A105670 a(1)=1 then bracketing n by powers of 2 as f(t)=2^t for f(t) < n <= f(t+1), a(n) = f(t+1) - a(n-f(t)).
1, 1, 3, 3, 7, 7, 5, 5, 15, 15, 13, 13, 9, 9, 11, 11, 31, 31, 29, 29, 25, 25, 27, 27, 17, 17, 19, 19, 23, 23, 21, 21, 63, 63, 61, 61, 57, 57, 59, 59, 49, 49, 51, 51, 55, 55, 53, 53, 33, 33, 35, 35, 39, 39, 37, 37, 47, 47, 45, 45, 41, 41, 43, 43, 127, 127, 125, 125, 121, 121, 123
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Programs
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Maple
A062383 := proc(n) ceil(log(n)/log(2)) ; 2^% ; end proc: A105670 := proc(n) option remember; if n = 1 then 1; else fn1 := A062383(n) ; fn := fn1/2 ; fn1-procname(n-fn) ; end if; end proc: seq(A105670(n),n=1..80) ; # R. J. Mathar, Nov 06 2011
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Mathematica
t[0] = 0; t[1] = 1; t[n_?EvenQ] := t[n] = t[n/2]; t[n_?OddQ] := t[n] = 1 - t[(n-1)/2]; a[1] = 1; a[n_?EvenQ] := a[n] = a[n - 1]; a[n_] := a[n] = 2*a[Ceiling[n/2]] - 1 + 2*t[Ceiling[n/2] - 1]; Table[a[n], {n, 1, 71}] (* Jean-François Alcover, Aug 13 2013 *)
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PARI
b(n,m)=if(n<2,1,m*m^floor(log(n-1)/log(m))-b(n-m^floor(log(n-1)/log(m)),m))
Formula
a(2n-1) = a(2n).
a(n) = 2*a(ceiling(n/2)) -1 + 2*t(ceiling(n/2)-1) where t(n) = A010060(n) is the Thue-Morse sequence.
a(2n-1) = a(2n) = 2*A006068(n-1)+1. - Jeffrey Shallit, Mar 15 2025
Extensions
Typo in data corrected by Jean-François Alcover, Aug 13 2013