cp's OEIS Frontend

A230640 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).

%I A230640 #27 Sep 05 2024 15:36:07
%S A230640 0,4,28,248,129140168,68630377364912,
%T A230640 2088595827392656793085408064780643444068898148936888424953199350296
%N A230640 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 A230640 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 A230640 f:=proc(n) option remember; local B, M;
%p A230640 if n<=1 then RETURN([0, 0]);
%p A230640 else
%p A230640 B:=(f(ceil(n/2))[2] + f(floor(n/2))[2] + 2)/2;
%p A230640 M:=3^B+f(floor(n/2))[2]+1; RETURN([B, M]); fi;
%p A230640 end proc;
%p A230640 [seq(f(n)[2], n=1..7)];
%Y A230640 Cf. A230639.
%Y A230640 Related base-3 sequences: A053735, A134451, A230641, A230642, A230643, A230853, A230854, A230855, A230856, A230639, A230640, A010063 (trajectory of 1)
%Y A230640 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 A230640 nonn,base
%O A230640 1,2
%A A230640 _N. J. A. Sloane_, Oct 31 2013