

A000523


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


189



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, n1 appears 2^(n1) 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
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
From Paul Weisenhorn, Sep 29 2010: (Start)
Arithmetic mean: m(1,c/(c1)) = (2c+1)/2c; harmonic mean: h(1,c/(c1)) = 2c/(2c1);
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. (End)
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
a(n) = Max_{k=1..n} A240857(n,k).  Reinhard Zumkeller, Apr 14 2014
Number of powers of 2 <= n.  RalphJoseph Tatt, Apr 23 2018


REFERENCES

R. Baumann, ComputerKnobelei, LOG IN Heft 159 (2009), 7477.  Paul Weisenhorn, Sep 29 2010
G. H. Hardy, Note on Dr. Vacca's series..., Quart. J. Pure Appl. Math. 43 (1912) 215216.
D. E. Knuth, The Art of Computer Programming, Vol. 1: Fundamental Algorithms, p. 400.
D. 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
GuoNiu Han, Enumeration of Standard Puzzles [Cached copy]
R. Stephan, Some divideandconquer sequences ...
R. Stephan, Table of generating functions


FORMULA

a(n) = if n > 1 then a(floor(n / 2)) + 1 else 0.  Reinhard Zumkeller, Oct 29 2001
G.f.: (1/(1x)) * Sum_{k>=1} x^2^k.  Ralf Stephan, Apr 13 2002
a(n)=k with 2^k <= n < 2^(k+1); a(n)=floor(log_2(n)).  Paul Weisenhorn, Sep 29 2010
a(n) = A113473(n)  1.  Filip Zaludek, Oct 29 2016


EXAMPLE

a(5)=2 because the binary expansion of 5 (=101) has three bits.
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.  Paul Weisenhorn, Sep 29 2010


MAPLE

A000523 := n>floor(simplify(log(n)/log(2)));
A000523 := proc(n) local nn, i; if(0 = n) then RETURN(infinity); fi; nn := n; for i from 1 to n do if(0 = nn) then RETURN(i); fi; nn := floor(nn/2); od; end;
for n from 1 to 1000 do a[n]:=floor(log[2](n)); end do; # Paul Weisenhorn, Sep 29 2010
A000523 := proc(n)
ilog2(n) ;
end proc: # R. J. Mathar, Nov 28 2016


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) = if( n<1, 0, floor(log(n) / log(2)))};
(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


CROSSREFS

Cf. A029837. Partial sums: A061168.
Cf. A000195, A000193, A004233.
a(n) = A070939(n)1 for n >= 1.
Sequence in context: A186437 A029835 A074280 * A124156 A072749 A066490
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



