=3^(2^3), a(12)=3^(3^2); and for n>6, a(2n)=3^a(n-1), a(2n-1)=2^a(n-1). - _Vladimir Reshetnikov_, Mar 19 2015

%e a(12) = 19683 because A248907(12) = 332, and 3^(3^2) = 19683.

%o (PARI) A256179(n)=A256229(A248907[n])) \\ where A248907 is assumed to be defined as vector. - _M. F. Hasler_, Mar 19 2015

%Y Cf. A185969, A248907, A075877.

%K nonn

%O 1,1

%A _Bob Selcoe_, Mar 18 2015

%E More terms from _M. F. Hasler_, Mar 19 2015

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Last modified March 28 13:25 EDT 2024. Contains 371254 sequences. (Running on oeis4.)