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A230303
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Let M(1)=0 and for n >= 2, let B(n)=M(ceiling(n/2))+M(floor(n/2))+2, M(n)=2^B(n)+M(floor(n/2))+1; sequence gives M(n).
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12
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OFFSET
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1,2
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COMMENTS
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M(n) is the smallest value of k such that A228085(k) = n. For example, 129 is the first time a 3 appears in A228085 (and is therefore the first term in A230092). M(4) = 4102 is the first time a 4 appears in A228085 (and is therefore the first term in A227915).
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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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FORMULA
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Define i by 2^(i-1) < n <= 2^i. Then it appears that
a(n) = 2^2^2^...^2^x
a tower of height i+3, containing i+2 2's, where x is in the range 0 < x <= 1.
For example, if n=7, i=3, and
a(18) = 2^4233+130 = 2^2^2^2^2^.88303276...
Note also that i+2 = A230864(a(n)).
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EXAMPLE
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The terms are 0, 2^2+0+1, 2^7+0+1, 2^12+5+1, 2^136+5+1, 2^160+129+1, 2^4233+129+1, 2^8206+4102+1, 2^k+4102+1 with k=2^136+4110, ... .
The length (in bits) of the n-th term is A230302(n)+1.
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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
if (n mod 2) = 0 then B:=2*f(n/2)[2]+2;
else B:=f((n+1)/2)[2]+f((n-1)/2)[2]+2; fi;
M:=2^B+f(floor(n/2))[2]+1; RETURN([B, M]); fi;
end proc;
[seq(f(n)[2], n=1..6)];
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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