

A230303


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).


12




OFFSET

1,2


COMMENTS

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).


REFERENCES

Max A. Alekseyev, Donovan Johnson and N. J. A. Sloane, On Kaprekar's Junction Numbers, in preparation, 2017.


LINKS

Table of n, a(n) for n=1..6.
Index entries for Colombian or self numbers and related sequences


FORMULA

Define i by 2^(i1) < 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)).


EXAMPLE

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 nth term is A230302(n)+1.


MAPLE

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((n1)/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)];


CROSSREFS

Cf. A228085, A230092, A227915, A230093, A230302 (for B(n)), A230864.
Smallest number m such that u + (sum of baseb digits of u) = m has exactly n solutions, for bases 2 through 10: A230303, A230640, A230638, A230867, A238840, A238841, A238842, A238843, A006064.
Sequence in context: A316986 A316392 A277259 * A094074 A012218 A012136
Adjacent sequences: A230300 A230301 A230302 * A230304 A230305 A230306


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Oct 24 2013; Mar 26 2014


STATUS

approved



