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a(1)=0; thereafter a(n) = 2^(a(ceiling(n/2)) + a(floor(n/2))).
2

%I #46 Sep 05 2024 15:43:58

%S 0,1,2,4,8,16,64,256,4096,65536,16777216,4294967296,

%T 1208925819614629174706176,340282366920938463463374607431768211456,

%U 2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576

%N a(1)=0; thereafter a(n) = 2^(a(ceiling(n/2)) + a(floor(n/2))).

%C a(16) = 2^512

%C = 134078079299425970995740249982058461274793658205923933777235\

%C 614437217640300735469768018742981669034276900318581864860508537538828119465\

%C 69946433649006084096.

%H Max A. Alekseyev and N. J. A. Sloane, <a href="https://arxiv.org/abs/2112.14365">On Kaprekar's Junction Numbers</a>, arXiv:2112.14365, 2021; Journal of Combinatorics and Number Theory 12:3 (2022), 115-155.

%F In general, for n >= 11, define i by 9*2^(i-1) < n <= 9*2^i. Then it appears that a(n) = 2^2^2^...^2^x, a tower of height i+5, containing i+4 2's, where x is in the range 0 < x <= 1.

%F For example, if n=18, i=1, and a(18) = 2^8192 = 2^2^2^2^2^0.91662699..., of height 6.

%F Note also that i+5 = A230864(a(n)).

%p f:=proc(n) option remember;

%p if n=1 then 0 else 2^(f(ceil(n/2))+f(floor(n/2))); fi; end;

%p [seq(f(n),n=1..16)];

%Y Cf. A230864, A230874, A230875.

%K nonn

%O 1,3

%A _N. J. A. Sloane_, Nov 02 2013; revised Mar 26 2014