OFFSET
1,3
COMMENTS
In the binary representation of n, swallow all zeros from the right, then add the number of swallowed zeros, and subtract 1. - Ralf Stephan, Aug 22 2013
LINKS
FORMULA
a(1) = 0; thereafter a(2*n) = a(n) + 1, a(2*n+1) = 2*n.
G.f.: sum(k>=0, (t^2+2t^3-t^4)/(1-t^2)^2, t=(x^2)^k).
a((2*n-1)*2^p) = p + 2*(n-1), p >= 0. - Johannes W. Meijer, Jan 23 2013
MAPLE
nmax:=73: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := p + 2*(n-1) od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Jan 23 2013
MATHEMATICA
a[n_] := With[{v = IntegerExponent[n, 2]}, v + n/2^v - 1];
Array[a, 100] (* Jean-François Alcover, Feb 28 2019 *)
PROG
(PARI) a(n) = valuation(n, 2) + n/2^valuation(n, 2) - 1
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ralf Stephan, Oct 30 2003
STATUS
approved