cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000120 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n).

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3
Offset: 0

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Author

Keywords

Comments

The binary weight of n is also called Hamming weight of n. [The term "Hamming weight" was named after the American mathematician Richard Wesley Hamming (1915-1998). - Amiram Eldar, Jun 16 2021]
a(n) is also the largest integer such that 2^a(n) divides binomial(2n, n) = A000984(n). - Benoit Cloitre, Mar 27 2002
To construct the sequence, start with 0 and use the rule: If k >= 0 and a(0), a(1), ..., a(2^k-1) are the first 2^k terms, then the next 2^k terms are a(0) + 1, a(1) + 1, ..., a(2^k-1) + 1. - Benoit Cloitre, Jan 30 2003
An example of a fractal sequence. That is, if you omit every other number in the sequence, you get the original sequence. And of course this can be repeated. So if you form the sequence a(0 * 2^n), a(1 * 2^n), a(2 * 2^n), a(3 * 2^n), ... (for any integer n > 0), you get the original sequence. - Christopher.Hills(AT)sepura.co.uk, May 14 2003
The n-th row of Pascal's triangle has 2^k distinct odd binomial coefficients where k = a(n) - 1. - Lekraj Beedassy, May 15 2003
Fixed point of the morphism 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, etc., starting from a(0) = 0. - Robert G. Wilson v, Jan 24 2006
a(n) is the number of times n appears among the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. - Jeremy Gardiner, Jan 25 2006
a(n) is the number of solutions of the Diophantine equation 2^m*k + 2^(m-1) + i = n, where m >= 1, k >= 0, 0 <= i < 2^(m-1); a(5) = 2 because only (m, k, i) = (1, 2, 0) [2^1*2 + 2^0 + 0 = 5] and (m, k, i) = (3, 0, 1) [2^3*0 + 2^2 + 1 = 5] are solutions. - Hieronymus Fischer, Jan 31 2006
The first appearance of k, k >= 0, is at a(2^k-1). - Robert G. Wilson v, Jul 27 2006
Sequence is given by T^(infinity)(0) where T is the operator transforming any word w = w(1)w(2)...w(m) into T(w) = w(1)(w(1)+1)w(2)(w(2)+1)...w(m)(w(m)+1). I.e., T(0) = 01, T(01) = 0112, T(0112) = 01121223. - Benoit Cloitre, Mar 04 2009
For n >= 2, the minimal k for which a(k(2^n-1)) is not multiple of n is 2^n + 3. - Vladimir Shevelev, Jun 05 2009
Triangle inequality: a(k+m) <= a(k) + a(m). Equality holds if and only if C(k+m, m) is odd. - Vladimir Shevelev, Jul 19 2009
a(k*m) <= a(k) * a(m). - Robert Israel, Sep 03 2023
The number of occurrences of value k in the first 2^n terms of the sequence is equal to binomial(n, k), and also equal to the sum of the first n - k + 1 terms of column k in the array A071919. Example with k = 2, n = 7: there are 21 = binomial(7,2) = 1 + 2 + 3 + 4 + 5 + 6 2's in a(0) to a(2^7-1). - Brent Spillner (spillner(AT)acm.org), Sep 01 2010, simplified by R. J. Mathar, Jan 13 2017
Let m be the number of parts in the listing of the compositions of n as lists of parts in lexicographic order, a(k) = n - length(composition(k)) for all k < 2^n and all n (see example); A007895 gives the equivalent for compositions into odd parts. - Joerg Arndt, Nov 09 2012
From Daniel Forgues, Mar 13 2015: (Start)
Just tally up row k (binary weight equal k) from 0 to 2^n - 1 to get the binomial coefficient C(n,k). (See A007318.)
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
Due to its fractal nature, the sequence is quite interesting to listen to.
(End)
The binary weight of n is a particular case of the digit sum (base b) of n. - Daniel Forgues, Mar 13 2015
The mean of the first n terms is 1 less than the mean of [a(n+1),...,a(2n)], which is also the mean of [a(n+2),...,a(2n+1)]. - Christian Perfect, Apr 02 2015
a(n) is also the largest part of the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2, 2, 2, 1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 20 2017
a(n) is also known as the population count of the binary representation of n. - Chai Wah Wu, May 19 2020

Examples

			Using the formula a(n) = a(floor(n / floor_pow4(n))) + a(n mod floor_pow4(n)):
  a(4) = a(1) + a(0) = 1,
  a(8) = a(2) + a(0) = 1,
  a(13) = a(3) + a(1) = 2 + 1 = 3,
  a(23) = a(1) + a(7) = 1 + a(1) + a(3) = 1 + 1 + 2 = 4.
_Gary W. Adamson_ points out (Jun 03 2009) that this can be written as a triangle:
  0,
  1,
  1,2,
  1,2,2,3,
  1,2,2,3,2,3,3,4,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
  1,2,2,3,2,3,...
where the rows converge to A063787.
From _Joerg Arndt_, Nov 09 2012: (Start)
Connection to the compositions of n as lists of parts (see comment):
[ #]:   a(n)  composition
[ 0]:   [0]   1 1 1 1 1
[ 1]:   [1]   1 1 1 2
[ 2]:   [1]   1 1 2 1
[ 3]:   [2]   1 1 3
[ 4]:   [1]   1 2 1 1
[ 5]:   [2]   1 2 2
[ 6]:   [2]   1 3 1
[ 7]:   [3]   1 4
[ 8]:   [1]   2 1 1 1
[ 9]:   [2]   2 1 2
[10]:   [2]   2 2 1
[11]:   [3]   2 3
[12]:   [2]   3 1 1
[13]:   [3]   3 2
[14]:   [3]   4 1
[15]:   [4]   5
(End)
		

References

  • Jean-Paul Allouche and Jeffrey Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 119.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589. - N. J. A. Sloane, Aug 03 2012
  • Manfred R. Schroeder, Fractals, Chaos, Power Laws. W.H. Freeman, 1991, p. 383.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The basic sequences concerning the binary expansion of n are this one, A000788, A000069, A001969, A023416, A059015, A007088.
Partial sums see A000788. For run lengths see A131534. See also A001792, A010062.
Number of 0's in n: A023416 and A080791.
a(n) = n - A011371(n).
Sum of digits of n written in bases 2-16: this sequence, A053735, A053737, A053824, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.
This is Guy Steele's sequence GS(3, 4) (see A135416).
Cf. A230952 (boustrophedon transform).
Cf. A070939 (length of binary representation of n).

Programs

  • Fortran
    c See link in A139351
    
  • Haskell
    import Data.Bits (Bits, popCount)
    a000120 :: (Integral t, Bits t) => t -> Int
    a000120 = popCount
    a000120_list = 0 : c [1] where c (x:xs) = x : c (xs ++ [x,x+1])
    -- Reinhard Zumkeller, Aug 26 2013, Feb 19 2012, Jun 16 2011, Mar 07 2011
    
  • Haskell
    a000120 = concat r
        where r = [0] : (map.map) (+1) (scanl1 (++) r)
    -- Luke Palmer, Feb 16 2014
    
  • Magma
    [Multiplicity(Intseq(n, 2), 1): n in [0..104]]; // Marius A. Burtea, Jan 22 2020
    
  • Magma
    [&+Intseq(n, 2):n in [0..104]]; // Marius A. Burtea, Jan 22 2020
  • Maple
    A000120 := proc(n) local w,m,i; w := 0; m := n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
    A000120 := n -> add(i, i=convert(n,base,2)): # Peter Luschny, Feb 03 2011
    with(Bits): p:=n->ilog2(n-And(n,n-1)): seq(p(binomial(2*n,n)),n=0..200) # Gary Detlefs, Jan 27 2019
  • Mathematica
    Table[DigitCount[n, 2, 1], {n, 0, 105}]
    Nest[Flatten[# /. # -> {#, # + 1}] &, {0}, 7] (* Robert G. Wilson v, Sep 27 2011 *)
    Table[Plus @@ IntegerDigits[n, 2], {n, 0, 104}]
    Nest[Join[#, # + 1] &, {0}, 7] (* IWABUCHI Yu(u)ki, Jul 19 2012 *)
    Log[2, Nest[Join[#, 2#] &, {1}, 14]] (* gives 2^14 term, Carlos Alves, Mar 30 2014 *)
  • PARI
    {a(n) = if( n<0, 0, 2*n - valuation((2*n)!, 2))};
    
  • PARI
    {a(n) = if( n<0, 0, subst(Pol(binary(n)), x ,1))};
    
  • PARI
    {a(n) = if( n<1, 0, a(n\2) + n%2)}; /* Michael Somos, Mar 06 2004 */
    
  • PARI
    a(n)=my(v=binary(n));sum(i=1,#v,v[i]) \\ Charles R Greathouse IV, Jun 24 2011
    
  • PARI
    a(n)=norml2(binary(n)) \\ better use {A000120=hammingweight}. - M. F. Hasler, Oct 09 2012, edited Feb 27 2020
    
  • PARI
    a(n)=hammingweight(n) \\ Michel Marcus, Oct 19 2013
    (Common Lisp) (defun floor-to-power (n pow) (declare (fixnum pow)) (expt pow (floor (log n pow)))) (defun enabled-bits (n) (if (< n 4) (n-th n (list 0 1 1 2)) (+ (enabled-bits (floor (/ n (floor-to-power n 4)))) (enabled-bits (mod n (floor-to-power n 4)))))) ; Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
    
  • Python
    def A000120(n): return bin(n).count('1') # Chai Wah Wu, Sep 03 2014
    
  • Python
    import numpy as np
    A000120 = np.array([0], dtype="uint8")
    for bitrange in range(25): A000120 = np.append(A000120, np.add(A000120, 1))
    print([A000120[n] for n in range(0, 105)]) # Karl-Heinz Hofmann, Nov 07 2022
    
  • Python
    def A000120(n): return n.bit_count() # Requires Python 3.10 or higher. - Pontus von Brömssen, Nov 08 2022
    
  • Python
    # Also see links.
    
  • SageMath
    def A000120(n):
        if n <= 1: return Integer(n)
        return A000120(n//2) + n%2
    [A000120(n) for n in range(105)]  # Peter Luschny, Nov 19 2012
    
  • SageMath
    def A000120(n) : return sum(n.digits(2)) # Eric M. Schmidt, Apr 26 2013
    
  • Scala
    (0 to 127).map(Integer.bitCount()) // _Alonso del Arte, Mar 05 2019
    

Formula

a(0) = 0, a(2*n) = a(n), a(2*n+1) = a(n) + 1.
a(0) = 0, a(2^i) = 1; otherwise if n = 2^i + j with 0 < j < 2^i, a(n) = a(j) + 1.
G.f.: Product_{k >= 0} (1 + y*x^(2^k)) = Sum_{n >= 0} y^a(n)*x^n. - N. J. A. Sloane, Jun 04 2009
a(n) = a(n-1) + 1 - A007814(n) = log_2(A001316(n)) = 2n - A005187(n) = A070939(n) - A023416(n). - Henry Bottomley, Apr 04 2001; corrected by Ralf Stephan, Apr 15 2002
a(n) = log_2(A000984(n)/A001790(n)). - Benoit Cloitre, Oct 02 2002
For n > 0, a(n) = n - Sum_{k=1..n} A007814(k). - Benoit Cloitre, Oct 19 2002
a(n) = n - Sum_{k>=1} floor(n/2^k) = n - A011371(n). - Benoit Cloitre, Dec 19 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^(2^k)/(1+x^(2^k)). - Ralf Stephan, Apr 19 2003
a(0) = 0, a(n) = a(n - 2^floor(log_2(n))) + 1. Examples: a(6) = a(6 - 2^2) + 1 = a(2) + 1 = a(2 - 2^1) + 1 + 1 = a(0) + 2 = 2; a(101) = a(101 - 2^6) + 1 = a(37) + 1 = a(37 - 2^5) + 2 = a(5 - 2^2) + 3 = a(1 - 2^0) + 4 = a(0) + 4 = 4; a(6275) = a(6275 - 2^12) + 1 = a(2179 - 2^11) + 2 = a(131 - 2^7) + 3 = a(3 - 2^1) + 4 = a(1 - 2^0) + 5 = 5; a(4129) = a(4129 - 2^12) + 1 = a(33 - 2^5) + 2 = a(1 - 2^0) + 3 = 3. - Hieronymus Fischer, Jan 22 2006
A fixed point of the mapping 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, ... With f(i) = floor(n/2^i), a(n) is the number of odd numbers in the sequence f(0), f(1), f(2), f(3), f(4), f(5), ... - Philippe Deléham, Jan 04 2004
When read mod 2 gives the Morse-Thue sequence A010060.
Let floor_pow4(n) denote n rounded down to the next power of four, floor_pow4(n) = 4 ^ floor(log4 n). Then a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 2, a(n) = a(floor(n / floor_pow4(n))) + a(n % floor_pow4(n)). - Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
a(n) = n - Sum_{k=2..n} Sum_{j|n, j >= 2} (floor(log_2(j)) - floor(log_2(j-1))). - Hieronymus Fischer, Jun 18 2007
a(n) = A138530(n, 2) for n > 1. - Reinhard Zumkeller, Mar 26 2008
a(A077436(n)) = A159918(A077436(n)); a(A000290(n)) = A159918(n). - Reinhard Zumkeller, Apr 25 2009
a(n) = A063787(n) - A007814(n). - Gary W. Adamson, Jun 04 2009
a(n) = A007814(C(2n, n)) = 1 + A007814(C(2n-1, n)). - Vladimir Shevelev, Jul 20 2009
For odd m >= 1, a((4^m-1)/3) = a((2^m+1)/3) + (m-1)/2 (mod 2). - Vladimir Shevelev, Sep 03 2010
a(n) - a(n-1) = { 1 - a(n-1) if and only if A007814(n) = a(n-1), 1 if and only if A007814(n) = 0, -1 for all other A007814(n) }. - Brent Spillner (spillner(AT)acm.org), Sep 01 2010
a(A001317(n)) = 2^a(n). - Vladimir Shevelev, Oct 25 2010
a(n) = A139351(n) + A139352(n) = Sum_k {A030308(n, k)}. - Philippe Deléham, Oct 14 2011
From Hieronymus Fischer, Jun 10 2012: (Start)
a(n) = Sum_{j = 1..m+1} (floor(n/2^j + 1/2) - floor(n/2^j)), where m = floor(log_2(n)).
General formulas for the number of digits >= d in the base p representation of n, where 1 <= d < p: a(n) = Sum_{j = 1..m+1} (floor(n/p^j + (p-d)/p) - floor(n/p^j)), where m=floor(log_p(n)); g.f.: g(x) = (1/(1-x))*Sum_{j>=0} (x^(d*p^j) - x^(p*p^j))/(1-x^(p*p^j)). (End)
a(n) = A213629(n, 1) for n > 0. - Reinhard Zumkeller, Jul 04 2012
a(n) = A240857(n,n). - Reinhard Zumkeller, Apr 14 2014
a(n) = log_2(C(2*n,n) - (C(2*n,n) AND C(2*n,n)-1)). - Gary Detlefs, Jul 10 2014
Sum_{n >= 1} a(n)/2n(2n+1) = (gamma + log(4/Pi))/2 = A344716, where gamma is Euler's constant A001620; see Sondow 2005, 2010 and Allouche, Shallit, Sondow 2007. - Jonathan Sondow, Mar 21 2015
For any integer base b >= 2, the sum of digits s_b(n) of expansion base b of n is the solution of this recurrence relation: s_b(n) = 0 if n = 0 and s_b(n) = s_b(floor(n/b)) + (n mod b). Thus, a(n) satisfies: a(n) = 0 if n = 0 and a(n) = a(floor(n/2)) + (n mod 2). This easily yields a(n) = Sum_{i = 0..floor(log_2(n))} (floor(n/2^i) mod 2). From that one can compute a(n) = n - Sum_{i = 1..floor(log_2(n))} floor(n/2^i). - Marek A. Suchenek, Mar 31 2016
Sum_{k>=1} a(k)/2^k = 2 * Sum_{k >= 0} 1/(2^(2^k)+1) = 2 * A051158. - Amiram Eldar, May 15 2020
Sum_{k>=1} a(k)/(k*(k+1)) = A016627 = log(4). - Bernard Schott, Sep 16 2020
a(m*(2^n-1)) >= n. Equality holds when 2^n-1 >= A000265(m), but also in some other cases, e.g., a(11*(2^2-1)) = 2 and a(19*(2^3-1)) = 3. - Pontus von Brömssen, Dec 13 2020
G.f.: A(x) satisfies A(x) = (1+x)*A(x^2) + x/(1-x^2). - Akshat Kumar, Nov 04 2023

A070939 Length of binary representation of n.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7
Offset: 0

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Author

N. J. A. Sloane, May 18 2002

Keywords

Comments

Zero is assumed to be represented as 0.
For n>1, n appears 2^(n-1) times. - Lekraj Beedassy, Apr 12 2006
a(n) is the permanent of the n X n 0-1 matrix whose (i,j) entry is 1 iff i=1 or i=j or i=2*j. For example, a(4)=3 is per([[1, 1, 1, 1], [1, 1, 0, 0], [0, 0, 1, 0], [0, 1, 0, 1]]). - David Callan, Jun 07 2006
a(n) is the number of different contiguous palindromic bit patterns in the binary representation of n; for examples, for 5=101_2 the bit patterns are 0, 1, 101; for 7=111_2 the corresponding patterns are 1, 11, 111; for 13=1101_2 the patterns are 0, 1, 11, 101. - Hieronymus Fischer, Mar 13 2012
A103586(n) = a(n + a(n)); a(A214489(n)) = A103586(A214489(n)). - Reinhard Zumkeller, Jul 21 2012
Number of divisors of 2^n that are <= n. - Clark Kimberling, Apr 21 2019

Examples

			8 = 1000 in binary has length 4.
		

References

  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • L. Levine, Fractal sequences and restricted Nim, Ars Combin. 80 (2006), 113-127.

Crossrefs

A029837(n+1) gives the length of binary representation of n without the leading zeros (i.e., when zero is represented as the empty sequence). For n > 0 this is equal to a(n).
This is Guy Steele's sequence GS(4, 4) (see A135416).
Cf. A083652 (partial sums).

Programs

  • Haskell
    a070939 n = if n < 2 then 1 else a070939 (n `div` 2) + 1
    a070939_list = 1 : 1 : l [1] where
       l bs = bs' ++ l bs' where bs' = map (+ 1) (bs ++ bs)
    -- Reinhard Zumkeller, Jul 19 2012, Jun 07 2011
    
  • Magma
    A070939:=func< n | n eq 0 select 1 else #Intseq(n, 2) >; [ A070939(n): n in [0..104] ]; // Klaus Brockhaus, Jan 13 2011
    
  • Maple
    A070939 := n -> `if`(n=0, 1, ilog2(2*n)):
    seq(A070939(n), n=0..104); # revised by Peter Luschny, Aug 10 2017
  • Mathematica
    Table[Length[IntegerDigits[n, 2]], {n, 0, 50}] (* Stefan Steinerberger, Apr 01 2006 *)
    Join[{1},IntegerLength[Range[110],2]] (* Harvey P. Dale, Aug 18 2013 *)
    a[ n_] := If[ n < 1, Boole[n == 0], BitLength[n]]; (* Michael Somos, Jul 10 2018 *)
  • PARI
    {a(n) = if( n<1, n==0, #binary(n))} /* Michael Somos, Aug 31 2012 */
    
  • PARI
    apply( {A070939(n)=exponent(n+!n)+1}, [0..99]) \\ works for negative n and is much faster than the above. - M. F. Hasler, Jan 04 2014, updated Feb 29 2020
    
  • Python
    def a(n): return len(bin(n)[2:])
    print([a(n) for n in range(105)]) # Michael S. Branicky, Jan 01 2021
    
  • Python
    def A070939(n): return 1 if n == 0 else n.bit_length() # Chai Wah Wu, May 12 2022
  • Sage
    def A070939(n) : return (2*n).exact_log(2) if n != 0 else 1
    [A070939(n) for n in range(100)] # Peter Luschny, Aug 08 2012
    

Formula

a(0) = 1; for n >= 1, a(n) = 1 + floor(log_2(n)) = 1 + A000523(n).
G.f.: 1 + 1/(1-x) * Sum(k>=0, x^2^k). - Ralf Stephan, Apr 12 2002
a(0)=1, a(1)=1 and a(n) = 1+a(floor(n/2)). - Benoit Cloitre, Dec 02 2003
a(n) = A000120(n) + A023416(n). - Lekraj Beedassy, Apr 12 2006
a(2^m + k) = m + 1, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Mar 14 2017
a(n) = A113473(n) if n>0.

Extensions

a(4) corrected by Antti Karttunen, Feb 28 2003

A007088 The binary numbers (or binary words, or binary vectors, or binary expansion of n): numbers written in base 2.

Original entry on oeis.org

0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111
Offset: 0

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Author

Keywords

Comments

List of binary numbers. (This comment is to assist people searching for that particular phrase. - N. J. A. Sloane, Apr 08 2016)
Or, numbers that are sums of distinct powers of 10.
Or, numbers having only digits 0 and 1 in their decimal representation.
Complement of A136399; A064770(a(n)) = a(n). - Reinhard Zumkeller, Dec 30 2007
From Rick L. Shepherd, Jun 25 2009: (Start)
Nonnegative integers with no decimal digit > 1.
Thus nonnegative integers n in base 10 such that kn can be calculated by normal addition (i.e., n + n + ... + n, with k n's (but not necessarily k + k + ... + k, with n k's)) or multiplication without requiring any carry operations for 0 <= k <= 9. (End)
For n > 1: A257773(a(n)) = 10, numbers that are Belgian-k for k=0..9. - Reinhard Zumkeller, May 08 2015
For any integer n>=0, find the binary representation and then interpret as decimal representation giving a(n). - Michael Somos, Nov 15 2015
N is in this sequence iff A007953(N) = A101337(N). A028897 is a left inverse. - M. F. Hasler, Nov 18 2019
For n > 0, numbers whose largest decimal digit is 1. - Stefano Spezia, Nov 15 2023

Examples

			a(6)=110 because (1/2)*((1-(-1)^6)*10^0 + (1-(-1)^3)*10^1 + (1-(-1)^1)*10^2) = 10 + 100.
G.f. = x + 10*x^2 + 11*x^3 + 100*x^4 + 101*x^5 + 110*x^6 + 111*x^7 + 1000*x^8 + ...
.
  000    The numbers < 2^n can be regarded as vectors with
  001    a fixed length n if padded with zeros on the left
  010    side. This represents the n-fold Cartesian product
  011    over the set {0, 1}. In the example on the left,
  100    n = 3. (See also the second Python program.)
  101    Binary vectors in this format can also be seen as a
  110    representation of the subsets of a set with n elements.
  111    - _Peter Luschny_, Jan 22 2024
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 21.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §2.8 Binary, Octal, Hexadecimal, p. 64.
  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W. H. Freeman, 1991, p. 383.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The basic sequences concerning the binary expansion of n are this one, A000120 (Hammingweight: sum of bits), A000788 (partial sums of A000120), A000069 (A000120 is odd), A001969 (A000120 is even), A023416 (number of bits 0), A059015 (partial sums). Bisections A099820 and A099821.
Cf. A028897 (convert binary to decimal).

Programs

  • Haskell
    a007088 0 = 0
    a007088 n = 10 * a007088 n' + m where (n',m) = divMod n 2
    -- Reinhard Zumkeller, Jan 10 2012
    
  • Maple
    A007088 := n-> convert(n, binary): seq(A007088(n), n=0..50); # R. J. Mathar, Aug 11 2009
  • Mathematica
    Table[ FromDigits[ IntegerDigits[n, 2]], {n, 0, 39}]
    Table[Sum[ (Floor[( Mod[f/2 ^n, 2])])*(10^n) , {n, 0, Floor[Log[2, f]]}], {f, 1, 100}] (* José de Jesús Camacho Medina, Jul 24 2014 *)
    FromDigits/@Tuples[{1,0},6]//Sort (* Harvey P. Dale, Aug 10 2017 *)
  • PARI
    {a(n) = subst( Pol( binary(n)), x, 10)}; /* Michael Somos, Jun 07 2002 */
    
  • PARI
    {a(n) = if( n<=0, 0, n%2 + 10*a(n\2))}; /* Michael Somos, Jun 07 2002 */
    
  • PARI
    a(n)=fromdigits(binary(n),10) \\ Charles R Greathouse IV, Apr 08 2015
    
  • Python
    def a(n): return int(bin(n)[2:])
    print([a(n) for n in range(40)]) # Michael S. Branicky, Jan 10 2021
    
  • Python
    from itertools import product
    n = 4
    for p in product([0, 1], repeat=n): print(''.join(str(x) for x in p))
    # Peter Luschny, Jan 22 2024

Formula

a(n) = Sum_{i=0..m} d(i)*10^i, where Sum_{i=0..m} d(i)*2^i is the base 2 representation of n.
a(n) = (1/2)*Sum_{i>=0} (1-(-1)^floor(n/2^i))*10^i. - Benoit Cloitre, Nov 20 2001
a(n) = A097256(n)/9.
a(2n) = 10*a(n), a(2n+1) = a(2n)+1.
G.f.: 1/(1-x) * Sum_{k>=0} 10^k * x^(2^k)/(1+x^(2^k)) - for sequence as decimal integers. - Franklin T. Adams-Watters, Jun 16 2006
a(A000290(n)) = A001737(n). - Reinhard Zumkeller, Apr 25 2009
a(n) = Sum_{k>=0} A030308(n,k)*10^k. - Philippe Deléham, Oct 19 2011
For n > 0: A054055(a(n)) = 1. - Reinhard Zumkeller, Apr 25 2012
a(n) = Sum_{k=0..floor(log_2(n))} floor((Mod(n/2^k, 2)))*(10^k). - José de Jesús Camacho Medina, Jul 24 2014

A055642 Number of digits in the decimal expansion of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
Offset: 0

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Author

Henry Bottomley, Jun 06 2000

Keywords

Comments

From Hieronymus Fischer, Jun 08 2012: (Start)
For n > 0 the first differences of A117804.
The total number of digits necessary to write down all the numbers 0, 1, 2, ..., n is A117804(n+1). (End)
Here a(0) = 1, but a different common convention is to consider that the expansion of 0 in any base b > 0 has 0 terms and digits. - M. F. Hasler, Dec 07 2018

Examples

			Examples:
999: 1 + floor(log_10(999)) = 1 + floor(2.x) = 1 + 2 = 3 or
      ceiling(log_10(999+1)) = ceiling(log_10(1000)) = ceiling(3) = 3;
1000: 1 + floor(log_10(1000)) = 1 + floor(3) = 1 + 3 = 4 or
      ceiling(log_10(1000+1)) = ceiling(log_10(1001)) = ceiling(3.x) = 4;
1001: 1 + floor(log_10(1001)) = 1 + floor(3.x) = 1 + 3 = 4 or
      ceiling(log_10(1001+1)) = ceiling(log_10(1002)) = ceiling(3.x) = 4;
		

Crossrefs

Programs

  • Haskell
    a055642 :: Integer -> Int
    a055642 = length . show  -- Reinhard Zumkeller, Feb 19 2012, Apr 26 2011
    
  • Magma
    [ #Intseq(n): n in [0..105] ];   //  Bruno Berselli, Jun 30 2011
    (Common Lisp) (defun A055642 (n) (if (zerop n) 1 (floor (log n 10)))) ; James Spahlinger, Oct 13 2012
    
  • Maple
    A055642 := proc(n)
            max(1,ilog10(n)+1) ;
    end proc:  # R. J. Mathar, Nov 30 2011
  • Mathematica
    Join[{1}, Array[ Floor[ Log[10, 10# ]] &, 104]] (* Robert G. Wilson v, Jan 04 2006 *)
    Join[{1},Table[IntegerLength[n],{n,104}]]
    IntegerLength[Range[0,120]] (* Harvey P. Dale, Jul 02 2016 *)
  • PARI
    a(n)=#Str(n) \\ M. F. Hasler, Nov 17 2008
    
  • PARI
    A055642(n)=logint(n+!n,10)+1 \\ Increasingly faster than the above, for larger n. (About twice as fast for n ~ 10^7.) - M. F. Hasler, Dec 07 2018
    
  • Python
    def a(n): return len(str(n))
    print([a(n) for n in range(121)]) # Michael S. Branicky, May 10 2022
    
  • Python
    def A055642(n): # Faster than len(str(n)) from ~ 50 digits on
        L = math.log10(n or 1)
        if L.is_integer() and 10**int(L)>n: return int(L or 1)
        return int(L)+1  # M. F. Hasler, Apr 08 2024

Formula

a(A046760(n)) < A050252(A046760(n)); a(A046759(n)) > A050252(A046759(n)). - Reinhard Zumkeller, Jun 21 2011
a(n) = A196563(n) + A196564(n).
a(n) = 1 + floor(log_10(n)) = 1 + A004216(n) = ceiling(log_10(n+1)) = A004218(n+1), if n >= 1. - Daniel Forgues, Mar 27 2014
a(A046758(n)) = A050252(A046758(n)). - Reinhard Zumkeller, Jun 21 2011
a(n) = A117804(n+1) - A117804(n), n > 0. - Hieronymus Fischer, Jun 08 2012
G.f.: g(x) = 1 + (1/(1-x))*Sum_{j>=0} x^(10^j). - Hieronymus Fischer, Jun 08 2012
a(n) = A262190(n) for n < 100; a(A262198(n)) != A262190(A262198(n)). - Reinhard Zumkeller, Sep 14 2015

A029931 If 2n = Sum 2^e_i, a(n) = Sum e_i.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 6, 7, 8, 9, 9, 10, 11, 12, 10, 11, 12, 13, 13, 14, 15, 16, 11, 12, 13, 14, 14, 15, 16, 17, 15, 16, 17, 18, 18, 19, 20, 21, 7, 8, 9, 10, 10, 11, 12, 13, 11, 12, 13, 14, 14, 15, 16
Offset: 0

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Author

Keywords

Comments

Write n in base 2, n = sum b(i)*2^(i-1), then a(n) = sum b(i)*i. - Benoit Cloitre, Jun 09 2002
May be regarded as a triangular array read by rows, giving weighted sum of compositions in standard order. The standard order of compositions is given by A066099. - Franklin T. Adams-Watters, Nov 06 2006
Sum of all positive integer roots m_i of polynomial {m,k} - see link [Shevelev]; see also A264613. - Vladimir Shevelev, Dec 13 2015
Also the sum of binary indices of n, where a binary index of n (A048793) is any position of a 1 in its reversed binary expansion. For example, the binary indices of 11 are {1,2,4}, so a(11) = 7. - Gus Wiseman, May 22 2024

Examples

			14 = 8+4+2 so a(7) = 3+2+1 = 6.
Composition number 11 is 2,1,1; 1*2+2*1+3*1 = 7, so a(11) = 7.
The triangle starts:
  0
  1
  2 3
  3 4 5 6
The reversed binary expansion of 18 is (0,1,0,0,1) with 1's at positions {2,5}, so a(18) = 2 + 5 = 7. - _Gus Wiseman_, Jul 22 2019
		

Crossrefs

Other sequences that are built by replacing 2^k in the binary representation with other numbers: A022290 (Fibonacci), A059590 (factorials), A073642, A089625 (primes), A116549, A326031.
Cf. A001793 (row sums), A011782 (row lengths), A059867, A066099, A124757.
Row sums of A048793 and A272020.
Contains exactly A000009(n) copies of n.
For length instead of sum we have A000120, complement A023416.
For minimum instead of sum we have A001511, opposite A000012.
For maximum instead of sum we have A029837 or A070939, opposite A070940.
For product instead of sum we have A096111.
The reverse version is A230877, row sums of A371572.
The reverse complement is A359359, row sums of A371571.
The complement is A359400, row sums of A368494.
Numbers k such that a(k) is prime are A372689.
A014499 lists binary indices of prime numbers.
A019565 gives Heinz number of binary indices, inverse A048675.
A372471 lists binary indices of primes, row-sums A372429.

Programs

  • Haskell
    a029931 = sum . zipWith (*) [1..] . a030308_row
    -- Reinhard Zumkeller, Feb 28 2014
    
  • Maple
    HammingWeight := n -> add(i, i = convert(n, base, 2)):
    a := proc(n) option remember; `if`(n = 0, 0,
    ifelse(n::even, a(n/2) + HammingWeight(n/2), a(n-1) + 1)) end:
    seq(a(n), n = 0..78); # Peter Luschny, Oct 30 2021
  • Mathematica
    a[n_] := (b = IntegerDigits[n, 2]).Reverse @ Range[Length @ b]; Array[a,78,0] (* Jean-François Alcover, Apr 28 2011, after B. Cloitre *)
  • PARI
    for(n=0,100,l=length(binary(n)); print1(sum(i=1,l, component(binary(n),i)*(l-i+1)),","))
    
  • PARI
    a(n) = my(b=binary(n)); b*-[-#b..-1]~; \\ Ruud H.G. van Tol, Oct 17 2023
    
  • Python
    def A029931(n): return sum(i if j == '1' else 0 for i, j in enumerate(bin(n)[:1:-1],1)) # Chai Wah Wu, Dec 20 2022
    (C#)
    ulong A029931(ulong n) {
        ulong result = 0, counter = 1;
        while(n > 0) {
            if (n % 2 == 1)
              result += counter;
            counter++;
            n /= 2;
        }
        return result;
    } // Frank Hollstein, Jan 07 2023

Formula

a(n) = a(n - 2^L(n)) + L(n) + 1 [where L(n) = floor(log_2(n)) = A000523(n)] = sum of digits of A048794 [at least for n < 512]. - Henry Bottomley, Mar 09 2001
a(0) = 0, a(2n) = a(n) + e1(n), a(2n+1) = a(2n) + 1, where e1(n) = A000120(n). a(n) = log_2(A029930(n)). - Ralf Stephan, Jun 19 2003
G.f.: (1/(1-x)) * Sum_{k>=0} (k+1)*x^2^k/(1+x^2^k). - Ralf Stephan, Jun 23 2003
a(n) = Sum_{k>=0} A030308(n,k)*A000027(k+1). - Philippe Deléham, Oct 15 2011
a(n) = sum of n-th row of the triangle in A213629. - Reinhard Zumkeller, Jun 17 2012
From Reinhard Zumkeller, Feb 28 2014: (Start)
a(A089633(n)) = n and a(m) != n for m < A089633(n).
a(n) = Sum_{k=1..A070939(n)} k*A030308(n,k-1). (End)
a(n) = A073642(n) + A000120(n). - Peter Kagey, Apr 04 2016

Extensions

More terms from Erich Friedman

A000069 Odious numbers: numbers with an odd number of 1's in their binary expansion.

Original entry on oeis.org

1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 21, 22, 25, 26, 28, 31, 32, 35, 37, 38, 41, 42, 44, 47, 49, 50, 52, 55, 56, 59, 61, 62, 64, 67, 69, 70, 73, 74, 76, 79, 81, 82, 84, 87, 88, 91, 93, 94, 97, 98, 100, 103, 104, 107, 109, 110, 112, 115, 117, 118, 121, 122, 124, 127, 128
Offset: 1

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Author

Keywords

Comments

This sequence and A001969 give the unique solution to the problem of splitting the nonnegative integers into two classes in such a way that sums of pairs of distinct elements from either class occur with the same multiplicities [Lambek and Moser]. Cf. A000028, A000379.
In French: les nombres impies.
Has asymptotic density 1/2, since exactly 2 of the 4 numbers 4k, 4k+1, 4k+2, 4k+3 have an even sum of bits, while the other 2 have an odd sum. - Jeffrey Shallit, Jun 04 2002
Nim-values for game of mock turtles played with n coins.
A115384(n) = number of odious numbers <= n; A000120(a(n)) = A132680(n). - Reinhard Zumkeller, Aug 26 2007
Indices of 1's in the Thue-Morse sequence A010060. - Tanya Khovanova, Dec 29 2008
For any positive integer m, the partition of the set of the first 2^m positive integers into evil ones E and odious ones O is a fair division for any polynomial sequence p(k) of degree less than m, that is, Sum_{k in E} p(k) = Sum_{k in O} p(k) holds for any polynomial p with deg(p) < m. - Pietro Majer, Mar 15 2009
For n>1 let b(n) = a(n-1). Then b(b(n)) = 2b(n). - Benoit Cloitre, Oct 07 2010
Lexicographically earliest sequence of distinct nonnegative integers with no term being the binary exclusive OR of any terms. The equivalent sequence for addition or for subtraction is A005408 (the odd numbers) and for multiplication is A026424. - Peter Munn, Jan 14 2018
Numbers of the form m XOR (2*m+1) for some m >= 0. - Rémy Sigrist, Apr 14 2022

Examples

			For k=2, x=0 and x=0.2 we respectively have 1^2 + 2^2 + 4^2 + 7^2 = 0^2 + 3^2 + 5^2 + 6^2 = 70;
(1.2)^2 + (2.2)^2 + (4.2)^2 + (7.2)^2 = (0.2)^2 + (3.2)^2 + (5.2)^2 + (6.2)^2 = 75.76;
for k=3, x=1.8, we have (2.8)^3 + (3.8)^3 + (5.8)^3 + (8.8)^3 + (9.8)^3 + (12.8)^3 + (14.8)^3 + (15.8)^3 = (1.8)^3 + (4.8)^3 + (6.8)^3 + (7.8)^3 + (10.8)^3 + (11.8)^3 + (13.8)^3 + (16.8)^3 = 11177.856. - _Vladimir Shevelev_, Jan 16 2012
		

References

  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 433.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 22.
  • Vladimir S. Shevelev, On some identities connected with the partition of the positive integers with respect to the Morse sequence, Izv. Vuzov of the North-Caucasus region, Nature sciences 4 (1997), 21-23 (in Russian).
  • N. J. A. Sloane, A handbook of Integer Sequences, Academic Press, 1973 (including this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015.
Complement of A001969 (the evil numbers). Cf. A133009.
a(n) = 2*n + 1 - A010060(n) = A001969(n) + (-1)^A010060(n).
First differences give A007413.
Note that A000079, A083420, A002042, A002089, A132679 are subsequences.
See A027697 for primes, also A230095.
Cf. A005408 (odd numbers), A006068, A026424.

Programs

  • Haskell
    a000069 n = a000069_list !! (n-1)
    a000069_list = [x | x <- [0..], odd $ a000120 x]
    -- Reinhard Zumkeller, Feb 01 2012
    
  • Magma
    [ n: n in [1..130] | IsOdd(&+Intseq(n, 2)) ]; // Klaus Brockhaus, Oct 07 2010
    
  • Maple
    s := proc(n) local i,j,k,b,sum,ans; ans := [ ]; j := 0; for i while jA000069 := n->t1[n]; # s(k) gives first k terms.
    is_A000069 := n -> type(add(i,i=convert(n,base,2)),odd):
    seq(`if`(is_A000069(i),i,NULL),i=0..40); # Peter Luschny, Feb 03 2011
  • Mathematica
    Select[Range[300], OddQ[DigitCount[ #, 2][[1]]] &] (* Stefan Steinerberger, Mar 31 2006 *)
    a[ n_] := If[ n < 1, 0, 2 n - 1 - Mod[ Total @ IntegerDigits[ n - 1, 2], 2]]; (* Michael Somos, Jun 01 2013 *)
  • PARI
    {a(n) = if( n<1, 0, 2*n - 1 - subst( Pol(binary( n-1)), x, 1) % 2)}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    {a(n) = if( n<2, n==1, if( n%2, a((n+1)/2) + n-1, -a(n/2) + 3*(n-1)))}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    a(n)=2*n-1-hammingweight(n-1)%2 \\ Charles R Greathouse IV, Mar 22 2013
    
  • Python
    [n for n in range(1, 201) if bin(n)[2:].count("1") % 2] # Indranil Ghosh, May 03 2017
    
  • Python
    def A000069(n): return ((m:=n-1)<<1)+(m.bit_count()&1^1) # Chai Wah Wu, Mar 03 2023

Formula

G.f.: 1 + Sum_{k>=0} (t*(2+2t+5t^2-t^4)/(1-t^2)^2) * Product_{j=0..k-1} (1-x^(2^j)), t=x^2^k. - Ralf Stephan, Mar 25 2004
a(n+1) = (1/2) * (4*n + 1 + (-1)^A000120(n)). - Ralf Stephan, Sep 14 2003
Numbers n such that A010060(n) = 1. - Benoit Cloitre, Nov 15 2003
a(2*n+1) + a(2*n) = A017101(n) = 8*n+3. a(2*n+1) - a(2*n) gives the Thue-Morse sequence (1, 3 version): 1, 3, 3, 1, 3, 1, 1, 3, 3, 1, 1, 3, 1, ... A001969(n) + A000069(n) = A016813(n) = 4*n+1. - Philippe Deléham, Feb 04 2004
(-1)^a(n) = 2*A010060(n)-1. - Benoit Cloitre, Mar 08 2004
a(1) = 1; for n > 1: a(2*n) = 6*n-3 -a(n), a(2*n+1) = a(n+1) + 2*n. - Corrected by Vladimir Shevelev, Sep 25 2011
For k >= 1 and for every real (or complex) x, we have Sum_{i=1..2^k} (a(i)+x)^s = Sum_{i=1..2^k} (A001969(i)+x)^s, s=0..k.
For x=0, s <= k-1, this is known as Prouhet theorem (see J.-P. Allouche and Jeffrey Shallit, The Ubiquitous Prouhet-Thue-Morse Sequence). - Vladimir Shevelev, Jan 16 2012
a(n+1) mod 2 = 1 - A010060(n) = A010059(n). - Robert G. Wilson v, Jan 18 2012
A005590(a(n)) > 0. - Reinhard Zumkeller, Apr 11 2012
A106400(a(n)) = -1. - Reinhard Zumkeller, Apr 29 2012
a(n+1) = A006068(n) XOR (2*A006068(n) + 1). - Rémy Sigrist, Apr 14 2022

A001969 Evil numbers: nonnegative integers with an even number of 1's in their binary expansion.

Original entry on oeis.org

0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39, 40, 43, 45, 46, 48, 51, 53, 54, 57, 58, 60, 63, 65, 66, 68, 71, 72, 75, 77, 78, 80, 83, 85, 86, 89, 90, 92, 95, 96, 99, 101, 102, 105, 106, 108, 111, 113, 114, 116, 119, 120, 123, 125, 126, 129
Offset: 1

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Author

Keywords

Comments

This sequence and A000069 give the unique solution to the problem of splitting the nonnegative integers into two classes in such a way that sums of pairs of distinct elements from either class occur with the same multiplicities [Lambek and Moser]. Cf. A000028, A000379.
In French: les nombres païens.
Theorem: First differences give A036585. (Observed by Franklin T. Adams-Watters.)
Proof from Max Alekseyev, Aug 30 2006 (edited by N. J. A. Sloane, Jan 05 2021): (Start)
Observe that if the last bit of a(n) is deleted, we get the nonnegative numbers 0, 1, 2, 3, ... in order.
The last bit in a(n+1) is 1 iff the number of bits in n is odd, that is, iff A010060(n+1) is 1.
So, taking into account the different offsets here and in A010060, we have a(n) = 2*(n-1) + A010060(n-1).
Therefore the first differences of the present sequence equal 2 + first differences of A010060, which equals A036585. QED (End)
Integers k such that A010060(k-1)=0. - Benoit Cloitre, Nov 15 2003
Indices of zeros in the Thue-Morse sequence A010060 shifted by 1. - Tanya Khovanova, Feb 13 2009
Conjecture, checked up to 10^6: a(n) is also the sequence of numbers k representable as k = ror(x) XOR rol(x) (for some integer x) where ror(x)=A038572(x) is x rotated one binary place to the right, rol(x)=A006257(x) is x rotated one binary place to the left, and XOR is the binary exclusive-or operator. - Alex Ratushnyak, May 14 2016
From Charlie Neder, Oct 07 2018: (Start)
Conjecture is true: ror(x) and rol(x) have an even number of 1 bits in total (= 2 * A000120(x)), and XOR preserves the parity of this total, so the resulting number must have an even number of 1 bits. An x can be constructed corresponding to a(n) like so:
If the number of bits in a(n) is even, add a leading 0 so a(n) is 2k+1 bits long.
Do an inverse shuffle on a(n), then "divide" by 11, rotate the result k bits to the right, and shuffle to get x. (End)
Numbers of the form m XOR (2*m) for some m >= 0. - Rémy Sigrist, Feb 07 2021
The terms "evil numbers" and "odious numbers" were coined by Richard K. Guy, c. 1976 (Haque and Shallit, 2016) and appeared in the book by Berlekamp et al. (Vol. 1, 1st ed., 1982). - Amiram Eldar, Jun 08 2021

References

  • Elwyn R. Berlekamp, John H. Conway, Richard K. Guy, Winning Ways for Your Mathematical Plays, Volume 1, 2nd ed., A K Peters, 2001, chapter 14, p. 110.
  • Hugh L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 208.
  • Donald J. Newman, A Problem Seminar, Springer; see Problem #89.
  • Vladimir S. Shevelev, On some identities connected with the partition of the positive integers with respect to the Morse sequence, Izv. Vuzov of the North-Caucasus region, Nature sciences 4 (1997), 21-23 (Russian).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Complement of A000069 (the odious numbers). Cf. A133009.
a(n)=2*n+A010060(n)=A000069(n)-(-1)^A010060(n). Cf. A018900.
The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015.
Cf. A036585 (differences), A010060, A006364.
For primes see A027699, also A130593.

Programs

  • Haskell
    a001969 n = a001969_list !! (n-1)
    a001969_list = [x | x <- [0..], even $ a000120 x]
    -- Reinhard Zumkeller, Feb 01 2012
    
  • Magma
    [ n : n in [0..129] | IsEven(&+Intseq(n,2)) ]; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006
    
  • Maple
    s := proc(n) local i,j,ans; ans := [ ]; j := 0; for i from 0 while jA001969 := n->t1[n]; # s(k) gives first k terms.
    # Alternative:
    seq(`if`(add(k, k=convert(n,base,2))::even, n, NULL), n=0..129); # Peter Luschny, Jan 15 2021
    # alternative for use outside this sequence
    isA001969 := proc(n)
        add(d,d=convert(n,base,2)) ;
        type(%,'even') ;
    end proc:
    A001969 := proc(n)
        option remember ;
        local a;
        if n = 0 then
            1;
        else
            for a from procname(n-1)+1 do
                if isA001969(a) then
                    return a;
                end if;
            end do:
        end if;
    end proc:
    seq(A001969(n),n=1..200) ; # R. J. Mathar, Aug 07 2022
  • Mathematica
    Select[Range[0,300], EvenQ[DigitCount[ #, 2][[1]]] &]
    a[ n_] := If[ n < 1, 0, With[{m = n - 1}, 2 m + Mod[-Total@IntegerDigits[m, 2], 2]]]; (* Michael Somos, Jun 09 2019 *)
  • PARI
    a(n)=n-=1; 2*n+subst(Pol(binary(n)),x,1)%2
    
  • PARI
    a(n)=if(n<1,0,if(n%2==0,a(n/2)+n,-a((n-1)/2)+3*n))
    
  • PARI
    a(n)=2*(n-1)+hammingweight(n-1)%2 \\ Charles R Greathouse IV, Mar 22 2013
    
  • Python
    def ok(n): return bin(n)[2:].count('1') % 2 == 0
    print(list(filter(ok, range(130)))) # Michael S. Branicky, Jun 02 2021
    
  • Python
    from itertools import chain, count, islice
    def A001969_gen(): # generator of terms
        return chain((0,),chain.from_iterable((sorted(n^ n<<1 for n in range(2**l,2**(l+1))) for l in count(0))))
    A001969_list = list(islice(A001969_gen(),30)) # Chai Wah Wu, Jun 29 2022
    
  • Python
    def A001969(n): return ((m:=n-1).bit_count()&1)+(m<<1) # Chai Wah Wu, Mar 03 2023

Formula

a(n+1) - A001285(n) = 2n-1 has been verified for n <= 400. - John W. Layman, May 16 2003 [This can be directly verified by comparing Ralf Stephan's formulas for this sequence (see below) and for A001285. - Jianing Song, Nov 04 2024]
Note that 2n+1 is in the sequence iff 2n is not and so this sequence has asymptotic density 1/2. - Franklin T. Adams-Watters, Aug 23 2006
a(n) = (1/2) * (4n - 3 - (-1)^A000120(n-1)). - Ralf Stephan, Sep 14 2003
G.f.: Sum_{k>=0} (t(3+2t+3t^2)/(1-t^2)^2) * Product_{l=0..k-1} (1-x^(2^l)), where t = x^2^k. - Ralf Stephan, Mar 25 2004
a(2*n+1) + a(2*n) = A017101(n-1) = 8*n-5.
a(2*n) - a(2*n-1) gives the Thue-Morse sequence (3, 1 version): 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, .... A001969(n) + A000069(n) = A016813(n-1) = 4*n-3. - Philippe Deléham, Feb 04 2004
a(1) = 0; for n > 1: a(n) = 3*n-3 - a(n/2) if n even, a(n) = a((n+1)/2)+n-1 if n odd.
Let b(n) = 1 if sum of digits of n is even, -1 if it is odd; then Shallit (1985) showed that Product_{n>=0} ((2n+1)/(2n+2))^b(n) = 1/sqrt(2).
a(n) = 2n - 2 + A010060(n-1). - Franklin T. Adams-Watters, Aug 28 2006
A005590(a(n-1)) <= 0. - Reinhard Zumkeller, Apr 11 2012
A106400(a(n-1)) = 1. - Reinhard Zumkeller, Apr 29 2012
a(n) = (a(n-1) + 2) XOR A010060(a(n-1) + 2). - Falk Hüffner, Jan 21 2022
a(n+1) = A006068(n) XOR (2*A006068(n)). - Rémy Sigrist, Apr 14 2022

Extensions

More terms from Robin Trew (trew(AT)hcs.harvard.edu)

A048675 If n = p_i^e_i * ... * p_k^e_k, p_i < ... < p_k primes (with p_i = prime(i)), then a(n) = (1/2) * (e_i * 2^i + ... + e_k * 2^k).

Original entry on oeis.org

0, 1, 2, 2, 4, 3, 8, 3, 4, 5, 16, 4, 32, 9, 6, 4, 64, 5, 128, 6, 10, 17, 256, 5, 8, 33, 6, 10, 512, 7, 1024, 5, 18, 65, 12, 6, 2048, 129, 34, 7, 4096, 11, 8192, 18, 8, 257, 16384, 6, 16, 9, 66, 34, 32768, 7, 20, 11, 130, 513, 65536, 8, 131072, 1025, 12, 6, 36, 19
Offset: 1

Views

Author

Antti Karttunen, Jul 14 1999

Keywords

Comments

The original motivation for this sequence was to encode the prime factorization of n in the binary representation of a(n), each such representation being unique as long as this map is restricted to A005117 (squarefree numbers, resulting a permutation of nonnegative integers A048672) or any of its subsequence, resulting an injective function like A048623 and A048639.
However, also the restriction to A260443 (not all terms of which are squarefree) results a permutation of nonnegative integers, namely A001477, the identity permutation.
When a polynomial with nonnegative integer coefficients is encoded with the prime factorization of n (e.g., as in A206296, A260443), then a(n) gives the evaluation of that polynomial at x=2.
The primitive completely additive integer sequence that satisfies a(n) = a(A225546(n)), n >= 1. By primitive, we mean that if b is another such sequence, then there is an integer k such that b(n) = k * a(n) for all n >= 1. - Peter Munn, Feb 03 2020
If the binary rank of an integer partition y is given by Sum_i 2^(y_i-1), and the Heinz number is Product_i prime(y_i), then a(n) is the binary rank of the integer partition with Heinz number n. Note the function taking a set s to Sum_i 2^(s_i-1) is the inverse of A048793 (binary indices), and the function taking a multiset m to Product_i prime(m_i) is the inverse of A112798 (prime indices). - Gus Wiseman, May 22 2024

Examples

			From _Gus Wiseman_, May 22 2024: (Start)
The A018819(7) = 6 cases of binary rank 7 are the following, together with their prime indices:
   30: {1,2,3}
   40: {1,1,1,3}
   54: {1,2,2,2}
   72: {1,1,1,2,2}
   96: {1,1,1,1,1,2}
  128: {1,1,1,1,1,1,1}
(End)
		

Crossrefs

Row 2 of A104244.
Similar logarithmic functions: A001414, A056239, A090880, A289506, A293447.
Left inverse of the following sequences: A000079, A019565, A038754, A068911, A134683, A260443, A332824.
A003961, A028234, A032742, A055396, A064989, A067029, A225546, A297845 are used to express relationship between terms of this sequence.
Cf. also A048623, A048676, A099884, A277896 and tables A277905, A285325.
Cf. A297108 (Möbius transform), A332813 and A332823 [= a(n) mod 3].
Pairs of sequences (f,g) that satisfy a(f(n)) = g(n), possibly with offset change: (A000203,A331750), (A005940,A087808), (A007913,A248663), (A007947,A087207), (A097248,A048675), (A206296,A000129), (A248692,A056239), (A283477,A005187), (A284003,A006068), (A285101,A028362), (A285102,A068052), (A293214,A001065), (A318834,A051953), (A319991,A293897), (A319992,A293898), (A320017,A318674), (A329352,A069359), (A332461,A156552), (A332462,A156552), (A332825,A000010) and apparently (A163511,A135529).
See comments/formulas in A277333, A331591, A331740 giving their relationship to this sequence.
The formula section details how the sequence maps the terms of A329050, A329332.
A277892, A322812, A322869, A324573, A324575 give properties of the n-th term of this sequence.
The term k appears A018819(k) times.
The inverse transformation is A019565 (Heinz number of binary indices).
The version for distinct prime indices is A087207.
Numbers k such that a(k) is prime are A277319, counts A372688.
Grouping by image gives A277905.
A014499 lists binary indices of prime numbers.
A061395 gives greatest prime index, least A055396.
A112798 lists prime indices, length A001222, reverse A296150, sum A056239.
Binary indices:
- listed A048793, sum A029931
- reversed A272020
- opposite A371572, sum A230877
- length A000120, complement A023416
- min A001511, opposite A000012
- max A070939, opposite A070940
- complement A368494, sum A359400
- opposite complement A371571, sum A359359

Programs

  • Maple
    nthprime := proc(n) local i; if(isprime(n)) then for i from 1 to 1000000 do if(ithprime(i) = n) then RETURN(i); fi; od; else RETURN(0); fi; end; # nthprime(2) = 1, nthprime(3) = 2, nthprime(5) = 3, etc. - this is also A049084.
    A048675 := proc(n) local s,d; s := 0; for d in ifactors(n)[ 2 ] do s := s + d[ 2 ]*(2^(nthprime(d[ 1 ])-1)); od; RETURN(s); end;
    # simpler alternative
    f:= n -> add(2^(numtheory:-pi(t[1])-1)*t[2], t=ifactors(n)[2]):
    map(f, [$1..100]); # Robert Israel, Oct 10 2016
  • Mathematica
    a[1] = 0; a[n_] := Total[ #[[2]]*2^(PrimePi[#[[1]]]-1)& /@ FactorInteger[n] ]; Array[a, 100] (* Jean-François Alcover, Mar 15 2016 *)
  • PARI
    a(n) = my(f = factor(n)); sum(k=1, #f~, f[k,2]*2^primepi(f[k,1]))/2; \\ Michel Marcus, Oct 10 2016
    
  • PARI
    \\ The following program reconstructs terms (e.g. for checking purposes) from the factorization file prepared by Hans Havermann:
    v048675sigs = readvec("a048675.txt");
    A048675(n) = if(n<=2,n-1,my(prsig=v048675sigs[n],ps=prsig[1],es=prsig[2]); prod(i=1,#ps,ps[i]^es[i])); \\ Antti Karttunen, Feb 02 2020
    
  • Python
    from sympy import factorint, primepi
    def a(n):
        if n==1: return 0
        f=factorint(n)
        return sum([f[i]*2**(primepi(i) - 1) for i in f])
    print([a(n) for n in range(1, 51)]) # Indranil Ghosh, Jun 19 2017

Formula

a(1) = 0, a(n) = 1/2 * (e1*2^i1 + e2*2^i2 + ... + ez*2^iz) if n = p_{i1}^e1*p_{i2}^e2*...*p_{iz}^ez, where p_i is the i-th prime. (e.g. p_1 = 2, p_2 = 3).
Totally additive with a(p^e) = e * 2^(PrimePi(p)-1), where PrimePi(n) = A000720(n). [Missing factor e added to the comment by Antti Karttunen, Jul 29 2015]
From Antti Karttunen, Jul 29 2015: (Start)
a(1) = 0; for n > 1, a(n) = 2^(A055396(n)-1) + a(A032742(n)). [Where A055396(n) gives the index of the smallest prime dividing n and A032742(n) gives the largest proper divisor of n.]
a(1) = 0; for n > 1, a(n) = (A067029(n) * (2^(A055396(n)-1))) + a(A028234(n)).
Other identities. For all n >= 0:
a(A019565(n)) = n.
a(A260443(n)) = n.
a(A206296(n)) = A000129(n).
a(A005940(n+1)) = A087808(n).
a(A007913(n)) = A248663(n).
a(A007947(n)) = A087207(n).
a(A283477(n)) = A005187(n).
a(A284003(n)) = A006068(n).
a(A285101(n)) = A028362(1+n).
a(A285102(n)) = A068052(n).
Also, it seems that a(A163511(n)) = A135529(n) for n >= 1. (End)
a(1) = 0, a(2n) = 1+a(n), a(2n+1) = 2*a(A064989(2n+1)). - Antti Karttunen, Oct 11 2016
From Peter Munn, Jan 31 2020: (Start)
a(n^2) = a(A003961(n)) = 2 * a(n).
a(A297845(n,k)) = a(n) * a(k).
a(n) = a(A225546(n)).
a(A329332(n,k)) = n * k.
a(A329050(n,k)) = 2^(n+k).
(End)
From Antti Karttunen, Feb 02-25 2020, Feb 01 2021: (Start)
a(n) = Sum_{d|n} A297108(d) = Sum_{d|A225546(n)} A297108(d).
a(n) = a(A097248(n)).
For n >= 2:
A001221(a(n)) = A322812(n), A001222(a(n)) = A277892(n).
A000203(a(n)) = A324573(n), A033879(a(n)) = A324575(n).
For n >= 1, A331750(n) = a(A000203(n)).
For n >= 1, the following chains hold:
A293447(n) >= a(n) >= A331740(n) >= A331591(n).
a(n) >= A087207(n) >= A248663(n).
(End)
a(n) = A087207(A097248(n)). - Flávio V. Fernandes, Jul 16 2025

Extensions

Entry revised by Antti Karttunen, Jul 29 2015
More linking formulas added by Antti Karttunen, Apr 18 2017

A016825 Positive integers congruent to 2 (mod 4): a(n) = 4*n+2, for n >= 0.

Original entry on oeis.org

2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154, 158, 162, 166, 170, 174, 178, 182, 186, 190, 194, 198, 202, 206, 210, 214, 218, 222, 226, 230, 234
Offset: 0

Views

Author

Keywords

Comments

Twice the odd numbers, also called singly even numbers.
Numbers having equal numbers of odd and even divisors: A001227(a(n)) = A000005(2*a(n)). - Reinhard Zumkeller, Dec 28 2003
Continued fraction for coth(1/2) = (e+1)/(e-1). The continued fraction for tanh(1/2) = (e-1)/(e+1) would be a(0) = 0, a(n) = A016825(n-1), n >= 1.
No solutions to a(n) = b^2 - c^2. - Henry Bottomley, Jan 13 2001
Sequence gives m such that 8 is the largest power of 2 dividing A003629(k)^m-1 for any k. - Benoit Cloitre, Apr 05 2002
k such that Sum_{d|k} (-1)^d = A048272(k) = 0. - Benoit Cloitre, Apr 15 2002
Also k such that Sum_{d|k} phi(d)*mu(k/d) = A007431(k) = 0. - Benoit Cloitre, Apr 15 2002
Also k such that Sum_{d|k} (d/A000005(d))*mu(k/d) = 0, k such that Sum_{d|k}(A000005(d)/d)*mu(k/d) = 0. - Benoit Cloitre, Apr 19 2002
Solutions to phi(x) = phi(x/2); primorial numbers are here. - Labos Elemer, Dec 16 2002
Together with 1, numbers that are not the leg of a primitive Pythagorean triangle. - Lekraj Beedassy, Nov 25 2003
For n > 0: complement of A107750 and A023416(a(n)-1) = A023416(a(n)) <> A023416(a(n)+1). - Reinhard Zumkeller, May 23 2005
Also the minimal value of Sum_{i=1..n+2} (p(i) - p(i+1))^2, where p(n+3) = p(1), as p ranges over all permutations of {1,2,...,n+2} (see the Mihai reference). Example: a(2)=10 because the values of the sum for the permutations of {1,2,3,4} are 10 (8 times), 12 (8 times) and 18 (8 times). - Emeric Deutsch, Jul 30 2005
Except for a(n)=2, numbers having 4 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005
A139391(a(n)) = A006370(a(n)) = A005408(n). - Reinhard Zumkeller, Apr 17 2008
Also a(n) = (n-1) + n + (n+1) + (n+2), so a(n) and -a(n) are all the integers that are sums of four consecutive integers. - Rick L. Shepherd, Mar 21 2009
The denominator in Pi/8 = 1/2 - 1/6 + 1/10 - 1/14 + 1/18 - 1/22 + .... - Mohammad K. Azarian, Oct 13 2011
This sequence gives the positive zeros of i^x + 1 = 0, x real, where i^x = exp(i*x*Pi/2). - Ilya Gutkovskiy, Aug 08 2015
Numbers k such that Sum_{j=1..k} j^3 is not a multiple of k. - Chai Wah Wu, Aug 23 2017
Numbers k such that Lucas(k) is a multiple of 3. - Bruno Berselli, Oct 17 2017
Also numbers k such that t^k == -1 (mod 5), where t is a term of A047221. - Bruno Berselli, Dec 28 2017
The even numbers form a ring, and these are the primes in that ring. Note that unique factorization into primes does not hold, since 60 = 2*30 = 6*10. - N. J. A. Sloane, Nov 11 2019
Also numbers ending with 10 in base 2. - John Keith, May 09 2022

Examples

			0.4621171572600097585023184... = 0 + 1/(2 + 1/(6 + 1/(10 + 1/(14 + ...)))), i.e., c.f. for tanh(1/2).
2.1639534137386528487700040... = 2 + 1/(6 + 1/(10 + 1/(14 + 1/(18 + ...)))), i.e., c.f. for coth(1/2).
		

References

  • H. Bass, Mathematics, Mathematicians and Mathematics Education, Bull. Amer. Math. Soc. (N.S.) 42 (2004), no. 4, 417-430.
  • Arthur Beiser, Concepts of Modern Physics, 2nd Ed., McGraw-Hill, 1973.
  • J. R. Goldman, The Queen of Mathematics, 1998, p. 70.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 262, 278.

Crossrefs

First differences of A001105.
Cf. A160327 (decimal expansion).
Subsequence of A042963.
Essentially the complement of A042965.

Programs

Formula

a(n) = 4*n + 2, for n >= 0.
a(n) = 2*A005408(n). - Lekraj Beedassy, Nov 28 2003
a(n) = A118413(n+1,2) for n>1. - Reinhard Zumkeller, Apr 27 2006
From Michael Somos, Apr 11 2007: (Start)
G.f.: 2*(1+x)/(1-x)^2.
E.g.f.: 2*(1+2*x)*exp(x).
a(n) = a(n-1) + 4.
a(-1-n) = -a(n). (End)
a(n) = 8*n - a(n-1) for n > 0, a(0)=2. - Vincenzo Librandi, Nov 20 2010
From Reinhard Zumkeller, Jun 11 2012, Jun 30 2012 and Jul 20 2012: (Start)
A080736(a(n)) = 0.
A007814(a(n)) = 1;
A037227(a(n)) = 3.
A214546(a(n)) = 0. (End)
a(n) = T(n+2) - T(n-2) where T(n) = n*(n+1)/2 = A000217(n). In general, if M(k,n) = 2*k*n + k, then M(k,n) = T(n+k) - T(n-k). - Charlie Marion, Feb 24 2020
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 1/sqrt(2-sqrt(2)) (A285871).
Product_{n>=1} (1 + (-1)^n/a(n)) = sqrt(1-1/sqrt(2)) (A154739). (End)

A054429 Simple self-inverse permutation of natural numbers: List each block of 2^n numbers (from 2^n to 2^(n+1) - 1) in reverse order.

Original entry on oeis.org

1, 3, 2, 7, 6, 5, 4, 15, 14, 13, 12, 11, 10, 9, 8, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 127, 126, 125, 124, 123, 122, 121
Offset: 1

Views

Author

Keywords

Comments

a(n) gives the position of the inverse of the n-th term in the full Stern-Brocot tree: A007305(a(n)+2) = A047679(n) and A047679(a(n)) = A007305(n+2). - Reinhard Zumkeller, Dec 22 2008
From Gary W. Adamson, Jun 21 2012: (Start)
The mapping and conversion rules are as follows:
By rows, we have ...
1;
3, 2;
7, 6, 5, 4;
15, 14, 13, 12, 11, 10, 9, 8;
... onto which we are to map one-half of the Stern-Brocot infinite Farey Tree:
1/2
1/3, 2/3
1/4, 2/5, 3/5, 3/4
1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5
...
The conversion rules are: Convert the decimal to binary, adding a duplicate of the rightmost binary term to its right. For example, 10 = 1010, which becomes 10100. Then, from the left, record the number of runs = [1,1,1,2], the continued fraction representation of 5/8. Check: 10 decimal corresponds to 5/8 as shown in the overlaid mapping. Take decimal 9 = 1001 which becomes 10011, with a continued fraction representation of [1,2,2] = 5/7. Check: 9 decimal corresponds to 5/7 in the Farey Tree map. (End)
From Indranil Ghosh, Jan 19 2017: (Start)
a(n) is the value generated when n is converted into its Elias gamma code, the 1's and 0's are interchanged and the resultant is converted back to its decimal value for all values of n > 1. For n = 1, A054429(n) = 1 but after converting 1 to Elias gamma code, interchanging the 1's and 0's and converting it back to decimal, the result produced is 0.
For example, let n = 10. The Elias gamma code for 10 is '1110010'. After interchanging the 1's and 0's it becomes "0001101" and 1101_2 = 13_10. So a(10) = 13. (End)
From Yosu Yurramendi, Mar 09 2017 (similar to Zumkeller's comment): (Start)
A002487(a(n)) = A002487(n+1), A002487(a(n)+1) = A002487(n), n > 0.
A162909(a(n)) = A162910(n), A162910(a(n)) = A162909(n), n > 0.
A162911(a(n)) = A162912(n), A162912(a(n)) = A162911(n), n > 0.
A071766(a(n)) = A245326(n), A245326(a(n)) = A071766(n), n > 0.
A229742(a(n)) = A245325(n), A245325(a(n)) = A229742(n), n > 0.
A020651(a(n)) = A245327(n), A245327(a(n)) = A020651(n), n > 0.
A020650(a(n)) = A245328(n), A245328(a(n)) = A020650(n), n > 0. (End)
From Yosu Yurramendi, Mar 29 2017: (Start)
A063946(a(n)) = a(A063946(n)) = A117120(n), n > 0.
A065190(a(n)) = a(A065190(n)) = A092569(n), n > 0.
A258746(a(n)) = a(A258746(n)) = A165199(n), n > 0.
A258996(a(n)) = a(A258996(n)), n > 0.
A117120(a(n)) = a(A117120(n)), n > 0.
A092569(a(n)) = a(A092569(n)), n > 0. (End)

Crossrefs

See also A054424, A054430.
{A000027, A054429, A059893, A059894} form a 4-group.
This is Guy Steele's sequence GS(6, 5) (see A135416).

Programs

  • Haskell
    a054429 n = a054429_list !! (n-1)
    a054429_list = f [1..] where
       f xs@(x:_) = reverse us ++ f vs where (us, vs) = splitAt x xs
    -- Reinhard Zumkeller, Jun 01 2015, Feb 21 2014
    
  • Maple
    A054429 := n -> 3*2^ilog2(n) - n - 1:
    seq(A054429(n), n = 1..70); # [Updated by Peter Luschny, Apr 24 2024]
  • Mathematica
    Flatten[Table[Range[2^(n+1)-1,2^n,-1],{n,0,6}]] (* Harvey P. Dale, Dec 17 2013 *)
  • PARI
    A054429(n)= 3<<#binary(n\2)-n-1 \\ M. F. Hasler, Aug 18 2014
    
  • Python
    from itertools import count, islice
    def A054429_gen(): # generator of terms
        return (m for n in count(0) for m in range((1<A054429_list = list(islice(A054429_gen(),30)) # Chai Wah Wu, Jul 27 2023
  • R
    maxblock <- 10 # by choice
    a <- NULL
    for(m in 0:maxblock) a <- c(a, rev(2^m:(2^(m+1)-1)))
    a
    # Yosu Yurramendi, Mar 10 2017
    

Formula

a(n) = ReflectBinTreePermutation(n).
a(n) = if n=1 then 1 else 2*a(floor(n/2)) + 1 - n mod 2. - Reinhard Zumkeller, Feb 18 2003
G.f.: 1/(1-x) * ((x-2x^2)/(1-x) + Sum_{k>=0} 3*2^k*x^2^k). - Ralf Stephan, Sep 15 2003
A000120(a(n)) = A000120(A059894(n)) = A023416(n) + 1. - Ralf Stephan, Oct 05 2003
A115310(n, 1) = a(n). - Reinhard Zumkeller, Jan 20 2006
a(1) = 1, a(2^(m+1) + k) = a(2^m+k) + 2^(m+1),
a(2^(m+1) + 2^m+k) = a(2^m+k) + 2^m, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Apr 06 2017
a(n) = A117120(A063946(n)) = A063946(A117120(n)) = A092569(A065190(n)) = A065190(A092569(n)), n > 0. - Yosu Yurramendi, Apr 10 2017
a(n) = 3*A053644(n) - n - 1. - Alan Michael Gómez Calderón, Feb 28 2025
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