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).
0, 5, 129, 4102, 87112285931760246646623899502532662132742, 1852673427797059126777135760139006525652319754650249024631321344126610074239106
Offset: 1
Keywords
Examples
The terms are a(1) = 0, a(2) = 2^2+0+1, a(3) = 2^7+0+1, a(4) = 2^12+5+1, a(5) = 2^136+5+1, a(6) = 2^160+129+1, a(7) = 2^4233+129+1, a(8) = 2^8206+4102+1, a(9) = 2^k+4102+1 with k=2^136+4110, ... . The length (in bits) of the n-th term is A230302(n)+1.
Links
- Max A. Alekseyev and N. J. A. Sloane, On Kaprekar's Junction Numbers, arXiv:2112.14365, 2021; Journal of Combinatorics and Number Theory 12:3 (2022), 115-155.
- Index entries for Colombian or self numbers and related sequences
Crossrefs
Programs
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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((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)];
Formula
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(7) = 2^4233+130 = 2^2^2^2^2^.88303276...
Note also that i+2 = A230864(a(n)).
Extensions
a(1)-a(8) were found by Donovan Johnson, Oct 22 2013.
Comments