A004755 Binary expansion starts 11.

3, 6, 7, 12, 13, 14, 15, 24, 25, 26, 27, 28, 29, 30, 31, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• base
• easy

Comments

a(n) is the smallest value > a(n-1) (or > 1 for n=1) for which A001511(a(n)) = A001511(n). - Franklin T. Adams-Watters, Oct 23 2006

Examples

			12 in binary is 1100, so 12 is in the sequence.
		

Links

• Kenny Lau, Table of n, a(n) for n = 1..16383 (first 1023 terms from T. D. Noe)
• Ralf Stephan, Some divide-and-conquer sequences ...
• Ralf Stephan, Table of generating functions

Crossrefs

Equals union of A079946 and A080565.

Cf. A004754 (10), A004756 (100), A004757 (101), A004758 (110), A004759 (111).

Cf. A004760, A053644, A062050, A076877.

Programs

Haskell

import Data.List (transpose)
a004755 n = a004755_list !! (n-1)
a004755_list = 3 : concat (transpose [zs, map (+ 1) zs])
                   where zs = map (* 2) a004755_list
-- Reinhard Zumkeller, Dec 04 2015

Maple

a:= proc(n) n+2*2^floor(log(n)/log(2)) end: seq(a(n),n=1..60); # Muniru A Asiru, Oct 16 2018

Mathematica

Flatten[Table[FromDigits[#,2]&/@(Join[{1,1},#]&/@Tuples[{0,1},n]),{n,0,5}]] (* Harvey P. Dale, Feb 05 2015 *)

PARI

a(n)=n+2*2^floor(log(n)/log(2))

PARI

is(n)=n>2 && binary(n)[2] \\ Charles R Greathouse IV, Sep 23 2012

Python

f = open('b004755.txt', 'w')
lo = 3
hi = 4
i = 1
while i<16384:
    for x in range(lo,hi):
        f.write(str(i)+" "+str(x)+"\n")
        i += 1
    lo <<= 1
    hi <<= 1
# Kenny Lau, Jul 05 2016

Python

def A004755(n): return n+(1<Chai Wah Wu, Jul 13 2022

Formula

a(2n) = 2*a(n), a(2n+1) = 2*a(n) + 1 + 2*[n==0].

a(n) = n + 2 * 2^floor(log_2(n)) = A004754(n) + A053644(n).

a(n) = 2n + A080079(n). - Benoit Cloitre, Feb 22 2003

G.f.: (1/(1+x)) * (1 + Sum_{k>=0, t=x^2^k} 2^k*(2t+t^2)/(1+t)).

a(n) = n + 2^(floor(log_2(n)) + 1) = n + A062383(n). - Franklin T. Adams-Watters, Oct 23 2006

a(2^m+k) = 2^(m+1) + 2^m + k, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Aug 08 2016

Extensions

Edited by Ralf Stephan, Oct 12 2003