OFFSET
0,3
FORMULA
Recurrence: a(2n) = a(n) + [(n+1)/2] + 1, a(2n+1) = 2n+1.
a(n) = A090740(n) + n - 1, for n >= 1. - Amiram Eldar, Sep 14 2024
PROG
(PARI) a(n)=if(n<1, 0, if(n%2==0, a(n/2)+2*floor((n+2)/4)+1, n))
CROSSREFS
KEYWORD
nonn
AUTHOR
Ralf Stephan, Mar 16 2004
STATUS
approved