

A000523


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


257



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
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OFFSET

1,4


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.  RalphJoseph Tatt, Apr 23 2018
a(n) + 1 is the minimum number of pairwise disjoint subsets of an nelement 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 nnode binary tree.  Yuchun Ji, Mar 22 2021


REFERENCES

Rüdeger Baumann, ComputerKnobelei, LOG IN Heft 159 (2009), 7477.  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. 215216.
Ernst Jacobsthal, Über die Eulersche konstante, MathematischNaturwissenschaftliche Blätter, Vol. 3, No. 9 (1906), pp. 153154.
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


LINKS

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
GuoNiu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
GuoNiu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
G. H. Hardy, Note on Dr. Vacca's series for gamma, Quart. J. Pure Appl. Math. 43 (1912), 215216. [Available only in the USA through the Hathi Trust.]
Ralf Stephan, Some divideandconquer sequences with (relatively) simple ordinary generating functions, 2004.
Ralf Stephan, Table of generating functions (ps file).
Ralf Stephan, Table of generating functions (pdf file).
G. Vacca, A new series for the Eulerian constant gamma=.577..., Quart. J. Pure Appl. Math., Vol. 41 (1910), pp. 363368.


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 nth number with no 0 in ternary representation = A081604(A032924(n)); A107680(n) = A003462(a(n+1)).  Reinhard Zumkeller, May 20 2005
a(n) = A152487(n1,0) = 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


EXAMPLE

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


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 *)


PROG

(Magma) [Ilog2(n) : n in [1..130] ];
(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
(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
(Python)
def A000523(n):
return len(bin(n))3 # Chai Wah Wu, Jul 09 2020


CROSSREFS

Cf. A000193, A000195, A001222, A001620, A003462, A004233, A029837, A032924, A061168 (partial sums), A070939, A081604, A107680, A113473, A152487, A240857.
Sequence in context: A345376 A029835 A074280 * A124156 A324965 A072749
Adjacent sequences: A000520 A000521 A000522 * A000524 A000525 A000526


KEYWORD

nonn,easy,nice,look


AUTHOR

N. J. A. Sloane


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


STATUS

approved



