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A256179
Sequence of power towers in ascending order, using all possible permutations of 2's and 3's.
5
2, 3, 4, 8, 9, 16, 27, 81, 256, 512, 6561, 19683, 65536, 43046721, 134217728, 7625597484987, 2417851639229258349412352, 443426488243037769948249630619149892803, 115792089237316195423570985008687907853269984665640564039457584007913129639936
OFFSET
1,1
COMMENTS
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...
LINKS
Pontus von Brömssen, Table of n, a(n) for n = 1..22
Vladimir Reshetnikov, 2-3 sequence puzzle, SeqFan list, Mar 18 2015.
FORMULA
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
EXAMPLE
a(12) = 19683 because A248907(12) = 332, and 3^(3^2) = 19683.
a(23) = 2^3^2^3 = 11423...73952 (1976 digits), because A248907(23) = 2323.
PROG
(PARI) A256179(n)=A256229(A248907[n]) \\ where A248907 is assumed to be defined as vector. - M. F. Hasler, Mar 19 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Bob Selcoe, Mar 18 2015
EXTENSIONS
More terms from M. F. Hasler, Mar 19 2015
STATUS
approved