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Let M(1)=0 and for n>1, B(n)=(M(ceiling(n/2))+M(floor(n/2))+2)/2, M(n)=3^B(n)+M(floor(n/2))+1. This sequence gives M(n).
20

%I #27 Sep 05 2024 15:36:07

%S 0,4,28,248,129140168,68630377364912,

%T 2088595827392656793085408064780643444068898148936888424953199350296

%N Let M(1)=0 and for n>1, B(n)=(M(ceiling(n/2))+M(floor(n/2))+2)/2, M(n)=3^B(n)+M(floor(n/2))+1. This sequence gives M(n).

%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.

%p f:=proc(n) option remember; local B, M;

%p if n<=1 then RETURN([0, 0]);

%p else

%p B:=(f(ceil(n/2))[2] + f(floor(n/2))[2] + 2)/2;

%p M:=3^B+f(floor(n/2))[2]+1; RETURN([B, M]); fi;

%p end proc;

%p [seq(f(n)[2], n=1..7)];

%Y Cf. A230639.

%Y Related base-3 sequences: A053735, A134451, A230641, A230642, A230643, A230853, A230854, A230855, A230856, A230639, A230640, A010063 (trajectory of 1)

%Y Smallest number m such that u + (sum of base-b digits of u) = m has exactly n solutions, for bases 2 through 10: A230303, A230640, A230638, A230867, A238840, A238841, A238842, A238843, A006064.

%K nonn,base

%O 1,2

%A _N. J. A. Sloane_, Oct 31 2013