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A230640
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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).
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20
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OFFSET
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1,2
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LINKS
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Max A. Alekseyev and N. J. A. Sloane, On Kaprekar's Junction Numbers, arXiv:2112.14365, 2021; Journal of Combinatorics and Number Theory, 2022 (to appear).
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MAPLE
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f:=proc(n) option remember; local B, M;
if n<=1 then RETURN([0, 0]);
else
B:=(f(ceil(n/2))[2] + f(floor(n/2))[2] + 2)/2;
M:=3^B+f(floor(n/2))[2]+1; RETURN([B, M]); fi;
end proc;
[seq(f(n)[2], n=1..7)];
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CROSSREFS
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Related base-3 sequences: A053735, A134451, A230641, A230642, A230643, A230853, A230854, A230855, A230856, A230639, A230640, A010063 (trajectory of 1)
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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