|
| |
|
|
A029837
|
|
Binary order of n: log_2(n) rounded up to next integer.
|
|
254
|
|
|
|
0, 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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,3
|
|
|
COMMENTS
|
Or, ceiling(log_2(n)).
Worst-case cost of binary search.
Equal to number of binary digits in n unless n is a power of 2 when it is one less.
Thus a(n) gives the length of the binary representation of n - 1 (n >= 2), which is also A070939(n - 1).
Let x(0) = n > 1 and x(k + 1) = x(k) - floor(x(k)/2), then a(n) is the smallest integer such that x(a(n)) = 1. - Benoit Cloitre, Aug 29 2002
Also number of division steps when going from n to 1 by process of adding 1 if odd, or dividing by 2 if even. - Cino Hilliard, Mar 25 2003
Number of ways to write n as (x + 2^y), x >= 0. Number of ways to write n + 1 as 2^x + 3^y (cf. A004050). - Benoit Cloitre, Mar 29 2003
The minimum number of cuts for dividing an object into n (possibly unequal) pieces. - Karl Ove Hufthammer (karl(AT)huftis.org), Mar 29 2010
|
|
|
REFERENCES
|
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, 1989, p. 70.
G. J. E. Rawlins, Compared to What? An Introduction to the Analysis of Algorithms, W. H. Freeman, 1992; see pp. 108, 118.
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = ceiling(log_2(n)).
a(1) = 0; for n > 1, a(2n) = a(n) + 1, a(2n + 1) = a(n) + 1. Alternatively, a(1) = 0; for n > 1, a(n) = a(ceiling(n/2)) + 1. [corrected by Ilya Gutkovskiy, Mar 21 2020]
a(n) = k such that n^(1/k - 1) > 2 > n^(1/k), or the least value of k for which floor n^(1/k) = 1. a(n) = k for all n such that 2^(k - 1) < n < 2^k. - Amarnath Murthy, May 06 2001
G.f.: x/(1 - x) * Sum_{k >= 0} x^2^k. - Ralf Stephan, Apr 13 2002
a(n+1) = -Sum_{k = 1..n} mu(2*k)*floor(n/k). - Benoit Cloitre, Oct 21 2009
|
|
|
EXAMPLE
|
a(1) = 0, since log_2(1) = 0.
a(2) = 1, since log_2(2) = 1.
a(3) = 2, since log_2(3) = 1.58...
a(n) = 7 for n = 65, 66, ..., 127, 128.
G.f. = x^2 + 2*x^3 + 2*x^4 + 3*x^5 + 3*x^6 + 3*x^7 + 3*x^8 + 4*x^9 + ... - Michael Somos, Jun 02 2019
|
|
|
MAPLE
|
a:= n-> (p-> p+`if`(2^p<n, 1, 0))(ilog2(n)):
|
|
|
MATHEMATICA
|
Table[IntegerLength[n - 1, 2], {n, 1, 105}] (* Peter Luschny, Dec 02 2017 *)
a[n_] := If[n < 1, 0, BitLength[n - 1]]; (* Michael Somos, Jul 10 2018 *)
|
|
|
PROG
|
(PARI) {a(n) = if( n<1, 0, ceil(log(n) / log(2)))};
(PARI) /* Set p = 1, then: */
xpcount(n, p) = for(x=1, n, p1 = x; ct=0; while(p1>1, if(p1%2==0, p1/=2; ct++, p1 = p1*p+1)); print1(ct, ", "))
(PARI) {a(n) = if( n<2, 0, exponent(n-1)+1)}; /* Michael Somos, Jul 10 2018 */
(Haskell)
a029837 n = a029837_list !! (n-1)
a029837_list = scanl1 (+) a209229_list
(Scala) (1 to 80).map(n => Math.ceil(Math.log(n)/Math.log(2)).toInt) // Alonso del Arte, Feb 19 2020
(Python)
s = bin(n)[2:]
return len(s) - (1 if s.count('1') == 1 else 0) # Chai Wah Wu, Jul 09 2020
(Python)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy,nice
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
