A136013 a(n) = floor(n/2) + 2*a(floor(n/2)), a(0) = 0.

0, 0, 1, 1, 4, 4, 5, 5, 12, 12, 13, 13, 16, 16, 17, 17, 32, 32, 33, 33, 36, 36, 37, 37, 44, 44, 45, 45, 48, 48, 49, 49, 80, 80, 81, 81, 84, 84, 85, 85, 92, 92, 93, 93, 96, 96, 97, 97, 112, 112, 113, 113, 116, 116, 117, 117, 124, 124, 125, 125, 128

Offset: 0

Author

Jack Preston (jpreston(AT)earthlink.net), Mar 20 2008

Keywords

• nonn
• easy

Comments

A recursive sequence that seems to be related to the ruler function.

It seems that a(2n) = a(2n+1) = A080277(n). - Emeric Deutsch, Mar 31 2008

It appears that if the binary expansion of n is n = Sum b_i*2^i (b_i=0 or 1), then a(n) = Sum i*b_i*2^(i-1). - Marc LeBrun, Sep 07 2015

The observations in the preceding two comments (by Emeric Deutsch and Marc LeBrun) follow from the formulas in A333979. - Pontus von Brömssen, Sep 06 2020

This sequence is a variant of the arithmetic derivative (A003415) based on powers of two instead of primes, because the relation a(m*n) = m*a(n) + n*a(m) holds. If we define the polynomial P(2) = bit0*2^0 + bit1*2^1 + bit2*2^2 + ... = n, and P'(2) is the derivative of P(2), then we will observe P'(2) = a(n). - Thomas Scheuerle, Aug 02 2022

Crossrefs

Cf. A080277, A003415, A333979, A135481 (first differences).

Programs

Maple

a:=proc(n) if n=0 then 0 else floor((1/2)*n)+2*a(floor((1/2)*n)) end if end proc: seq(a(n),n=0..60); # Emeric Deutsch, Mar 31 2008

Mathematica

a = {0}; Do[AppendTo[a, Floor[n/2] + 2*a[[Floor[n/2] + 1]]], {n, 1, 100}]; a (* Stefan Steinerberger, Mar 24 2008 *)
Table[Sum[2^(k-1)*Floor[n*2^-k], {k, 1, Log[2, n]}], {n, 0, 100}] (* Federico Provvedi, Aug 17 2013 *)

PARI

a(n) = fromdigits(Vec(Pol(binary(n))'),2); \\ Kevin Ryde, Apr 29 2021

Python

def A136013(n): return sum(map(lambda x:(x[0]+1)*(1<Chai Wah Wu, Jul 06 2022

Formula

a(n) = A333979(n,2). - Pontus von Brömssen, Sep 06 2020

Extensions

More terms from Stefan Steinerberger and Emeric Deutsch, Mar 24 2008

Spelling corrected by Jason G. Wurtzel, Aug 30 2010