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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A323234 Lexicographically earliest sequence such that a(i) = a(j) => f(i) = f(j), where f(1) = 0, and for n > 1, f(n) = ordered pair [A053645(n), A079944(n-2)], where A053645(n) gives n without its most significant bit, while A079944(n-2) gives the second most significant bit of n.

Original entry on oeis.org

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

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Author

Antti Karttunen, Jan 08 2019

Keywords

Comments

Also the restricted growth sequence transform of function f(1) = 0, f(n) = [A053645(n), A278222(n)] for n > 1.
For all i, j:
a(i) = a(j) => A286622(i) = A286622(j),
a(i) = a(j) => A323235(i) = A323235(j),
a(i) = a(j) => A323236(i) = A323236(j).

Crossrefs

Cf. also A300226 (an analogous filter sequence for prime factorization).

Programs

  • PARI
    up_to = 65537;
    rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
    A053644(n) = { my(k=1); while(k<=n, k<<=1); (k>>1); }; \\ From A053644
    A053645(n) = (n-A053644(n));
    A079944off0(n) = (1==binary(2+n)[2]);
    A323234aux(n) = if(1==n,0,[A053645(n), A079944off0(n-2)]);
    v323234 = rgs_transform(vector(up_to,n,A323234aux(n)));
    A323234(n) = v323234[n];

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

A007814 Exponent of highest power of 2 dividing n, a.k.a. the binary carry sequence, the ruler sequence, or the 2-adic valuation of n.

Original entry on oeis.org

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

Views

Author

John Tromp, Dec 11 1996

Keywords

Comments

This sequence is an exception to my usual rule that when every other term of a sequence is 0 then those 0's should be omitted. In this case we would get A001511. - N. J. A. Sloane
To construct the sequence: start with 0,1, concatenate to get 0,1,0,1. Add + 1 to last term gives 0,1,0,2. Concatenate those 4 terms to get 0,1,0,2,0,1,0,2. Add + 1 to last term etc. - Benoit Cloitre, Mar 06 2003
The sequence is invariant under the following two transformations: increment every element by one (1, 2, 1, 3, 1, 2, 1, 4, ...), put a zero in front and between adjacent elements (0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ...). The intermediate result is A001511. - Ralf Hinze (ralf(AT)informatik.uni-bonn.de), Aug 26 2003
Fixed point of the morphism 0->01, 1->02, 2->03, 3->04, ..., n->0(n+1), ..., starting from a(1) = 0. - Philippe Deléham, Mar 15 2004
Fixed point of the morphism 0->010, 1->2, 2->3, ..., n->(n+1), .... - Joerg Arndt, Apr 29 2014
a(n) is also the number of times to repeat a step on an even number in the hailstone sequence referenced in the Collatz conjecture. - Alex T. Flood (whiteangelsgrace(AT)gmail.com), Sep 22 2006
Let F(n) be the n-th Fermat number (A000215). Then F(a(r-1)) divides F(n)+2^k for r = k mod 2^n and r != 1. - T. D. Noe, Jul 12 2007
The following relation holds: 2^A007814(n)*(2*A025480(n-1)+1) = A001477(n) = n. (See functions hd, tl and cons in [Paul Tarau 2009].)
a(n) is the number of 0's at the end of n when n is written in base 2.
a(n+1) is the number of 1's at the end of n when n is written in base 2. - M. F. Hasler, Aug 25 2012
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 0). That is, A003188(n) XOR A003188(n+1) == 2^A007814(n). - Russ Cox, Dec 04 2010
The sequence is squarefree (in the sense of not containing any subsequence of the form XX) [Allouche and Shallit]. Of course it contains individual terms that are squares (such as 4). - Comment expanded by N. J. A. Sloane, Jan 28 2019
a(n) is the number of zero coefficients in the n-th Stern polynomial, A125184. - T. D. Noe, Mar 01 2011
Lemma: For n < m with r = a(n) = a(m) there exists n < k < m with a(k) > r. Proof: We have n=b2^r and m=c2^r with b < c both odd; choose an even i between them; now a(i2^r) > r and n < i2^r < m. QED. Corollary: Every finite run of consecutive integers has a unique maximum 2-adic valuation. - Jason Kimberley, Sep 09 2011
a(n-2) is the 2-adic valuation of A000166(n) for n >= 2. - Joerg Arndt, Sep 06 2014
a(n) = number of 1's in the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} p_j-th prime (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(24)=3; indeed, the partition having Heinz number 24 = 2*2*2*3 is [1,1,1,2]. - Emeric Deutsch, Jun 04 2015
a(n+1) is the difference between the two largest parts in the integer partition having viabin number n (0 is assumed to be a part). Example: a(20) = 2. Indeed, we have 19 = 10011_2, leading to the Ferrers board of the partition [3,1,1]. For the definition of viabin number see the comment in A290253. - Emeric Deutsch, Aug 24 2017
Apart from being squarefree, as noted above, the sequence has the property that every consecutive subsequence contains at least one number an odd number of times. - Jon Richfield, Dec 20 2018
a(n+1) is the 2-adic valuation of Sum_{e=0..n} u^e = (1 + u + u^2 + ... + u^n), for any u of the form 4k+1 (A016813). - Antti Karttunen, Aug 15 2020
{a(n)} represents the "first black hat" strategy for the game of countably infinitely many hats, with a probability of success of 1/3; cf. the Numberphile link below. - Frederic Ruget, Jun 14 2021
a(n) is the least nonnegative integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022

Examples

			2^3 divides 24, so a(24)=3.
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
  0;
  1,0;
  2,0,1,0;
  3,0,1,0,2,0,1,0;
  4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  5,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
  6,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,0,1,0,2,...
(End)
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 27.
  • K. Atanassov, On the 37th and the 38th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 2, 83-85.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

Cf. A011371 (partial sums), A094267 (first differences), A001511 (bisection), A346070 (mod 4).
Bisection of A050605 and |A088705|. Pairwise sums are A050603 and A136480. Difference of A285406 and A281264.
This is Guy Steele's sequence GS(1, 4) (see A135416). Cf. A053398(1,n). Column/row 1 of table A050602.
Cf. A007949 (3-adic), A235127 (4-adic), A112765 (5-adic), A122841 (6-adic), A214411 (7-adic), A244413 (8-adic), A122840 (10-adic).
Cf. A086463 (Dgf at s=2).

Programs

  • Haskell
    a007814 n = if m == 0 then 1 + a007814 n' else 0
                where (n', m) = divMod n 2
    -- Reinhard Zumkeller, Jul 05 2013, May 14 2011, Apr 08 2011
    
  • Haskell
    a007814 n | odd n = 0 | otherwise = 1 + a007814 (n `div` 2)
    --  Walt Rorie-Baety, Mar 22 2013
    
  • Magma
    [Valuation(n, 2): n in [1..120]]; // Bruno Berselli, Aug 05 2013
    
  • Maple
    ord := proc(n) local i,j; if n=0 then return 0; fi; i:=0; j:=n; while j mod 2 <> 1 do i:=i+1; j:=j/2; od: i; end proc: seq(ord(n), n=1..111);
    A007814 := n -> padic[ordp](n,2): seq(A007814(n), n=1..111); # Peter Luschny, Nov 26 2010
  • Mathematica
    Table[IntegerExponent[n, 2], {n, 64}] (* Eric W. Weisstein *)
    IntegerExponent[Range[64], 2] (* Eric W. Weisstein, Feb 01 2024 *)
    p=2; Array[ If[ Mod[ #, p ]==0, Select[ FactorInteger[ # ], Function[ q, q[ [ 1 ] ]==p ], 1 ][ [ 1, 2 ] ], 0 ]&, 96 ]
    DigitCount[BitXor[x, x - 1], 2, 1] - 1; a different version based on the same concept: Floor[Log[2, BitXor[x, x - 1]]] (* Jaume Simon Gispert (jaume(AT)nuem.com), Aug 29 2004 *)
    Nest[Join[ #, ReplacePart[ #, Length[ # ] -> Last[ # ] + 1]] &, {0, 1}, 5] (* N. J. Gunther, May 23 2009 *)
    Nest[ Flatten[# /. a_Integer -> {0, a + 1}] &, {0}, 7] (* Robert G. Wilson v, Jan 17 2011 *)
  • PARI
    A007814(n)=valuation(n,2);
    
  • Python
    import math
    def a(n): return int(math.log(n - (n & n - 1), 2)) # Indranil Ghosh, Apr 18 2017
    
  • Python
    def A007814(n): return (~n & n-1).bit_length() # Chai Wah Wu, Jul 01 2022
    
  • R
    sapply(1:100,function(x) sum(gmp::factorize(x)==2)) # Christian N. K. Anderson, Jun 20 2013
    
  • Scheme
    (define (A007814 n) (let loop ((n n) (e 0)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = A001511(n) - 1.
a(2*n) = A050603(2*n) = A001511(n).
a(n) = A091090(n-1) + A036987(n-1) - 1.
a(n) = 0 if n is odd, otherwise 1 + a(n/2). - Reinhard Zumkeller, Aug 11 2001
Sum_{k=1..n} a(k) = n - A000120(n). - Benoit Cloitre, Oct 19 2002
G.f.: A(x) = Sum_{k>=1} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Apr 10 2002
G.f. A(x) satisfies A(x) = A(x^2) + x^2/(1-x^2). A(x) = B(x^2) = B(x) - x/(1-x), where B(x) is the g.f. for A001151. - Franklin T. Adams-Watters, Feb 09 2006
Totally additive with a(p) = 1 if p = 2, 0 otherwise.
Dirichlet g.f.: zeta(s)/(2^s-1). - Ralf Stephan, Jun 17 2007
Define 0 <= k <= 2^n - 1; binary: k = b(0) + 2*b(1) + 4*b(2) + ... + 2^(n-1)*b(n-1); where b(x) are 0 or 1 for 0 <= x <= n - 1; define c(x) = 1 - b(x) for 0 <= x <= n - 1; Then: a(k) = c(0) + c(0)*c(1) + c(0)*c(1)*c(2) + ... + c(0)*c(1)...c(n-1); a(k+1) = b(0) + b(0)*b(1) + b(0)*b(1)*b(2) + ... + b(0)*b(1)...b(n-1). - Arie Werksma (werksma(AT)tiscali.nl), May 10 2008
a(n) = floor(A002487(n - 1) / A002487(n)). - Reikku Kulon, Oct 05 2008
Sum_{k=1..n} (-1)^A000120(n-k)*a(k) = (-1)^(A000120(n)-1)*(A000120(n) - A000035(n)). - Vladimir Shevelev, Mar 17 2009
a(A001147(n) + A057077(n-1)) = a(2*n). - Vladimir Shevelev, Mar 21 2009
For n>=1, a(A004760(n+1)) = a(n). - Vladimir Shevelev, Apr 15 2009
2^(a(n)) = A006519(n). - Philippe Deléham, Apr 22 2009
a(n) = A063787(n) - A000120(n). - Gary W. Adamson, Jun 04 2009
a(C(n,k)) = A000120(k) + A000120(n-k) - A000120(n). - Vladimir Shevelev, Jul 19 2009
a(n!) = n - A000120(n). - Vladimir Shevelev, Jul 20 2009
v_{2}(n) = Sum_{r>=1} (r / 2^(r+1)) Sum_{k=0..2^(r+1)-1} e^(2(k*Pi*i(n+2^r))/(2^(r+1))). - A. Neves, Sep 28 2010, corrected Oct 04 2010
a(n) mod 2 = A096268(n-1). - Robert G. Wilson v, Jan 18 2012
a(A005408(n)) = 1; a(A016825(n)) = 3; A017113(a(n)) = 5; A051062(a(n)) = 7; a(n) = (A037227(n)-1)/2. - Reinhard Zumkeller, Jun 30 2012
a((2*n-1)*2^p) = p, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 04 2013
a(n) = A067255(n,1). - Reinhard Zumkeller, Jun 11 2013
a(n) = log_2(n - (n AND n-1)). - Gary Detlefs, Jun 13 2014
a(n) = 1 + A000120(n-1) - A000120(n), where A000120 is the Hamming weight function. - Stanislav Sykora, Jul 14 2014
A053398(n,k) = a(A003986(n-1,k-1)+1); a(n) = A053398(n,1) = A053398(n,n) = A053398(2*n-1,n) = Min_{k=1..n} A053398(n,k). - Reinhard Zumkeller, Aug 04 2014
a((2*x-1)*2^n) = a((2*y-1)*2^n) for positive n, x and y. - Juri-Stepan Gerasimov, Aug 04 2016
a(n) = A285406(n) - A281264(n). - Ralf Steiner, Apr 18 2017
a(n) = A000005(n)/(A000005(2*n) - A000005(n)) - 1. - conjectured by Velin Yanev, Jun 30 2017, proved by Nicholas Stearns, Sep 11 2017
Equivalently to above formula, a(n) = A183063(n) / A001227(n), i.e., a(n) is the number of even divisors of n divided by number of odd divisors of n. - Franklin T. Adams-Watters, Oct 31 2018
a(n)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Feb 16 2019
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1. - Amiram Eldar, Jul 11 2020
a(n) = 2*Sum_{j=1..floor(log_2(n))} frac(binomial(n, 2^j)*2^(j-1)/n). - Dario T. de Castro, Jul 08 2022
a(n) = A070939(n) - A070939(A030101(n)). - Andrew T. Porter, Dec 16 2022
a(n) = floor((gcd(n, 2^n)^(n+1) mod (2^(n+1)-1)^2)/(2^(n+1)-1)) (see Lemma 3.4 from Mazzanti's 2002 article). - Lorenzo Sauras Altuzarra, Mar 10 2024
a(n) = 1 - A088705(n). - Chai Wah Wu, Sep 18 2024

Extensions

Formula index adapted to the offset of A025480 by R. J. Mathar, Jul 20 2010
Edited by Ralf Stephan, Feb 08 2014

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

Views

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

A000265 Remove all factors of 2 from n; or largest odd divisor of n; or odd part of n.

Original entry on oeis.org

1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 1, 17, 9, 19, 5, 21, 11, 23, 3, 25, 13, 27, 7, 29, 15, 31, 1, 33, 17, 35, 9, 37, 19, 39, 5, 41, 21, 43, 11, 45, 23, 47, 3, 49, 25, 51, 13, 53, 27, 55, 7, 57, 29, 59, 15, 61, 31, 63, 1, 65, 33, 67, 17, 69, 35, 71, 9, 73, 37, 75, 19, 77
Offset: 1

Views

Author

Keywords

Comments

When n > 0 is written as k*2^j with k odd then k = A000265(n) and j = A007814(n), so: when n is written as k*2^j - 1 with k odd then k = A000265(n+1) and j = A007814(n+1), when n > 1 is written as k*2^j + 1 with k odd then k = A000265(n-1) and j = A007814(n-1).
Also denominator of 2^n/n (numerator is A075101(n)). - Reinhard Zumkeller, Sep 01 2002
Slope of line connecting (o, a(o)) where o = (2^k)(n-1) + 1 is 2^k and (by design) starts at (1, 1). - Josh Locker (joshlocker(AT)macfora.com), Apr 17 2004
Numerator of n/2^(n-1). - Alexander Adamchuk, Feb 11 2005
From Marco Matosic, Jun 29 2005: (Start)
"The sequence can be arranged in a table:
1
1 3 1
1 5 3 7 1
1 9 5 11 3 13 7 15 1
1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31 1
Every new row is the previous row interspaced with the continuation of the odd numbers.
Except for the ones; the terms (t) in each column are t+t+/-s = t_+1. Starting from the center column of threes and working to the left the values of s are given by A000265 and working to the right by A000265." (End)
This is a fractal sequence. The odd-numbered elements give the odd natural numbers. If these elements are removed, the original sequence is recovered. - Kerry Mitchell, Dec 07 2005
2k + 1 is the k-th and largest of the subsequence of k terms separating two successive equal entries in a(n). - Lekraj Beedassy, Dec 30 2005
It's not difficult to show that the sum of the first 2^n terms is (4^n + 2)/3. - Nick Hobson, Jan 14 2005
In the table, for each row, (sum of terms between 3 and 1) - (sum of terms between 1 and 3) = A020988. - Eric Desbiaux, May 27 2009
This sequence appears in the analysis of A160469 and A156769, which resemble the numerator and denominator of the Taylor series for tan(x). - Johannes W. Meijer, May 24 2009
Indices n such that a(n) divides 2^n - 1 are listed in A068563. - Max Alekseyev, Aug 25 2013
From Alexander R. Povolotsky, Dec 17 2014: (Start)
With regard to the tabular presentation described in the comment by Marco Matosic: in his drawing, starting with the 3rd row, the first term in the row, which is equal to 1 (or, alternatively the last term in the row, which is also equal to 1), is not in the actual sequence and is added to the drawing as a fictitious term (for the sake of symmetry); an actual A000265(n) could be considered to be a(j,k) (where j >= 1 is the row number and k>=1 is the column subscript), such that a(j,1) = 1:
1
1 3
1 5 3 7
1 9 5 11 3 13 7 15
1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31
and so on ... .
The relationship between k and j for each row is 1 <= k <= 2^(j-1). In this corrected tabular representation, Marco's notion that "every new row is the previous row interspaced with the continuation of the odd numbers" remains true. (End)
Partitions natural numbers to the same equivalence classes as A064989. That is, for all i, j: a(i) = a(j) <=> A064989(i) = A064989(j). There are dozens of other such sequences (like A003602) for which this also holds: In general, all sequences for which a(2n) = a(n) and the odd bisection is injective. - Antti Karttunen, Apr 15 2017
From Paul Curtz, Feb 19 2019: (Start)
This sequence is the truncated triangle:
1, 1;
3, 1, 5;
3, 7, 1, 9;
5, 11, 3, 13, 7;
15, 1, 17, 9, 19, 5;
21, 11, 23, 3, 25, 13, 27;
7, 29, 15, 31, 1, 33, 17, 35;
...
The first column is A069834. The second column is A213671. The main diagonal is A236999. The first upper diagonal is A125650 without 0.
c(n) = ((n*(n+1)/2))/A069834 = 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 8, 8, 1, 1, ... for n > 0. n*(n+1)/2 is the rank of A069834. (End)
As well as being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 27 2019
a(n) is also the map n -> A026741(n) applied at least A007814(n) times. - Federico Provvedi, Dec 14 2021

Examples

			G.f. = x + x^2 + 3*x^3 + x^4 + 5*x^5 + 3*x^6 + 7*x^7 + x^8 + 9*x^9 + 5*x^10 + 11*x^11 + ...
		

References

  • 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

Cf. A049606 (partial products), A135013 (partial sums), A099545 (mod 4), A326937 (Dirichlet inverse).
Cf. A026741 (map), A001511 (converging steps), A038550 (prime index).
Cf. A195056 (Dgf at s=3).

Programs

  • Haskell
    a000265 = until odd (`div` 2)
    -- Reinhard Zumkeller, Jan 08 2013, Apr 08 2011, Oct 14 2010
    
  • Java
    int A000265(n){
        while(n%2==0) n>>=1;
        return n;
    }
    /* Aidan Simmons, Feb 24 2019 */
    
  • Julia
    using IntegerSequences
    [OddPart(n) for n in 1:77] |> println  # Peter Luschny, Sep 25 2021
    
  • Magma
    A000265:= func< n | n/2^Valuation(n,2) >;
    [A000265(n): n in [1..120]]; // G. C. Greubel, Jul 31 2024
    
  • Maple
    A000265:=proc(n) local t1,d; t1:=1; for d from 1 by 2 to n do if n mod d = 0 then t1:=d; fi; od; t1; end: seq(A000265(n), n=1..77);
    A000265 := n -> n/2^padic[ordp](n,2): seq(A000265(n), n=1..77); # Peter Luschny, Nov 26 2010
  • Mathematica
    a[n_Integer /; n > 0] := n/2^IntegerExponent[n, 2]; Array[a, 77] (* Josh Locker *)
    a[ n_] := If[ n == 0, 0, n / 2^IntegerExponent[ n, 2]]; (* Michael Somos, Dec 17 2014 *)
  • PARI
    {a(n) = n >> valuation(n, 2)}; /* Michael Somos, Aug 09 2006, edited by M. F. Hasler, Dec 18 2014 */
    
  • Python
    from _future_ import division
    def A000265(n):
        while not n % 2:
            n //= 2
        return n # Chai Wah Wu, Mar 25 2018
    
  • Python
    def a(n):
        while not n&1: n >>= 1
        return n
    print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Jun 26 2025
    
  • SageMath
    def A000265(n): return n//2^valuation(n,2)
    [A000265(n) for n in (1..121)] # G. C. Greubel, Jul 31 2024
  • Scheme
    (define (A000265 n) (let loop ((n n)) (if (odd? n) n (loop (/ n 2))))) ;; Antti Karttunen, Apr 15 2017
    

Formula

a(n) = if n is odd then n, otherwise a(n/2). - Reinhard Zumkeller, Sep 01 2002
a(n) = n/A006519(n) = 2*A025480(n-1) + 1.
Multiplicative with a(p^e) = 1 if p = 2, p^e if p > 2. - David W. Wilson, Aug 01 2001
a(n) = Sum_{d divides n and d is odd} phi(d). - Vladeta Jovovic, Dec 04 2002
G.f.: -x/(1 - x) + Sum_{k>=0} (2*x^(2^k)/(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))). - Ralf Stephan, Sep 05 2003
(a(k), a(2k), a(3k), ...) = a(k)*(a(1), a(2), a(3), ...) In general, a(n*m) = a(n)*a(m). - Josh Locker (jlocker(AT)mail.rochester.edu), Oct 04 2005
a(n) = Sum_{k=0..n} A127793(n,k)*floor((k+2)/2) (conjecture). - Paul Barry, Jan 29 2007
Dirichlet g.f.: zeta(s-1)*(2^s - 2)/(2^s - 1). - Ralf Stephan, Jun 18 2007
a(A132739(n)) = A132739(a(n)) = A132740(n). - Reinhard Zumkeller, Aug 27 2007
a(n) = 2*A003602(n) - 1. - Franklin T. Adams-Watters, Jul 02 2009
a(n) = n/gcd(2^n,n). (This also shows that the true offset is 0 and a(0) = 0.) - Peter Luschny, Nov 14 2009
a(-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2011
From Reinhard Zumkeller, May 01 2012: (Start)
A182469(n, k) = A027750(a(n), k), k = 1..A001227(n).
a(n) = A182469(n, A001227(n)). (End)
a((2*n-1)*2^p) = 2*n - 1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
G.f.: G(0)/(1 - 2*x^2 + x^4) - 1/(1 - x), where G(k) = 1 + 1/(1 - x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2)))/(x^(2^k)*(1 - 2*x^(2^(k+1)) + x^(2^(k+2))) + (1 - 2*x^(2^(k+2)) + x^(2^(k+3)))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 06 2013
a(n) = A003961(A064989(n)). - Antti Karttunen, Apr 15 2017
Completely multiplicative with a(2) = 1 and a(p) = p for prime p > 2, i.e., the sequence b(n) = a(n) * A008683(n) for n > 0 is the Dirichlet inverse of a(n). - Werner Schulte, Jul 08 2018
From Peter Bala, Feb 27 2019: (Start)
O.g.f.: F(x) - F(x^2) - F(x^4) - F(x^8) - ..., where F(x) = x/(1 - x)^2 is the generating function for the positive integers.
O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8) + ..., where L(x) = log(1/(1 - x)).
Sum_{n >= 1} x^n/a(n) = 1/2*log(G(x)), where G(x) = 1 + 2*x + 4*x^2 + 6*x^3 + 10*x^4 + ... is the o.g.f. of A000123. (End)
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(2*n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019
a(n) = A049606(n) / A049606(n-1). - Flávio V. Fernandes, Dec 08 2020
a(n) = numerator of n/2^(floor(n/2)). - Federico Provvedi, Dec 14 2021
a(n) = Sum_{d divides n} (-1)^(d+1)*phi(2*n/d). - Peter Bala, Jan 14 2024
a(n) = A030101(A030101(n)). - Darío Clavijo, Sep 19 2024

Extensions

Additional comments from Henry Bottomley, Mar 02 2000
More terms from Larry Reeves (larryr(AT)acm.org), Mar 14 2000
Name clarified by David A. Corneth, Apr 15 2017

A000695 Moser-de Bruijn sequence: sums of distinct powers of 4.

Original entry on oeis.org

0, 1, 4, 5, 16, 17, 20, 21, 64, 65, 68, 69, 80, 81, 84, 85, 256, 257, 260, 261, 272, 273, 276, 277, 320, 321, 324, 325, 336, 337, 340, 341, 1024, 1025, 1028, 1029, 1040, 1041, 1044, 1045, 1088, 1089, 1092, 1093, 1104, 1105, 1108, 1109, 1280, 1281, 1284, 1285
Offset: 0

Views

Author

Keywords

Comments

Although this is a list, it has offset 0 for both historical and mathematical reasons.
Numbers whose set of base-4 digits is a subset of {0,1}. - Ray Chandler, Aug 03 2004, corrected by M. F. Hasler, Oct 16 2018
Numbers k such that the sum of the base-2 digits of k = sum of the base-4 digits of k. - Clark Kimberling
Numbers having the same representation in both binary and negabinary (A039724). - Eric W. Weisstein
This sequence has many other interesting and useful properties. Every term k corresponds to a unique pair i,j with k = a(i) + 2*a(j) (i=A059905(n), j=A059906(n)) -- see A126684. Every list of numbers L = [L1,L2,L3,...] can be encoded uniquely by "recursive binary interleaving", where f(L) = a(L1) + 2*a(f([L2,L3,...])) with f([])=0. - Marc LeBrun, Feb 07 2001
This may be described concisely using the "rebase" notation b[n]q, which means "replace b with q in the expansion of n", thus "rebasing" n from base b into base q. The present sequence is 2[n]4. Many interesting operations (e.g., 10[n](1/10) = digit reverse, shifted) are nicely expressible this way. Note that q[n]b is (roughly) inverse to b[n]q. It's also natural to generalize the idea of "basis" so as to cover the likes of F[n]2, the so-called "fibbinary" numbers (A003714) and provide standard ready-made images of entities obeying other arithmetics, say like GF2[n]2 (e.g., primes = A014580, squares = the present sequence, etc.). - Marc LeBrun, Mar 24 2005
a(n) is also equal to the product n X n formed using carryless binary multiplication (A059729, A063010). - Henry Bottomley, Jul 03 2001
Numbers k such that A004117(k) is odd. - Pontus von Brömssen, Nov 25 2008
Fixed point of the morphism: 0 -> 01; 1 -> 45; 2 -> 89; ...; n -> (4n)(4n+1), starting from a(0)=0. - Philippe Deléham, Oct 22 2011
If n is even and present, so is n+1. - Robert G. Wilson v, Oct 24 2014
Also: interleave binary digits of n with 0's. (Equivalent to the "rebase" interpretation above.) - M. F. Hasler, Oct 16 2018
Named after the Austrian-Canadian mathematician Leo Moser (1921-1970) and the Dutch mathematician Nicolaas Govert de Bruijn (1918-2012). - Amiram Eldar, Jun 19 2021
Conjecture: The sums of distinct powers of k > 2 can be constructed as the following (k-1)-ary rooted tree. For each n the tree grows and a(n) is then the total number of nodes. For n = 1, the root of the tree is added. For n > 1, if n is odd one leaf of depth n-2 grows one child. If n is even all leaves of depth >= (n - 1 - A000225(A001511(n/2))) grow the maximum number of children. An illustration is provided in the links. - John Tyler Rascoe, Oct 09 2022

Examples

			G.f.: x + 4*x^2 + 5*x^3 + 16*x^4 + 17*x^5 + 20*x^6 + 21*x^7 + 64*x^8 + ...
If n=27, then b_0=1, b_1=1, b_2=0, b_3=1, b_4=1. Therefore a(27) = 4^4 + 4^3 + 4 + 1 = 325; k = b_0 + b_2*2 + b_4*2^2 = 5, l = b_1 + b_3*2 = 3, such that a(5)=17, a(3)=5 and 27 = 17 + 2*5. - _Vladimir Shevelev_, Nov 10 2008
		

References

  • 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

For generating functions Product_{k>=0} (1 + a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
Main diagonal of A048720, second column of A048723.
A062880(n) = 2*a(n); A001196(n) = 3*a(n).
Row 4 of array A104257.

Programs

  • C
    uint32_t a_next(uint32_t a_n) { return (a_n + 0xaaaaaaab) & 0x55555555; } /* Falk Hüffner, Jan 24 2022 */
  • Haskell
    a000695 n = if n == 0 then 0 else 4 * a000695 n' + b
                where (n',b) = divMod n 2
    -- Reinhard Zumkeller, Feb 21 2014, Dec 03 2011
    
  • Julia
    function a(n)
        m, r, b = n, 0, 1
        while m > 0
            m, q = divrem(m, 2)
            r += b * q
            b *= 4
        end
    r end; [a(n) for n in 0:51] |> println # Peter Luschny, Jan 03 2021
    
  • Magma
    m:=60; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!( (&+[4^k*x^(2^k)/(1+x^(2^k)): k in [0..20]])/(1-x) )); // G. C. Greubel, Dec 06 2018
    
  • Maple
    a:= proc(n) local m, r, b; m, r, b:= n, 0, 1;
          while m>0 do r:= r+b*irem(m, 2, 'm'); b:= b*4 od; r
        end:
    seq(a(n), n=0..100);  # Alois P. Heinz, Mar 16 2013
  • Mathematica
    Table[FromDigits[Riffle[IntegerDigits[n, 2], 0], 2], {n, 0, 51}] (* Jacob A. Siehler, Jun 30 2010 *)
    Table[FromDigits[IntegerDigits[n, 2], 4], {n, 0, 51}] (* IWABUCHI Yu(u)ki, Apr 06 2013 *)
    Union@ Flatten@ NestList[ Join[ 4#, 4# + 1] &, {0}, 6] (* Robert G. Wilson v, Aug 30 2014 *)
    Select[ Range[0, 1320], Total@ IntegerDigits[#, 2] == Total@ IntegerDigits[#, 4] &] (* Robert G. Wilson v, Oct 24 2014 *)
    Union[FromDigits[#,4]&/@Flatten[Table[Tuples[{0,1},n],{n,6}],1]] (* Harvey P. Dale, Oct 03 2015 *)
    a[ n_] := Which[n < 1, 0, EvenQ[n], a[n/2] 4, True, a[n - 1] + 1]; (* Michael Somos, Nov 30 2016 *)
  • PARI
    a(n)=n=binary(n);sum(i=1,#n,n[i]*4^(#n-i)) \\ Charles R Greathouse IV, Mar 04 2013
    
  • PARI
    {a(n) = if( n<1, 0, n%2, a(n-1) + 1, a(n/2) * 4)}; /* Michael Somos, Nov 30 2016 */
    
  • PARI
    A000695(n)=fromdigits(binary(n),4) \\ M. F. Hasler, Oct 16 2018
    
  • Python
    def a(n):
        n = bin(n)[2:]
        x = len(n)
        return sum(int(n[i]) * 4**(x - 1 - i) for i in range(x))
    [a(n) for n in range(101)] # Indranil Ghosh, Jun 25 2017
    
  • Python
    def a():
        x = 0
        while True:
            yield x
            y = ~(x << 1)
            x = (x - y) & y # Falk Hüffner, Dec 21 2021
    
  • Python
    from itertools import count, islice
    def A000695_gen(): # generator of terms
        yield (a:=0)
        for n in count(1):
            yield (a := a+((1<<((~n & n-1).bit_length()<<1)+1)+1)//3)
    A000695_list = list(islice(A000695_gen(),30)) # Chai Wah Wu, Feb 22 2023
    
  • Python
    def A000695(n): return int(bin(n)[2:],4) # Chai Wah Wu, Aug 21 2023
    
  • Sage
    s=(sum(4^k*x^(2^k)/(1+x^(2^k)) for k in range(10))/(1-x)).series(x, 60); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 06 2018
    

Formula

G.f.: 1/(1-x) * Sum_{k>=0} 4^k*x^2^k/(1+x^2^k). - Ralf Stephan, Apr 27 2003
Numbers k such that the coefficient of x^k is > 0 in Product_{n>=0} 1+x^(4^n). - Benoit Cloitre, Jul 29 2003
For n >= 1, a(n) = a(n-1) + (4^t+2)/6, where t is such that 2^t||2n,or t=A007814(2n). a(n) = (A145812(n+1) - 1)/2. - Vladimir Shevelev, Nov 07 2008
To get a(n), write n as Sum b_j*2^j, then a(n) = Sum b_j*2^(2j). The Diophantine equation a(k)+2a(l)=n has the unique solution: k=Sum b_(2j)*2^j, l=Sum b_(2j+1)*2^j. - Vladimir Shevelev, Nov 10 2008
If a(k)*a(l)=a(m), then k*l=m (the inverse, generally speaking, is not true). - Vladimir Shevelev, Nov 21 2008
Let F(x) be the generating function, then F(x)*F(x^2) = 1/(1-x). - Joerg Arndt, May 12 2010
a(n+1) = (a(n) + 1/3) & -1/3, where & is bitwise AND, -1/3 is represented as the infinite dyadic ...010101 (just as -1 is ...111111 in two's complement) and +1/3 is ...101011. - Marc LeBrun, Sep 30 2010
a(n) = Sum_{k>=0} {A030308(n,k)*b(k)} with b(k) = 4^k = A000302(k). - Philippe Deléham, Oct 18 2011
A182560(6*a(n)) = 0. - Reinhard Zumkeller, May 05 2012
G.f.: x/(1-x^2) + 4*x^2/((1-x)*(W(0) - 4*x - 4*x^2)), where W(k) = 1 + 4*x^(2^k) + 5*x^(2^(k+1)) - 4*x^(2^(k+1))*(1 + x^(2^(k+1)))^2/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 04 2014
liminf a(n)/n^2 = 1/3 and limsup a(n)/n^2 = 1. - Gheorghe Coserea, Sep 15 2015
Let f(x) = (Sum_{k=-oo..oo} floor(x*2^k)/4^k)/2. Then f(x) is a real-valued extension of a(n), which a(n) approximates in the sense that f(x) = lim_{k->oo} a(floor(x*2^k))/a(2^k). - Velin Yanev, Nov 28 2016
G.f. A(x) satisfies x/(1 - x^2) = A(x) - 4 * (1+x) * A(x^2). - Michael Somos, Nov 30 2016
a(2^k) = 4^k = A000302(k). a(n + 2^k) = a(n) + a(2^k) for 2^k > n >= 1. - David A. Corneth, Oct 16 2018
Sum_{n>=1} 1/a(n) = 1.886176434476107244547259512076353532930680508099044818673061351780360211128... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022

A001511 The ruler function: exponent of the highest power of 2 dividing 2n. Equivalently, the 2-adic valuation of 2n.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

Number of 2's dividing 2*n.
a(n) is equivalently the exponent of the smallest power of 2 which does not divide n. - David James Sycamore, Oct 02 2023
a(n) - 1 is the number of trailing zeros in the binary expansion of n.
If you are counting in binary and the least significant bit is numbered 1, the next bit is 2, etc., a(n) is the bit that is incremented when increasing from n-1 to n. - Jud McCranie, Apr 26 2004
Number of steps to reach an integer starting with (n+1)/2 and using the map x -> x*ceiling(x) (cf. A073524).
a(n) is the number of the disk to be moved at the n-th step of the optimal solution to Towers of Hanoi problem (comment from Andreas M. Hinz).
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 1). This is essentially equivalent to Hinz's comment. - Adam Kertesz, Jul 28 2001
a(n) is the Hamming distance between n and n-1 (in binary). This is equivalent to Kertesz's comments above. - Tak-Shing Chan (chan12(AT)alumni.usc.edu), Feb 25 2003
Let S(0) = {1}, S(n) = {S(n-1), S(n-1)-{x}, x+1} where x = last term of S(n-1); sequence gives S(infinity). - Benoit Cloitre, Jun 14 2003
The sum of all terms up to and including the first occurrence of m is 2^m-1. - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003
m appears every 2^m terms starting with the 2^(m-1)th term. - Donald Sampson (marsquo(AT)hotmail.com), Dec 08 2003
Sequence read mod 4 gives A092412. - Philippe Deléham, Mar 28 2004
If q = 2n/2^A001511(n) and if b(m) is defined by b(0)=q-1 and b(m)=2*b(m-1)+1, then 2n = b(A001511(n)) + 1. - Gerald McGarvey, Dec 18 2004
Repeating pattern ABACABADABACABAE ... - Jeremy Gardiner, Jan 16 2005
Relation to C(n) = Collatz function iteration using only odd steps: a(n) is the number of right bits set in binary representation of A004767(n) (numbers of the form 4*m+3). So for m=A004767(n) it follows that there are exactly a(n) recursive steps where m
Between every two instances of any positive integer m there are exactly m distinct values (1 through m-1 and one value greater than m). - Franklin T. Adams-Watters, Sep 18 2006
Number of divisors of n of the form 2^k. - Giovanni Teofilatto, Jul 25 2007
Every prefix up to (but not including) the first occurrence of some k >= 2 is a palindrome. - Gary W. Adamson, Sep 24 2008
1 interleaved with (2 interleaved with (3 interleaved with ( ... ))). - Eric D. Burgess (ericdb(AT)gmail.com), Oct 17 2009
A054525 (Möbius transform) * A001511 = A036987 = A047999^(-1) * A001511. - Gary W. Adamson, Oct 26 2009
Equals A051731 * A036987, (inverse Möbius transform of the Fredholm-Rueppel sequence) = A047999 * A036987. - Gary W. Adamson, Oct 26 2009
Cf. A173238, showing links between generalized ruler functions and A000041. - Gary W. Adamson, Feb 14 2010
Given A000041, P(x) = A(x)/A(x^2) with P(x) = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), A(x) = (1 + x + 3x^2 + 4x^3 + 10x^4 + 13x^5 + ...), A(x^2) = (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...), where A092119 = (1, 1, 3, 4, 10, ...) = Euler transform of the ruler sequence, A001511. - Gary W. Adamson, Feb 11 2010
Subtracting 1 from every term and deleting any 0's yields the same sequence, A001511. - Ben Branman, Dec 28 2011
In the listing of the compositions of n as lists in lexicographic order, a(k) is the last part of composition(k) for all k <= 2^(n-1) and all n, see example. - Joerg Arndt, Nov 12 2012
According to Hinz, et al. (see links), this sequence was studied by Louis Gros in his 1872 pamphlet "Théorie du Baguenodier" and has therefore been called the Gros sequence.
First n terms comprise least squarefree word of length n using positive integers, where "squarefree" means that the word contains no consecutive identical subwords; e.g., 1 contains no square; 11 contains a square but 12 does not; 121 contains no square; both 1211 and 1212 have squares but 1213 does not; etc. - Clark Kimberling, Sep 05 2013
Length of 0-run starting from 2 (10, 100, 110, 1000, 1010, ...), or length of 1-run starting from 1 (1, 11, 101, 111, 1001, 1011, ...) of every second number, from right to left in binary representation. - Armands Strazds, Apr 13 2017
a(n) is also the frequency of the largest part in 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 24 2017
As A000005(n) equals the number of even divisors of 2n and A001227(n) = A001227(2n), the formula A001511(n) = A000005(n)/A001227(n) might be read as "The number of even divisors of 2n is always divisible by the number of odd divisors of 2n" (where number of divisors means sum of zeroth powers of divisors). Conjecture: For any nonnegative integer k, the sum of the k-th powers of even divisors of n is always divisible by the sum of the k-th powers of odd divisors of n. - Ivan N. Ianakiev, Jul 06 2019
From Benoit Cloitre, Jul 14 2022: (Start)
To construct the sequence, start from 1's separated by a place 1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,...
Then put the 2's in every other remaining place
1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,...
Then the 3's in every other remaining place
1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,...
Then the 4's in every other remaining place
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,4,1,2,1,...
By iterating this process, we get the ruler function 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,... (End)
a(n) is the least positive integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022
a(n) is the smallest positive integer that does not occur in the coincidences of the sequence so far a(1..n-1) and its reverse. - Neal Gersh Tolunsky, Jan 18 2023
The geometric mean of this sequence approaches the Somos constant (A112302). - Jwalin Bhatt, Jan 31 2025

Examples

			For example, 2^1|2, 2^2|4, 2^1|6, 2^3|8, 2^1|10, 2^2|12, ... giving the initial terms 1, 2, 1, 3, 1, 2, ...
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
1;
2,1;
3,1,2,1;
4,1,2,1,3,1,2,1;
5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
6,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
7,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6,1,2,1,3,...
(End)
S(0) = {} S(1) = 1 S(2) = 1, 2, 1 S(3) = 1, 2, 1, 3, 1, 2, 1 S(4) = 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
From _Joerg Arndt_, Nov 12 2012: (Start)
The 16 compositions of 5 as lists in lexicographic order:
[ n]  a(n)  composition
[ 1]  [ 1]  [ 1 1 1 1 1 ]
[ 2]  [ 2]  [ 1 1 1 2 ]
[ 3]  [ 1]  [ 1 1 2 1 ]
[ 4]  [ 3]  [ 1 1 3 ]
[ 5]  [ 1]  [ 1 2 1 1 ]
[ 6]  [ 2]  [ 1 2 2 ]
[ 7]  [ 1]  [ 1 3 1 ]
[ 8]  [ 4]  [ 1 4 ]
[ 9]  [ 1]  [ 2 1 1 1 ]
[10]  [ 2]  [ 2 1 2 ]
[11]  [ 1]  [ 2 2 1 ]
[12]  [ 3]  [ 2 3 ]
[13]  [ 1]  [ 3 1 1 ]
[14]  [ 2]  [ 3 2 ]
[15]  [ 1]  [ 4 1 ]
[16]  [ 5]  [ 5 ]
a(n) is the last part in each list.
(End)
From _Omar E. Pol_, Aug 20 2013: (Start)
Also written as a triangle in which the right border gives A000027 and row lengths give A011782 and row sums give A000079 the sequence begins:
1;
2;
1,3;
1,2,1,4;
1,2,1,3,1,2,1,5;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,7;
(End)
G.f. = x + 2*x^2 + x^3 + 3*x^4 + x^5 + 2*x^6 + x^7 + 4*x^8 + x^9 + 2*x^10 + ...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 2nd ed., 2001-2003; see Dim- and Dim+ on p. 98; Dividing Rulers, on pp. 436-437; The Ruler Game, pp. 469-470; Ruler Fours, Fives, ... Fifteens on p. 470.
  • L. Gros, Théorie du Baguenodier, Aimé Vingtrinier, Lyon, 1872.
  • R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E22.
  • A. M. Hinz, The Tower of Hanoi, in Algebras and combinatorics (Hong Kong, 1997), 277-289, Springer, Singapore, 1999.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589.
  • Andrew Schloss, "Towers of Hanoi" composition, in The Digital Domain. Elektra/Asylum Records 9 60303-2, 1983. Works by Jaffe (Finale to "Silicon Valley Breakdown"), McNabb ("Love in the Asylum"), Schloss ("Towers of Hanoi"), Mattox ("Shaman"), Rush, Moorer ("Lions are Growing") and others.
  • 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

Column 1 of table A050600.
Sequence read mod 2 gives A035263.
Sequence is bisection of A007814, A050603, A050604, A067029, A089309.
This is Guy Steele's sequence GS(4, 2) (see A135416).
Cf. A005187 (partial sums), A085058 (bisection), A112302 (geometric mean), A171977 (2^a(n)).
Cf. A287896, A002487, A209229 (Mobius trans.), A092673 (Dirichlet inv.).
Cf. generalized ruler functions for k=3,4,5: A051064, A115362, A055457.

Programs

  • Haskell
    a001511 n = length $ takeWhile ((== 0) . (mod n)) a000079_list
    -- Reinhard Zumkeller, Sep 27 2011
    
  • Haskell
    a001511 n | odd n = 1 | otherwise = 1 + a001511 (n `div` 2)
    -- Walt Rorie-Baety, Mar 22 2013
    
  • MATLAB
    nmax=5;r=1;for n=2:nmax;r=[r n r];end % Adriano Caroli, Feb 26 2016
    
  • Magma
    [Valuation(2*n,2): n in [1..105]]; // Bruno Berselli, Nov 23 2015
    
  • Maple
    A001511 := n->2-wt(n)+wt(n-1); # where wt is defined in A000120
    # This is the binary logarithm of the denominator of (256^n-1)B_{8n}/n, in Maple parlance a := n -> log[2](denom((256^n-1)*bernoulli(8*n)/n)). - Peter Luschny, May 31 2009
    A001511 := n -> padic[ordp](2*n,2): seq(A001511(n), n=1..105);  # Peter Luschny, Nov 26 2010
    a:= n-> ilog2((Bits[Xor](2*n, 2*n-1)+1)/2): seq(a(n), n=1..50);  # Gary Detlefs, Dec 13 2018
  • Mathematica
    Array[ If[ Mod[ #, 2] == 0, FactorInteger[ # ][[1, 2]], 0] &, 105] + 1 (* or *)
    Nest[ Flatten[ # /. a_Integer -> {1, a + 1}] &, {1}, 7] (* Robert G. Wilson v, Mar 04 2005 *)
    IntegerExponent[2*n, 2] (* Alexander R. Povolotsky, Aug 19 2011 *)
    myHammingDistance[n_, m_] := Module[{g = Max[m, n], h = Min[m, n]}, b1 = IntegerDigits[g, 2]; b2 = IntegerDigits[h, 2, Length[b1]]; HammingDistance[b1, b2]] (* Vladimir Shevelev A206853 *) Table[ myHammingDistance[n, n - 1], {n, 111}] (* Robert G. Wilson v, Apr 05 2012 *)
    Table[Position[Reverse[IntegerDigits[n,2]],1,1,1],{n,110}]//Flatten (* Harvey P. Dale, Aug 18 2017 *)
  • PARI
    a(n) = sum(k=0,floor(log(n)/log(2)),floor(n/2^k)-floor((n-1)/2^k)) /* Ralf Stephan */
    
  • PARI
    a(n)=if(n%2,1,factor(n)[1,2]+1) /* Jon Perry, Jun 06 2004 */
    
  • PARI
    {a(n) = if( n, valuation(n, 2) + 1, 0)}; /* Michael Somos, Sep 30 2006 */
    
  • PARI
    {a(n)=if(n==1,1,polcoeff(x-sum(k=1, n-1, a(k)*x^k*(1-x^k)*(1-x+x*O(x^n))), n))} /* Paul D. Hanna, Jun 22 2007 */
    
  • Python
    def a(n): return bin(n)[2:][::-1].index("1") + 1 # Indranil Ghosh, May 11 2017
    
  • Python
    A001511 = lambda n: (n&-n).bit_length() # M. F. Hasler, Apr 09 2020
    
  • Python
    def A001511(n): return (~n & n-1).bit_length()+1 # Chai Wah Wu, Jul 01 2022
    
  • Sage
    [valuation(2*n,2) for n in (1..105)]  # Bruno Berselli, Nov 23 2015
    
  • Scheme
    (define (A001511 n) (let loop ((n n) (e 1)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

Formula

a(n) = A007814(n) + 1 = A007814(2*n).
a(2*n+1) = 1; a(2*n) = 1 + a(n). - Philippe Deléham, Dec 08 2003
a(n) = 2 - A000120(n) + A000120(n-1), n >= 1. - Daniele Parisse
a(n) = 1 + log_2(abs(A003188(n) - A003188(n-1))).
Multiplicative with a(p^e) = e+1 if p = 2; 1 if p > 2. - David W. Wilson, Aug 01 2001
For any real x > 1/2: lim_{N->infinity} (1/N)*Sum_{n=1..N} x^(-a(n)) = 1/(2*x-1); also lim_{N->infinity} (1/N)*Sum_{n=1..N} 1/a(n) = log(2). - Benoit Cloitre, Nov 16 2001
s(n) = Sum_{k=1..n} a(k) is asymptotic to 2*n since s(n) = 2*n - A000120(n). - Benoit Cloitre, Aug 31 2002
For any n >= 0, for any m >= 1, a(2^m*n + 2^(m-1)) = m. - Benoit Cloitre, Nov 24 2002
a(n) = Sum_{d divides n and d is odd} mu(d)*tau(n/d). - Vladeta Jovovic, Dec 04 2002
G.f.: A(x) = Sum_{k>=0} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Dec 24 2002
a(1) = 1; for n > 1, a(n) = a(n-1) + (-1)^n*a(floor(n/2)). - Vladeta Jovovic, Apr 25 2003
A fixed point of the mapping 1->12; 2->13; 3->14; 4->15; 5->16; ... . - Philippe Deléham, Dec 13 2003
Product_{k>0} (1+x^k)^a(k) is g.f. for A000041(). - Vladeta Jovovic, Mar 26 2004
G.f. A(x) satisfies A(x) = A(x^2) + x/(1-x). - Franklin T. Adams-Watters, Feb 09 2006
a(A118413(n,k)) = A002260(n,k); = a(A118416(n,k)) = A002024(n,k); a(A014480(n)) = A003602(A014480(n)). - Reinhard Zumkeller, Apr 27 2006
Ordinal transform of A003602. - Franklin T. Adams-Watters, Aug 28 2006 (The ordinal transform of a sequence b_0, b_1, b_2, ... is the sequence a_0, a_1, a_2, ... where a_n is the number of times b_n has occurred in {b_0 ... b_n}.)
Could be extended to n <= 0 using a(-n) = a(n), a(0) = 0, a(2*n) = a(n)+1 unless n=0. - Michael Somos, Sep 30 2006
A094267(2*n) = A050603(2*n) = A050603(2*n + 1) = a(n). - Michael Somos, Sep 30 2006
Sequence = A129360 * A000005 = M*V, where M = an infinite lower triangular matrix and V = d(n) as a vector: [1, 2, 2, 3, 2, 4, ...]. - Gary W. Adamson, Apr 15 2007
Row sums of triangle A130093. - Gary W. Adamson, May 13 2007
Dirichlet g.f.: zeta(s)*2^s/(2^s-1). - Ralf Stephan, Jun 17 2007
a(n) = -Sum_{d divides n} mu(2*d)*tau(n/d). - Benoit Cloitre, Jun 21 2007
G.f.: x/(1-x) = Sum_{n>=1} a(n)*x^n*( 1 - x^n ). - Paul D. Hanna, Jun 22 2007
2*n = 2^a(n)* A000265(n). - Eric Desbiaux, May 14 2009 [corrected by Alejandro Erickson, Apr 17 2012]
Multiplicative with a(2^k) = k + 1, a(p^k) = 1 for any odd prime p. - Franklin T. Adams-Watters, Jun 09 2009
With S(n): 2^n - 1 first elements of the sequence then S(0) = {} (empty list) and if n > 0, S(n) = S(n-1), n, S(n-1). - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
a(n) = log_2(A046161(n)/A046161(n-1)). - Johannes W. Meijer, Nov 04 2012
a((2*n-1)*2^p) = p+1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
a(n+1) = 1 + Sum_{j=0..ceiling(log_2(n+1))} (j * (1 - abs(sign((n mod 2^(j + 1)) - 2^j + 1)))). - Enrico Borba, Oct 01 2015
Conjecture: a(n) = A181988(n)/A003602(n). - L. Edson Jeffery, Nov 21 2015
a(n) = log_2(A006519(n)) + 1. - Doug Bell, Jun 02 2017
Inverse Moebius transform of A209229. - Andrew Howroyd, Aug 04 2018
a(n) = 1 + (A183063(n)/A001227(n)). - Omar E. Pol, Nov 06 2018 (after Franklin T. Adams-Watters)
a(n) = log_2((Xor(2*n,2*n-1)+1)/2). - Gary Detlefs, Dec 13 2018
(2^(a(n)-1)-1)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Dec 14 2018
a(n) = A000005(n)/A001227(n). - Ivan N. Ianakiev, Jul 05 2019
a(n) = Sum_{j=1..r} (j/2^j)*(Product_{k=1..j} (1 - (-1)^floor( (n+2^(j-1))/2^(k-1) ))), for n < a predefined 2^r. - Adriano Caroli, Sep 30 2019

Extensions

Name edited following suggestion by David James Sycamore, Oct 05 2023

A006519 Highest power of 2 dividing n.

Original entry on oeis.org

1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 32, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 64, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 32, 1, 2, 1, 4, 1, 2
Offset: 1

Comments

Least positive k such that m^k + 1 divides m^n + 1 (with fixed base m). - Vladimir Baltic, Mar 25 2002
To construct the sequence: start with 1, concatenate 1, 1 and double last term gives 1, 2. Concatenate those 2 terms, 1, 2, 1, 2 and double last term 1, 2, 1, 2 -> 1, 2, 1, 4. Concatenate those 4 terms: 1, 2, 1, 4, 1, 2, 1, 4 and double last term -> 1, 2, 1, 4, 1, 2, 1, 8, etc. - Benoit Cloitre, Dec 17 2002
a(n) = gcd(seq(binomial(2*n, 2*m+1)/2, m = 0 .. n - 1)) (odd numbered entries of even numbered rows of Pascal's triangle A007318 divided by 2), where gcd() denotes the greatest common divisor of a set of numbers. Due to the symmetry of the rows it suffices to consider m = 0 .. floor((n-1)/2). - Wolfdieter Lang, Jan 23 2004
Equals the continued fraction expansion of a constant x (cf. A100338) such that the continued fraction expansion of 2*x interleaves this sequence with 2's: contfrac(2*x) = [2; 1, 2, 2, 2, 1, 2, 4, 2, 1, 2, 2, 2, 1, 2, 8, 2, ...].
Simon Plouffe observes that this sequence and A003484 (Radon function) are very similar, the difference being all zeros except for every 16th term (see A101119 for nonzero differences). Dec 02 2004
This sequence arises when calculating the next odd number in a Collatz sequence: Next(x) = (3*x + 1) / A006519, or simply (3*x + 1) / BitAnd(3*x + 1, -3*x - 1). - Jim Caprioli, Feb 04 2005
a(n) = n if and only if n = 2^k. This sequence can be obtained by taking a(2^n) = 2^n in place of a(2^n) = n and using the same sequence building approach as in A001511. - Amarnath Murthy, Jul 08 2005
Also smallest m such that m + n - 1 = m XOR (n - 1); A086799(n) = a(n) + n - 1. - Reinhard Zumkeller, Feb 02 2007
Number of 1's between successive 0's in A159689. - Philippe Deléham, Apr 22 2009
Least number k such that all coefficients of k*E(n, x), the n-th Euler polynomial, are integers (cf. A144845). - Peter Luschny, Nov 13 2009
In the binary expansion of n, delete everything left of the rightmost 1 bit. - Ralf Stephan, Aug 22 2013
The equivalent sequence for partitions is A194446. - Omar E. Pol, Aug 22 2013
Also the 2-adic value of 1/n, n >= 1. See the Mahler reference, definition on p. 7. This is a non-archimedean valuation. See Mahler, p. 10. Sometimes called 2-adic absolute value of 1/n. - Wolfdieter Lang, Jun 28 2014
First 2^(k-1) - 1 terms are also the heights of the successive rectangles and squares of width 2 that are adjacent to any of the four sides of the toothpick structure of A139250 after 2^k stages, with k >= 2. For example: if k = 5 the heights after 32 stages are [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1] respectively, the same as the first 15 terms of this sequence. - Omar E. Pol, Dec 29 2020

Examples

			2^3 divides 24, but 2^4 does not divide 24, so a(24) = 8.
2^0 divides 25, but 2^1 does not divide 25, so a(25) = 1.
2^1 divides 26, but 2^2 does not divide 26, so a(26) = 2.
Per _Marc LeBrun_'s 2000 comment, a(n) can also be determined with bitwise operations in two's complement. For example, given n = 48, we see that n in binary in an 8-bit byte is 00110000 while -n is 11010000. Then 00110000 AND 11010000 = 00010000, which is 16 in decimal, and therefore a(48) = 16.
G.f. = x + 2*x^2 + x^3 + 4*x^4 + x^5 + 2*x^6 + x^7 + 8*x^8 + x^9 + ...
		

References

  • Kurt Mahler, p-adic numbers and their functions, second ed., Cambridge University Press, 1981.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Partial sums are in A006520, second partial sums in A022560.
Sequences used in definitions of this sequence: A000079, A001511, A004198, A007814.
Sequences with related definitions: A038712, A171977, A135481 (GS(1, 6)).
This is Guy Steele's sequence GS(5, 2) (see A135416).
Related to A007913 via A225546.
A059897 is used to express relationship between sequence terms.
Cf. A091476 (Dgf at s=2).

Programs

  • Haskell
    import Data.Bits ((.&.))
    a006519 n = n .&. (-n) :: Integer
    -- Reinhard Zumkeller, Mar 11 2012, Dec 29 2011
    
  • Julia
    using IntegerSequences
    [EvenPart(n) for n in 1:102] |> println  # Peter Luschny, Sep 25 2021
    
  • Magma
    [2^Valuation(n, 2): n in [1..100]]; // Vincenzo Librandi, Mar 27 2015
    
  • Maple
    with(numtheory): for n from 1 to 200 do if n mod 2 = 1 then printf(`%d,`,1) else printf(`%d,`,2^ifactors(n)[2][1][2]) fi; od:
    A006519 := proc(n) if type(n,'odd') then 1 ; else for f in ifactors(n)[2] do if op(1,f) = 2 then return 2^op(2,f) ; end if; end do: end if; end proc: # R. J. Mathar, Oct 25 2010
    A006519 := n -> 2^padic[ordp](n,2): # Peter Luschny, Nov 26 2010
  • Mathematica
    lowestOneBit[n_] := Block[{k = 0}, While[Mod[n, 2^k] == 0, k++]; 2^(k - 1)]; Table[lowestOneBit[n], {n, 102}] (* Robert G. Wilson v Nov 17 2004 *)
    Table[2^IntegerExponent[n, 2], {n, 128}] (* Jean-François Alcover, Feb 10 2012 *)
    Table[BitAnd[BitNot[i - 1], i], {i, 1, 102}] (* Peter Luschny, Oct 10 2019 *)
  • PARI
    {a(n) = 2^valuation(n, 2)};
    
  • PARI
    a(n)=1<Joerg Arndt, Jun 10 2011
    
  • PARI
    a(n)=bitand(n,-n); \\ Joerg Arndt, Jun 10 2011
    
  • PARI
    a(n)=direuler(p=2,n,if(p==2,1/(1-2*X),1/(1-X)))[n] \\ Ralf Stephan, Mar 27 2015
    
  • Python
    def A006519(n): return n&-n # Chai Wah Wu, Jul 06 2022
  • Scala
    (1 to 128).map(Integer.lowestOneBit()) // _Alonso del Arte, Mar 04 2020
    

Formula

a(n) = n AND -n (where "AND" is bitwise, and negative numbers are represented in two's complement in a suitable bit width). - Marc LeBrun, Sep 25 2000, clarified by Alonso del Arte, Mar 16 2020
Also: a(n) = gcd(2^n, n). - Labos Elemer, Apr 22 2003
Multiplicative with a(p^e) = p^e if p = 2; 1 if p > 2. - David W. Wilson, Aug 01 2001
G.f.: Sum_{k>=0} 2^k*x^2^k/(1 - x^2^(k+1)). - Ralf Stephan, May 06 2003
Dirichlet g.f.: zeta(s)*(2^s - 1)/(2^s - 2) = zeta(s)*(1 - 2^(-s))/(1 - 2*2^(-s)). - Ralf Stephan, Jun 17 2007
a(n) = 2^floor(A002487(n - 1) / A002487(n)). - Reikku Kulon, Oct 05 2008
a(n) = 2^A007814(n). - R. J. Mathar, Oct 25 2010
a((2*k - 1)*2^e) = 2^e, k >= 1, e >= 0. - Johannes W. Meijer, Jun 07 2011
a(n) = denominator of Euler(n-1, 1). - Arkadiusz Wesolowski, Jul 12 2012
a(n) = A011782(A001511(n)). - Omar E. Pol, Sep 13 2013
a(n) = (n XOR floor(n/2)) XOR (n-1 XOR floor((n-1)/2)) = n - (n AND n-1) (where "AND" is bitwise). - Gary Detlefs, Jun 12 2014
a(n) = ((n XOR n-1)+1)/2. - Gary Detlefs, Jul 02 2014
a(n) = A171977(n)/2. - Peter Kern, Jan 04 2017
a(n) = 2^(A001511(n)-1). - Doug Bell, Jun 02 2017
a(n) = abs(A003188(n-1) - A003188(n)). - Doug Bell, Jun 02 2017
Conjecture: a(n) = (1/(A000203(2*n)/A000203(n)-2)+1)/2. - Velin Yanev, Jun 30 2017
a(n) = (n-1) o n where 'o' is the bitwise converse nonimplication. 'o' is not commutative. n o (n+1) = A135481(n). - Peter Luschny, Oct 10 2019
From Peter Munn, Dec 13 2019: (Start)
a(A225546(n)) = A225546(A007913(n)).
a(A059897(n,k)) = A059897(a(n), a(k)). (End)
Sum_{k=1..n} a(k) ~ (1/(2*log(2)))*n*log(n) + (3/4 + (gamma-1)/(2*log(2)))*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 15 2022
a(n) = n / A000265(n). - Amiram Eldar, May 22 2025

Extensions

More terms from James Sellers, Jun 20 2000

A000523 a(n) = floor(log_2(n)).

Original entry on oeis.org

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

Keywords

Comments

Or, n >= 0 appears 2^n times. - Jon Perry, Sep 21 2002
a(n) + 1 = number of bits in binary expansion of n.
Largest power of 2 dividing lcm(1..n): A007814(A003418(n)).
log_2(0) = -infinity.
Also Max_{k=1..n} Omega(k), where Omega(n) = A001222(n), number of prime factors with repetition; see A080613. - Reinhard Zumkeller, Feb 25 2003
From Paul Weisenhorn, Sep 29 2010, updated Aug 11 2020: (Start)
Arithmetic mean: m(1,(c+1)/c) = (2*c+1)/(2*c); harmonic mean: h(1,(c+1)/c) = 2*(c+1)/(2*c+1);
a(n) is the number of means to reach (n+1)/n from 2/1; with m for 0 and h for 1, the inverse binary expansion of n, without the leading 1, gives the sequence of means.
For example, n=20; inverse binary expansion without the leading 1: 0010 ---> m m h m or m(1, m(1, h(1, m(1, 2)))) = 21/20.
The 4 twofold means for n from 4 to 7:
m(1,m(1,2)) = m(1,3/2) = 5/4,
h(1,m(1,2)) = h(1,3/2) = 6/5,
m(1,h(1,2)) = m(1,4/3) = 7/6,
h(1,h(1,2)) = h(1,4/3) = 8/7. (End) [Edited by Petros Hadjicostas, Jul 23 2020]
As function of the absolute value, defines the minimal Euclidean function v on Z\{0}. A ring R is Euclidean if for some function v : R\{0}->N a division by nonzero b can be defined with remainder r satisfying either r=0 or v(r) < v(b). For the integers taking v(n)=|n| works, but v(n) = floor(log_2(|n|)) works as well; moreover it is the possibility with smallest possible values. For division by b>0 one can always choose |r| <= floor(b/2); this sequence satisfies a(1) = 0 and recursively a(n) = 1 + max(a(1), ..., a(floor(n/2))) for n > 1. - Marc A. A. van Leeuwen, Feb 16 2011
Maximum number of guesses required to find any k in a range of 1..n, with 'higher', 'lower' and 'correct' as answers. - Jon Perry, Nov 02 2013
Number of powers of 2 <= n. - Ralph-Joseph Tatt, Apr 23 2018
a(n) + 1 is the minimum number of pairwise disjoint subsets of an n-element set such that for each k from 1 to n there is a set with cardinality k which is the union of some of those subsets. - Wojciech Raszka, Apr 15 2019
Minimum height of an n-node binary tree. - Yuchun Ji, Mar 22 2021

Examples

			a(5)=2 because the binary expansion of 5 (=101) has three bits.
		

References

  • Rüdeger Baumann, Computer-Knobelei, LOG IN Heft 159 (2009), 74-77. - Paul Weisenhorn, Sep 29 2010
  • G. H. Hardy, Note on Dr. Vacca's series for gamma, Quart. J. Pure Appl. Math., Vol. 43 (1912), pp. 215-216.
  • Ernst Jacobsthal, Über die Eulersche konstante, Mathematisch-Naturwissenschaftliche Blätter, Vol. 3, No. 9 (1906), pp. 153-154.
  • Donald E. Knuth, The Art of Computer Programming, Vol. 1: Fundamental Algorithms, p. 400.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589. - From N. J. A. Sloane, Aug 03 2012

Programs

  • Haskell
    a000523 1 = 0
    a000523 n = 1 + a000523 (div n 2)
    a000523_list = 0 : f [0] where
       f xs = ys ++ f ys where ys = map (+ 1) (xs ++ xs)
    -- Reinhard Zumkeller, Dec 31 2012, Feb 04 2012, Mar 18 2011
    
  • Magma
    [Ilog2(n) : n in [1..130] ];
    
  • Maple
    A000523 := proc(n)
        ilog2(n) ;
    end proc: # R. J. Mathar, Nov 28 2016
    seq(A000523(n), n=1..90);
  • Mathematica
    Floor[Log[2,Range[110]]] (* Harvey P. Dale, Jul 16 2012 *)
    a[ n_] := If[ n < 1, 0, BitLength[n] - 1]; (* Michael Somos, Jul 10 2018 *)
  • PARI
    {a(n) = floor(log(n) / log(2))} \\ Likely to yield incorrect results for many if not almost all n. Better use most recent code.
    
  • PARI
    {a(n) = if( n<1, 0, #binary(n) - 1)}; /* Michael Somos, May 28 2014 */
    
  • PARI
    a(n)=logint(n,2) \\ Charles R Greathouse IV, Sep 01 2015
    
  • PARI
    a(n)=exponent(n) \\ Charles R Greathouse IV, Nov 09 2017
    
  • Python
    def A000523(n):
        return len(bin(n))-3 # Chai Wah Wu, Jul 09 2020
    
  • Python
    def a(n): return n.bit_length() - 1
    print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Apr 18 2023

Formula

a(n) = A070939(n) - 1 for n >= 1.
a(n) = if n > 1, then a(floor(n / 2)) + 1; else 0. - Reinhard Zumkeller, Oct 29 2001
G.f.: (1/(1 - x)) * Sum_{k>=1} x^2^k. - Ralf Stephan, Apr 13 2002
a(n+1) = number of digits of n-th number with no 0 in ternary representation = A081604(A032924(n)); A107680(n) = A003462(a(n+1)). - Reinhard Zumkeller, May 20 2005
a(n) = A152487(n-1,0) = A152487(n,1). - Reinhard Zumkeller, Dec 06 2008
a(n) = k with 2^k <= n < 2^(k+1); a(n) = floor(log_2(n)). - Paul Weisenhorn, Sep 29 2010
a(n) = Max_{k=1..n} A240857(n,k). - Reinhard Zumkeller, Apr 14 2014
a(n) = A113473(n) - 1. - Filip Zaludek, Oct 29 2016
Sum_{n>=2} (-1)^n*a(n)/n = gamma = A001620 (Jacobsthal, 1906; Vacca, 1910). - Amiram Eldar, Jun 12 2021
a(n) = floor(Sum_{k=1..n-1} (n+1)^(n-2^k)) mod n. - Joseph M. Shunia, Jul 19 2024

Extensions

Error in 4th term, pointed out by Joe Keane (jgk(AT)jgk.org), has been corrected.
More terms from Michael Somos, Aug 02 2002

A023416 Number of 0's in binary expansion of n.

Original entry on oeis.org

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

Keywords

Comments

Another version (A080791) has a(0) = 0.

Crossrefs

The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015, A070939, A083652. Partial sums see A059015.
With initial zero and shifted right, same as A080791.
Cf. A055641 (for base 10), A188859.

Programs

  • Haskell
    a023416 0 = 1
    a023416 1 = 0
    a023416 n = a023416 n' + 1 - m where (n', m) = divMod n 2
    a023416_list = 1 : c [0] where c (z:zs) = z : c (zs ++ [z+1,z])
    -- Reinhard Zumkeller, Feb 19 2012, Jun 16 2011, Mar 07 2011
    
  • Maple
    A023416 := proc(n)
        if n = 0 then
            1;
        else
            add(1-e,e=convert(n,base,2)) ;
        end if;
    end proc: # R. J. Mathar, Jul 21 2012
  • Mathematica
    Table[ Count[ IntegerDigits[n, 2], 0], {n, 0, 100} ]
    DigitCount[Range[0,110],2,0] (* Harvey P. Dale, Jan 10 2013 *)
  • PARI
    a(n)=if(n==0,1,n=binary(n); sum(i=1, #n, !n[i])) \\ Charles R Greathouse IV, Jun 10 2011
    
  • PARI
    a(n)=if(n==0,1,#binary(n)-hammingweight(n)) \\ Charles R Greathouse IV, Nov 20 2012
    
  • PARI
    a(n) = if(n == 0, 1, 1+logint(n,2) - hammingweight(n))  \\ Gheorghe Coserea, Sep 01 2015
    
  • Python
    def A023416(n): return n.bit_length()-n.bit_count() if n else 1 # Chai Wah Wu, Mar 13 2023

Formula

a(n) = 1, if n = 0; 0, if n = 1; a(n/2)+1 if n even; a((n-1)/2) if n odd.
a(n) = 1 - (n mod 2) + a(floor(n/2)). - Marc LeBrun, Jul 12 2001
G.f.: 1 + 1/(1-x) * Sum_{k>=0} x^(2^(k+1))/(1+x^2^k). - Ralf Stephan, Apr 15 2002
a(n) = A070939(n) - A000120(n).
a(n) = A008687(n+1) - 1.
a(n) = A000120(A035327(n)).
From Hieronymus Fischer, Jun 12 2012: (Start)
a(n) = m + 1 + Sum_{j=1..m+1} (floor(n/2^j) - floor(n/2^j + 1/2)), where m=floor(log_2(n)).
General formulas for the number of digits <= d in the base p representation n, where 0 <= d < p.
a(n) = m + 1 + Sum_{j=1..m+1} (floor(n/p^j) - floor(n/p^j + (p-d-1)/p)), where m=floor(log_p(n)).
G.f.: 1 + (1/(1-x))*Sum_{j>=0} ((1-x^(d*p^j))*x^p^j + (1-x^p^j)*x^p^(j+1)/(1-x^p^(j+1))). (End)
Product_{n>=1} ((2*n)/(2*n+1))^((-1)^a(n)) = sqrt(2)/2 (A010503) (see Allouche & Shallit link). - Michel Marcus, Aug 31 2014
Sum_{n>=1} a(n)/(n*(n+1)) = 2 - 2*log(2) (A188859) (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021
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