1,3

In other words, number of steps to reach 1 starting from n and using the process : x->x-1 if n is odd and x->x/2 otherwise

a(n)=floor(log_2(n)) + number of 1's in binary representation of n.

Let u(1)=1, u(2n)=u(n)+1, u(2n+1)=u(2n)+1; then a(1)=0 and a(n)=u(n-1) - Benoit Cloitre (benoit7848c(AT)orange.fr), Dec 19 2002

G.f.: -2/(1-x) + 1/(1-x) * sum(k>=0, (2x^2^k+x^2^(k+1))/(1+x^2^k)). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Aug 15 2003

5 -> 4 -> 2 -> 1 so 3 steps are needed to reach 1 hence a(5)=3; 9 -> 8 -> 4 -> 2 -> 1 hence a(9)=4

A014701 := proc(n) local j, k; j := n; k := 0; while(j>1) do if j mod 2=1 then j := j-1 else j := j/2 fi; k := k+1 od end;

Cf. A003313.

a(n) = A056792(n) - 1 = A056791(n) - 2.

Adjacent sequences: A014698 A014699 A014700 this_sequence A014702 A014703 A014704

Sequence in context: A117497 A117498 A064097 this_sequence A056239 A100197 A057022

easy,nonn

James Kilfiger (jamesk(AT)maths.warwick.ac.uk)