A139352 -id:A139352 - cp's OEIS frontend

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

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

N. J. A. Sloane

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

			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)

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).

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Franklin T. Adams-Watters, and Frank Ruskey, Generating Functions for the Digital Sum and Other Digit Counting Sequences, JIS, Vol. 12 (2009), Article 09.5.6.
J.-P. Allouche, On an Inequality in a 1970 Paper of R. L. Graham, INTEGERS 21A (2021), #A2.
Jean-Paul Allouche and Jeffrey Shallit, The ring of k-regular sequences, Theoretical Computer Sci., Vol. 98 (1992), pp. 163-197. (PS file on author's web page.)
Jean-Paul Allouche, Jeffrey Shallit and Jonathan Sondow, Summation of Series Defined by Counting Blocks of Digits, arXiv:math/0512399 [math.NT], 2005-2006.
Jean-Paul Allouche, Jeffrey Shallit and Jonathan Sondow, Summation of series defined by counting blocks of digits, J. Number Theory, Vol. 123 (2007), pp. 133-143.
Richard Bellman and Harold N. Shapiro, On a problem in additive number theory, Annals Math., Vol. 49, No. 2 (1948), pp. 333-340. - _N. J. A. Sloane_, Mar 12 2009
C. Cobeli and A. Zaharescu, A game with divisors and absolute differences of exponents, Journal of Difference Equations and Applications, Vol. 20, #11 (2014) pp. 1489-1501, DOI: 10.1080/10236198.2014.940337. Also available as arXiv:1411.1334 [math.NT], 2014. See delta_n.
Jean Coquet, Power sums of digital sums, J. Number Theory, Vol. 22, No. 2 (1986), pp. 161-176.
F. M. Dekking, The Thue-Morse Sequence in Base 3/2, J. Int. Seq., Vol. 26 (2023), Article 23.2.3.
Karl Dilcher and Larry Ericksen, Hyperbinary expansions and Stern polynomials, Elec. J. Combin, Vol. 22 (2015), #P2.24.
Josef Eschgfäller and Andrea Scarpante, Dichotomic random number generators, arXiv:1603.08500 [math.CO], 2016.
Emmanuel Ferrand, Deformations of the Taylor Formula, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.7.
Steven R. Finch, Pascal Sebah and Zai-Qiao Bai, Odd Entries in Pascal's Trinomial Triangle, arXiv:0802.2654 [math.NT], 2008.
Philippe Flajolet, Peter Grabner, Peter Kirschenhofer, Helmut Prodinger and Robert F. Tichy, Mellin Transforms And Asymptotics: Digital Sums, Theoret. Computer Sci., Vol. 123, No. 2 (1994), pp. 291-314.
Michael Gilleland, Some Self-Similar Integer Sequences.
P. J. Grabner, P. Kirschenhofer, H. Prodinger and R. F. Tichy, On the moments of the sum-of-digits function, Applications of Fibonacci numbers, Vol. 5 (St. Andrews, 1992), Kluwer Acad. Publ., Dordrecht, 1993, 263-271.
Ronald L. Graham, On primitive graphs and optimal vertex assignments, Internat. Conf. Combin. Math. (New York, 1970), Annals of the NY Academy of Sciences, Vol. 175, 1970, pp. 170-186.
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Vacca-Type Series for Values of the Generalized Euler Constant Function and its Derivative, J. Integer Sequences, Vol. 13 (2010), Article 10.7.3.
Nick Hobson, Python program for this sequence.
Kathy Q. Ji and Herbert S. Wilf, Extreme Palindromes, arXiv:math/0611465 [math.CO], 2006.
Guy Louchard and Helmut Prodinger, The Largest Missing Value in a Composition of an Integer and Some Allouche-Shallit-Type Identities, J. Int. Seq., Vol. 16 (2013), Article 13.2.2.
J.-L. Mauclaire and Leo Murata, On q-additive functions, I, Proc. Japan Acad. Ser. A Math. Sci., Vol. 59, No. 6 (1983), pp. 274-276.
J.-L. Mauclaire and Leo Murata, On q-additive functions, II, Proc. Japan Acad. Ser. A Math. Sci., Vol. 59, No. 9 (1983), pp. 441-444.
M. D. McIlroy, The number of 1's in binary integers: bounds and extremal properties, SIAM J. Comput., Vol. 3 (1974), pp. 255-261.
Kerry Mitchell, Spirolateral-Type Images from Integer Sequences, 2013.
Sam Northshield, Stern's Diatomic Sequence 0,1,1,2,1,3,2,3,1,4,..., Amer. Math. Month., Vol. 117, No. 7 (2010), pp. 581-598.
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Carlo Sanna, On Arithmetic Progressions of Integers with a Distinct Sum of Digits, Journal of Integer Sequences, Vol. 15 (2012), Article 12.8.1. - _N. J. A. Sloane_, Dec 29 2012
Nanci Smith, Problem B-82, Fib. Quart., Vol. 4, No. 4 (1966), pp. 374-375.
Jonathan Sondow, New Vacca-type rational series for Euler's constant and its "alternating" analog ln 4/Pi, arXiv:math/0508042 [math.NT] 2005; Additive Number Theory, D. and G. Chudnovsky, eds., Springer, 2010, pp. 331-340.
Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions, 2004.
Ralf Stephan, Table of generating functions.
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Kenneth B. Stolarsky, Power and exponential sums of digital sums related to binomial coefficient parity, SIAM J. Appl. Math., Vol. 32 (1977), pp. 717-730. See B(n). - _N. J. A. Sloane_, Apr 05 2014
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Wikipedia, Hamming weight.
Wolfram Research, Numbers in Pascal's triangle.
Index entries for "core" sequences
Index entries for sequences related to binary expansion of n
Index entries for Colombian or self numbers and related sequences
Index entries for sequences that are fixed points of mappings

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.
Cf. A001620, A344716, A007814, A063787.
This is Guy Steele's sequence GS(3, 4) (see A135416).
Cf. A055640, A055641, A102669-A102685, A117804, A122840, A122841, A160093, A160094, A196563, A196564 (for base 10).
Cf. A230952 (boustrophedon transform).
Cf. A070939 (length of binary representation of n).

Fortran
```
c See link in A139351
```

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

a000120 = concat r
    where r = [0] : (map.map) (+1) (scanl1 (++) r)
-- Luke Palmer, Feb 16 2014

[Multiplicity(Intseq(n, 2), 1): n in [0..104]]; // Marius A. Burtea, Jan 22 2020

[&+Intseq(n, 2):n in [0..104]]; // Marius A. Burtea, Jan 22 2020

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

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 *)

{a(n) = if( n<0, 0, 2*n - valuation((2*n)!, 2))};

{a(n) = if( n<0, 0, subst(Pol(binary(n)), x ,1))};

{a(n) = if( n<1, 0, a(n\2) + n%2)}; /* Michael Somos, Mar 06 2004 */

a(n)=my(v=binary(n));sum(i=1,#v,v[i]) \\ Charles R Greathouse IV, Jun 24 2011

a(n)=norml2(binary(n)) \\ better use {A000120=hammingweight}. - M. F. Hasler, Oct 09 2012, edited Feb 27 2020

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

def A000120(n): return bin(n).count('1') # Chai Wah Wu, Sep 03 2014

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

def A000120(n): return n.bit_count() # Requires Python 3.10 or higher. - Pontus von Brömssen, Nov 08 2022

Python
```
# Also see links.
```

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

def A000120(n) : return sum(n.digits(2)) # Eric M. Schmidt, Apr 26 2013

(0 to 127).map(Integer.bitCount()) // _Alonso del Arte, Mar 05 2019

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

A065359 Alternating bit sum for n: replace 2^k with (-1)^k in binary expansion of n.

Original entry on oeis.org

0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2, -1, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2, -1, 0, 1, -1, 0, -2, -1, -3, -2, -1, 0, -2, -1, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2, -1, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 3, 4, 2, 3, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2

Offset: 0

Marc LeBrun, Oct 31 2001

Notation: (2)[n](-1)
From David W. Wilson and Ralf Stephan, Jan 09 2007: (Start)
a(n) is even iff n in A001969; a(n) is odd iff n in A000069.
a(n) == 0 (mod 3) iff n == 0 (mod 3).
a(n) == 0 (mod 6) iff (n == 0 (mod 3) and n/3 not in A036556).
a(n) == 3 (mod 6) iff (n == 0 (mod 3) and n/3 in A036556). (End)
a(n) = A030300(n) - A083905(n). - Ralf Stephan, Jul 12 2003
From Robert G. Wilson v, Feb 15 2011: (Start)
First occurrence of k and -k: 0, 1, 2, 5, 10, 21, 42, 85, ..., (A000975); i.e., first 0 occurs for 0, first 1 occurs for 1, first -1 occurs at 2, first 2 occurs for 5, etc.;
  a(n)=-3 only if n mod 3 = 0,
  a(n)=-2 only if n mod 3 = 1,
  a(n)=-1 only if n mod 3 = 2,
  a(n)= 0 only if n mod 3 = 0,
  a(n)= 1 only if n mod 3 = 1,
  a(n)= 2 only if n mod 3 = 2,
  a(n)= 3 only if n mod 3 = 0, ..., . (End)
a(n) modulo 2 is the Prouhet-Thue-Morse sequence A010060. - Philippe Deléham, Oct 20 2011
In the Koch curve, number the segments starting with n=0 for the first segment. The net direction (i.e., the sum of the preceding turns) of segment n is a(n)*60 degrees. This is since in the curve each base-4 digit 0,1,2,3 of n is a sub-curve directed respectively 0, +60, -60, 0 degrees, which is the net 0,+1,-1,0 of two bits in the sum here. - Kevin Ryde, Jan 24 2020

			Alternating bit sum for 11 = 1011 in binary is 1 - 1 + 0 - 1 = -1, so a(11) = -1.

Antti Karttunen, Table of n, a(n) for n = 0..65535 (terms 0..1000 from Harry J. Smith)
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197. [Preprint.]
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
H.-K. Hwang, S. Janson and T.-H. Tsai. Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications. Preprint, 2016.
H.-K. Hwang, S. Janson and T.-H. Tsai. Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications. ACM Transactions on Algorithms, 13:4 (2017), #47. DOI:10.1145/3127585.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 45.
Helge von Koch, Une Méthode Géométrique Élémentaire pour l'Étude de Certaines Questions de la Théorie des Courbes Planes, Acta Arithmetica, volume 30, 1906, pages 145-174.
William Paulsen, wpaulsen(AT)csm.astate.edu, Partitioning the [prime] maze
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions

Cf. A005536 (partial sums), A056832 (abs first differences), A010060 (mod 2), A039004 (indices of 0's).
Cf. A000120, A065360, A065081, A019565, A195017.
Cf. also A004718.
Cf. analogous sequences for bases 3-10: A065368, A346688, A346689, A346690, A346691, A346731, A346732, A055017 and also A373605 (for primorial base).

a065359 0 = 0
a065359 n = - a065359 n' + m where (n', m) = divMod n 2
-- Reinhard Zumkeller, Mar 20 2015

A065359 := proc(n) local dgs ; dgs := convert(n,base,2) ; add( -op(i,dgs)*(-1)^i,i=1..nops(dgs)) ; end proc: # R. J. Mathar, Feb 04 2011

f[0]=0; f[n_] := Plus @@ (-(-1)^Range[ Floor[ Log2@ n + 1]] Reverse@ IntegerDigits[n, 2]); Array[ f, 107, 0]

a(n) = my(s=0, u=1); for(k=0,#binary(n)-1,s+=bittest(n,k)*u;u=-u);s /* Washington Bomfim, Jan 18 2011 */

a(n) = my(b=binary(n)); b*[(-1)^k|k<-[-#b+1..0]]~; \\ Ruud H.G. van Tol, Oct 16 2023

a(n) = if(n==0, 0, 2*hammingweight(bitand(n, ((4<<(2*logint(n,4)))-1)/3)) - hammingweight(n)) \\ Andrew Howroyd, Dec 14 2024

def a(n):
    return sum((-1)**k for k, bi in enumerate(bin(n)[2:][::-1]) if bi=='1')
print([a(n) for n in range(107)]) # Michael S. Branicky, Jul 13 2021

from sympy.ntheory import digits
def A065359(n): return sum((0,1,-1,0)[i] for i in digits(n,4)[1:]) # Chai Wah Wu, Jul 19 2024

G.f.: (1/(1-x)) * Sum_{k>=0} (-1)^k*x^2^k/(1+x^2^k). - Ralf Stephan, Mar 07 2003
a(0) = 0, a(2n) = -a(n), a(2n+1) = 1-a(n). - Ralf Stephan, Mar 07 2003
a(n) = Sum_{k>=0} A030308(n,k)*(-1)^k. - Philippe Deléham, Oct 20 2011
a(n) = -a(floor(n/2)) + n mod 2. - Reinhard Zumkeller, Mar 20 2015
a(n) = A139351(n) - A139352(n). - Kevin Ryde, Jan 24 2020
G.f. A(x) satisfies: A(x) = x / (1 - x^2) - (1 + x) * A(x^2). - Ilya Gutkovskiy, Jul 28 2021
a(n) = A195017(A019565(n)). - Antti Karttunen, Jun 19 2024

More terms from Ralf Stephan, Jul 12 2003

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A139351 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives e(n).

Original entry on oeis.org

0, 1, 0, 1, 1, 2, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 1, 2, 2, 3, 2, 3, 0, 1, 0, 1, 1, 2, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 1, 2, 2, 3, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 2, 3, 3, 4, 3, 4, 1, 2, 1

Offset: 0

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n)+o(n) = A000120(n), the binary weight of n.

a(n) is also number of 1's and 3's in 4-ary representation of n. - Frank Ruskey, May 02 2009

			For n = 43 = 2^0 + 2^1 + 2^3 + 2^5, e(43)=1, o(43)=3.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Franklin T. Adams-Watters and Frank Ruskey, Generating Functions for the Digital Sum and Other Digit Counting Sequences, JIS 12 (2009), Article 09.5.6.
N. J. A. Sloane, Fortran program for this and related sequences.

Cf. A000120, A030308, A059841, A139352, A139353, A139354, A139355, A039004, A139370, A139371, A139372, A139373.

Fortran
```
See Sloane link.
```

import Data.List (unfoldr)
a139351 = sum . map (`mod` 2) .
   unfoldr (\x -> if x == 0 then Nothing else Just (x, x `div` 4))
-- Reinhard Zumkeller, Apr 22 2011

A139351 := proc(n)
    local a,bdgs,r;
    a := 0 ;
    bdgs := convert(n,base,2) ;
    for r from 1 to nops(bdgs) by 2 do
        if op(r,bdgs) = 1 then
            a := a+1 ;
        end if;
    end do:
    a;
end proc: # R. J. Mathar, Jul 21 2016

terms = 99; s = (1/(1-z))*Sum[z^(4^m)/(1+z^(4^m)), {m, 0, Log[4, terms] // Ceiling}] + O[z]^terms; CoefficientList[s, z] (* Jean-François Alcover, Jul 21 2017 *)
a[0] = 0; a[n_] := a[n] = a[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

a(n)=if(n>3,a(n\4))+n%2 \\ Charles R Greathouse IV, Apr 21 2016

a(n) + A139352(n) = A000120(n).
G.f.: (1/(1-z))*Sum_{m>=0} (z^(4^m)/(1+z^(4^m))). - Frank Ruskey, May 03 2009
Recurrence relation: a(0)=0, a(4m) = a(4m+2) = a(m), a(4m+1) = a(4m+3) = 1+a(m). - Frank Ruskey, May 11 2009
a(n) = Sum_{k} A030308(n,k)*A059841(k). - Philippe Deléham, Oct 14 2011

Typo in example fixed by Reinhard Zumkeller, Apr 22 2011

A139370 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence lists n such that e(n) < o(n).

Original entry on oeis.org

2, 8, 10, 11, 14, 26, 32, 34, 35, 38, 40, 41, 42, 43, 44, 46, 47, 50, 56, 58, 59, 62, 74, 98, 104, 106, 107, 110, 122, 128, 130, 131, 134, 136, 137, 138, 139, 140, 142, 143, 146, 152, 154, 155, 158, 160, 161, 162, 163, 164, 166, 167, 168, 169, 170, 171

Offset: 1

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

nonn

e(n)+o(n) = A000120(n), the binary weight of n. For e(n) = o(n) see A039004.

Primes of this sequence are in A065049; but A065049 contains also other primes (see A152715). [Vladimir Shevelev, Dec 11 2008]

Cf. A000120, A139351, A139352, A139353, A139354, A139355.

Cf. A039004, A139371, A139372, A139373.

Fortran
```
c See link in A139351
```

aQ[n_] := Module[{d = Reverse[IntegerDigits[n,2]]}, Total@d[[1;;-1;;2]] < Total@d[[2;;-1;;2]]]; Select[Range[180], aQ] (* Amiram Eldar, Dec 15 2018 *)

isok(n) = {my(irb = Vec(select(x->(x%2), Vecrev(binary(n)), 1))); #select(x->(x%2), irb) < #irb/2;} \\ Michel Marcus, Dec 15 2018

A139353 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives e(n)*o(n).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 2, 0, 1, 0, 2, 1, 2, 2, 4, 0, 0, 1, 2, 0, 0, 2, 3, 1, 2, 2, 4, 2, 3, 4, 6, 0, 1, 0, 2, 1, 2, 2, 4, 0, 2, 0, 3, 2, 4, 3, 6, 1, 2, 2, 4, 2, 3, 4, 6, 2, 4, 3, 6, 4, 6, 6, 9, 0, 0, 1, 2, 0, 0, 2, 3, 1, 2, 2, 4, 2, 3, 4, 6, 0, 0, 2, 3, 0, 0, 3, 4, 2, 3, 4, 6, 3, 4, 6, 8, 1, 2, 2

Offset: 0

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n) + o(n) = A000120(n), the binary weight of n.

			If n = 43 = 2^0+2^2+2^3+2^5, e(43)=1, o(43)=3.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000120, A139351, A139352, A139354, A139355, A039004, A139370, A139371, A139372, A139373.

Fortran
```
c See link in A139351
```

e[0] = 0; e[n_] := e[n] = e[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0];
o[0] = 0; o[n_] := o[n] = o[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0];
a[n_] := e[n] * o[n]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

a(n) = A139351(n) * A139352(n). - Amiram Eldar, Jul 18 2023

A139354 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives min{e(n), o(n)}.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 0, 1, 0, 1, 1, 1, 1, 2, 0, 1, 0, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2, 2, 3, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1

Offset: 0

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n) + o(n) = A000120(n), the binary weight of n.

			If n = 43 = 2^0+2^2+2^3+2^5, e(43)=1, o(43)=3.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000120, A139351, A139352, A139353, A139355, A039004, A139370, A139371, A139372, A139373.

Fortran
```
c See link in A139351
```

e[0] = 0; e[n_] := e[n] = e[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0];
o[0] = 0; o[n_] := o[n] = o[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0];
a[n_] := Min[e[n], o[n]]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

a(n) = min(A139351(n), A139352(n)). - Amiram Eldar, Jul 18 2023

A139355 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives max{e(n), o(n)}.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 3, 3, 2, 2, 3, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 2, 3, 3, 2, 3, 3, 3, 1, 2, 1, 2, 2, 3, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 3, 3, 4, 3, 4, 2, 3, 2, 3, 3, 4, 3, 4, 1, 2, 2

Offset: 0

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n) + o(n) = A000120(n), the binary weight of n.

			If n = 43 = 2^0+2^2+2^3+2^5, e(43)=1, o(43)=3.

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000120, A139351, A139352, A139353, A139354, A039004, A139370, A139371, A139372, A139373.

Fortran
```
c See link in A139351
```

e[0] = 0; e[n_] := e[n] = e[Floor[n/4]] + If[OddQ[Mod[n, 4]], 1, 0];
o[0] = 0; o[n_] := o[n] = o[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0];
a[n_] := Max[e[n], o[n]]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

a(n) = max(A139351(n), A139352(n)). - Amiram Eldar, Jul 18 2023

A139371 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence lists n such that e(n) <= o(n).

Original entry on oeis.org

0, 2, 3, 6, 8, 9, 10, 11, 12, 14, 15, 18, 24, 26, 27, 30, 32, 33, 34, 35, 36, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 54, 56, 57, 58, 59, 60, 62, 63, 66, 72, 74, 75, 78, 90, 96, 98, 99, 102, 104, 105, 106, 107, 108, 110, 111, 114, 120, 122, 123, 126

Offset: 1

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n)+o(n) = A000120(n), the binary weight of n. For e(n) = o(n) see A039004.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000120, A139351, A139352, A139353, A139354, A139355, A039004, A139370, A139372, A139373.

Fortran
```
c See link in A139351
```

q[n_] := Module[{d = Reverse[IntegerDigits[n, 2]]}, Total@ d[[1;; -1;; 2]] <= Total@ d[[2;; -1;; 2]]]; Select[Range[0, 130], q] (* Amiram Eldar, Aug 31 2023 *)

A139372 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence lists n such that e(n) >= o(n).

Original entry on oeis.org

0, 1, 3, 4, 5, 6, 7, 9, 12, 13, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 33, 36, 37, 39, 45, 48, 49, 51, 52, 53, 54, 55, 57, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91

Offset: 1

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

e(n)+o(n) = A000120(n), the binary weight of n. For e(n) = o(n) see A039004.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000120, A139351, A139352, A139353, A139354, A139355, A039004, A139370, A139371, A139373.

Fortran
```
c See link in A139351
```

q[n_] := Module[{d = Reverse[IntegerDigits[n, 2]]}, Total@ d[[1;; -1;; 2]] >= Total@ d[[2;; -1;; 2]]]; Select[Range[0, 100], q] (* Amiram Eldar, Aug 31 2023 *)

cp's OEIS Frontend

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

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A065359 Alternating bit sum for n: replace 2^k with (-1)^k in binary expansion of n.

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A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

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A139351 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives e(n).

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