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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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

A055642 Number of digits in the decimal expansion of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
Offset: 0

Views

Author

Henry Bottomley, Jun 06 2000

Keywords

Comments

From Hieronymus Fischer, Jun 08 2012: (Start)
For n > 0 the first differences of A117804.
The total number of digits necessary to write down all the numbers 0, 1, 2, ..., n is A117804(n+1). (End)
Here a(0) = 1, but a different common convention is to consider that the expansion of 0 in any base b > 0 has 0 terms and digits. - M. F. Hasler, Dec 07 2018

Examples

			Examples:
999: 1 + floor(log_10(999)) = 1 + floor(2.x) = 1 + 2 = 3 or
      ceiling(log_10(999+1)) = ceiling(log_10(1000)) = ceiling(3) = 3;
1000: 1 + floor(log_10(1000)) = 1 + floor(3) = 1 + 3 = 4 or
      ceiling(log_10(1000+1)) = ceiling(log_10(1001)) = ceiling(3.x) = 4;
1001: 1 + floor(log_10(1001)) = 1 + floor(3.x) = 1 + 3 = 4 or
      ceiling(log_10(1001+1)) = ceiling(log_10(1002)) = ceiling(3.x) = 4;
		

Crossrefs

Programs

  • Haskell
    a055642 :: Integer -> Int
    a055642 = length . show  -- Reinhard Zumkeller, Feb 19 2012, Apr 26 2011
    
  • Magma
    [ #Intseq(n): n in [0..105] ];   //  Bruno Berselli, Jun 30 2011
    (Common Lisp) (defun A055642 (n) (if (zerop n) 1 (floor (log n 10)))) ; James Spahlinger, Oct 13 2012
    
  • Maple
    A055642 := proc(n)
            max(1,ilog10(n)+1) ;
    end proc:  # R. J. Mathar, Nov 30 2011
  • Mathematica
    Join[{1}, Array[ Floor[ Log[10, 10# ]] &, 104]] (* Robert G. Wilson v, Jan 04 2006 *)
    Join[{1},Table[IntegerLength[n],{n,104}]]
    IntegerLength[Range[0,120]] (* Harvey P. Dale, Jul 02 2016 *)
  • PARI
    a(n)=#Str(n) \\ M. F. Hasler, Nov 17 2008
    
  • PARI
    A055642(n)=logint(n+!n,10)+1 \\ Increasingly faster than the above, for larger n. (About twice as fast for n ~ 10^7.) - M. F. Hasler, Dec 07 2018
    
  • Python
    def a(n): return len(str(n))
    print([a(n) for n in range(121)]) # Michael S. Branicky, May 10 2022
    
  • Python
    def A055642(n): # Faster than len(str(n)) from ~ 50 digits on
        L = math.log10(n or 1)
        if L.is_integer() and 10**int(L)>n: return int(L or 1)
        return int(L)+1  # M. F. Hasler, Apr 08 2024

Formula

a(A046760(n)) < A050252(A046760(n)); a(A046759(n)) > A050252(A046759(n)). - Reinhard Zumkeller, Jun 21 2011
a(n) = A196563(n) + A196564(n).
a(n) = 1 + floor(log_10(n)) = 1 + A004216(n) = ceiling(log_10(n+1)) = A004218(n+1), if n >= 1. - Daniel Forgues, Mar 27 2014
a(A046758(n)) = A050252(A046758(n)). - Reinhard Zumkeller, Jun 21 2011
a(n) = A117804(n+1) - A117804(n), n > 0. - Hieronymus Fischer, Jun 08 2012
G.f.: g(x) = 1 + (1/(1-x))*Sum_{j>=0} x^(10^j). - Hieronymus Fischer, Jun 08 2012
a(n) = A262190(n) for n < 100; a(A262198(n)) != A262190(A262198(n)). - Reinhard Zumkeller, Sep 14 2015

A000788 Total number of 1's in binary expansions of 0, ..., n.

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 9, 12, 13, 15, 17, 20, 22, 25, 28, 32, 33, 35, 37, 40, 42, 45, 48, 52, 54, 57, 60, 64, 67, 71, 75, 80, 81, 83, 85, 88, 90, 93, 96, 100, 102, 105, 108, 112, 115, 119, 123, 128, 130, 133, 136, 140, 143, 147, 151, 156, 159, 163, 167, 172, 176, 181, 186
Offset: 0

Views

Author

Keywords

Comments

Partial sums of A000120.
The graph of this sequence is a version of the Takagi curve: see Lagarias (2012), Section 9, especially Theorem 9.1. - N. J. A. Sloane, Mar 12 2016
a(n-1) is the largest possible number of ordered pairs (a,b) such that a/b is a prime in a subset of the positive integers with n elements. - Yifan Xie, Feb 21 2025

References

  • J.-P. Allouche & J. Shallit, Automatic sequences, Cambridge University Press, 2003, p. 94
  • R. Bellman and H. N. Shapiro, On a problem in additive number theory, Annals Math., 49 (1948), 333-340. See Eq. 1.9. [From N. J. A. Sloane, Mar 12 2009]
  • L. E. Bush, An asymptotic formula for the average sums of the digits of integers, Amer. Math. Monthly, 47 (1940), pp. 154-156. [From the bibliography of Stolarsky, 1977]
  • P. Cheo and S. Yien, A problem on the k-adic representation of positive integers (Chinese; English summary), Acta Math. Sinica, 5 (1955), pp. 433-438. [From the bibliography of Stolarsky, 1977]
  • M. P. Drazin and J. S. Griffith, On the decimal representation of integers, Proc. Cambridge Philos. Soc., (4), 48 (1952), pp. 555-565. [From the bibliography of Stolarsky, 1977]
  • E. N. Gilbert, Games of identification or convergence, SIAM Review, 4 (1962), 16-24.
  • Grabner, P. J.; Kirschenhofer, P.; Prodinger, H.; Tichy, R. F.; On the moments of the sum-of-digits function. Applications of Fibonacci numbers, Vol. 5 (St. Andrews, 1992), 263-271, Kluwer Acad. Publ., Dordrecht, 1993.
  • R. L. Graham, On primitive graphs and optimal vertex assignments, pp. 170-186 of Internat. Conf. Combin. Math. (New York, 1970), Annals of the NY Academy of Sciences, Vol. 175, 1970.
  • E. Grosswald, Properties of some arithmetic functions, J. Math. Anal. Appl., 28 (1969), pp.405-430.
  • Donald E. Knuth, The Art of Computer Programming, volume 3 Sorting and Searching, section 5.3.4, subsection Bitonic sorting, with C'(p) = a(p-1).
  • Hiu-Fai Law, Spanning tree congestion of the hypercube, Discrete Math., 309 (2009), 6644-6648 (see p(m) on page 6647).
  • Z. Li and E. M. Reingold, Solution of a divide-and-conquer maximin recurrence, SIAM J. Comput., 18 (1989), 1188-1200.
  • B. Lindström, On a combinatorial problem in number theory, Canad. Math. Bull., 8 (1965), 477-490.
  • Mauclaire, J.-L.; Murata, Leo; On q-additive functions. I. Proc. Japan Acad. Ser. A Math. Sci. 59 (1983), no. 6, 274-276.
  • Mauclaire, J.-L.; Murata, Leo; On q-additive functions. II. Proc. Japan Acad. Ser. A Math. Sci. 59 (1983), no. 9, 441-444.
  • M. D. McIlroy, The number of 1's in binary integers: bounds and extremal properties, SIAM J. Comput., 3 (1974), 255-261.
  • L. Mirsky, A theorem on representations of integers in the scale of r, Scripta Math., 15 (1949), pp. 11-12.
  • I. Shiokawa, On a problem in additive number theory, Math. J. Okayama Univ., 16 (1974), pp.167-176. [From the bibliography of Stolarsky, 1977]
  • 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).
  • K. B. Stolarsky, Power and exponential sums of digital sums related to binomial coefficient parity, SIAM J. Appl. Math., 32 (1977), 717-730.
  • Trollope, J. R. An explicit expression for binary digital sums. Math. Mag. 41 1968 21-25.

Crossrefs

For number of 0's in binary expansion of 0, ..., n see A059015.
The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015, A070939, A083652.

Programs

  • Haskell
    a000788_list = scanl1 (+) A000120_list
    -- Walt Rorie-Baety, Jun 30 2012
    
  • Haskell
    {a000788 0 = 0; a00788 n = a000788 n2 + a000788 (n-n2-1) + (n-n2) where n2 = n `div` 2}
    -- Walt Rorie-Baety, Jul 15 2012
    
  • Maple
    a:= proc(n) option remember; `if`(n=0, 0, a(n-1)+add(i, i=Bits[Split](n))) end:
    seq(a(n), n=0..62);  # Alois P. Heinz, Nov 11 2024
  • Mathematica
    a[n_] := Count[ Table[ IntegerDigits[k, 2], {k, 0, n}], 1, 2]; Table[a[n], {n, 0, 62}] (* Jean-François Alcover, Dec 16 2011 *)
    Table[Plus@@Flatten[IntegerDigits[Range[n], 2]], {n, 0, 62}] (* Alonso del Arte, Dec 16 2011 *)
    Accumulate[DigitCount[Range[0,70],2,1]] (* Harvey P. Dale, Jun 08 2013 *)
  • PARI
    A000788(n)={ n<3 && return(n); if( bittest(n,0) \\
    , n+1 == 1<A000788(n>>1)*2+n>>1+1 \\
    , n == 1<A000788(n>>=1)+A000788(n-1)+n )} \\ M. F. Hasler, Nov 22 2009
    
  • PARI
    a(n)=sum(k=1,n,hammingweight(k)) \\ Charles R Greathouse IV, Oct 04 2013
    
  • PARI
    a(n) = if (n==0, 0, m = logint(n, 2); r = n % 2^m; m*2^(m-1) + r + 1 + a(r)); \\ Michel Marcus, Mar 27 2018
    
  • PARI
    a(n)={n++; my(t, i, s); c=n; while(c!=0, i++; c\=2); for(j=1, i, d=(n\2^(i-j))%2; t+=(2^(i-j)*(s*d+d*(i-j)/2)); s+=d); t} \\ David A. Corneth, Nov 26 2024
    (C++) /* See David W. Wilson link. */
    
  • Python
    def A000788(n): return sum(i.bit_count() for i in range(1,n+1)) # Chai Wah Wu, Mar 01 2023
    
  • Python
    def A000788(n): return (n+1)*n.bit_count()+(sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1))>>1) # Chai Wah Wu, Nov 11 2024

Formula

McIlroy (1974) gives bounds and recurrences. - N. J. A. Sloane, Mar 24 2014
Stolarsky (1977) studies the asymptotics, and gives at least nine references to earlier work on the problem. I have added all the references that were not here already. - N. J. A. Sloane, Apr 06 2014
a(n) = Sum_{k=1..n} A000120(k). - Benoit Cloitre, Dec 19 2002
a(0) = 0, a(2n) = a(n)+a(n-1)+n, a(2n+1) = 2a(n)+n+1. - Ralf Stephan, Sep 13 2003
a(n) = n*log_2(n)/2 + O(n); a(2^n)=n*2^(n-1)+1. - Benoit Cloitre, Sep 25 2003 (The first result is due to Bellman and Shapiro, - N. J. A. Sloane, Mar 24 2014)
a(n) = n*log_2(n)/2+n*F(log_2(n)) where F is a nowhere differentiable continuous function of period 1 (see Allouche & Shallit). - Benoit Cloitre, Jun 08 2004
G.f.: (1/(1-x)^2) * Sum_{k>=0} x^2^k/(1+x^2^k). - Ralf Stephan, Apr 19 2003
a(2^n-1) = A001787(n) = n*2^(n-1). - M. F. Hasler, Nov 22 2009
a(4^n-2) = n(4^n-2).
For real n, let f(n) = [n]/2 if [n] even, n-[n+1]/2 otherwise. Then a(n) = Sum_{k>=0} 2^k*f((n+1)/2^k).
a(A000225(n)) = A173921(A000225(n)) = A001787(n); a(A000079(n)) = A005183(n). - Reinhard Zumkeller, Mar 04 2010
From Hieronymus Fischer, Jun 10 2012: (Start)
a(n) = (1/2)*Sum_{j=1..m+1} (floor(n/2^j + 1/2)*(2n + 2 - floor(n/2^j + 1/2))*2^j - floor(n/2^j)*(2n + 2 - (1 + floor(n/2^j)) * 2^j)), where m=floor(log_2(n)).
a(n) = (n+1)*A000120(n) - 2^(m-1) + 1/4 + (1/2)*Sum_{j=1..m+1} ((floor(n/2^j) + 1/2)^2 - floor(n/2^j + 1/2)^2)*2^j, where m=floor(log_2(n)).
a(2^m-1) = m*2^(m-1).
(This is the total number of '1' digits occurring in all the numbers with <= m bits.)
Generic formulas for the number of digits >= d in the base p representations of all integers from 0 to n, where 1<= d < p.
a(n) = (1/2)*Sum_{j=1..m+1} (floor(n/p^j + (p-d)/p)*(2n + 2 + ((p-2*d)/p - floor(n/p^j + (p-d)/p))*p^j) - floor(n/p^j)*(2n + 2 - (1+floor(n/p^j)) * p^j)), where m=floor(log_p(n)).
a(n) = (n+1)*F(n,p,d) + (1/2)*Sum_{j=1..m+1} ((((p-2*d)/p)*floor(n/p^j+(p-d)/p) + floor(n/p^j))*p^j - (floor(n/p^j+(p-d)/p)^2 - floor(n/p^j)^2)*p^j), where m=floor(log_p(n)) and F(n,p,d) = number of digits >= d in the base p representation of n.
a(p^m-1) = (p-d)*m*p^(m-1).
(This is the total number of digits >= d occurring in all the numbers with <= m digits in base p representation.)
G.f.: g(x) = (1/(1-x)^2)*Sum_{j>=0} (x^(d*p^j) - x^(p*p^j))/(1-x^(p*p^j)). (End)
a(n) = Sum_{k=1..n} A000120(A240857(n,k)). - Reinhard Zumkeller, Apr 14 2014
For n > 0, if n is written as 2^m + r with 0 <= r < 2^m, then a(n) = m*2^(m-1) + r + 1 + a(r). - Shreevatsa R, Mar 20 2018
a(n) = n*(n+1)/2 + Sum_{k=1..floor(n/2)} ((2k-1)((g(n,k)-1)*2^(g(n,k) + 1) + 2) - (n+1)*(g(n,k)+1)*g(n,k)/2), where g(n,k) = floor(log_2(n/(2k-1))). - Fabio Visonà, Mar 17 2020
From Jeffrey Shallit, Aug 07 2021: (Start)
A 2-regular sequence, satisfying the identities
a(4n+1) = -a(2n) + a(2n+1) + a(4n)
a(4n+2) = -2a(2n) + 2a(2n+1) + a(4n)
a(4n+3) = -4a(n) + 4a(2n+1)
a(8n) = 4a(n) - 8a(2n) + 5a(4n)
a(8n+4) = -9a(2n) + 5a(2n+1) + 4a(4n)
for n>=0. (End)
a(n) = Sum_{k=0..floor(log_2(n+1))} k * A360189(n,k). - Alois P. Heinz, Mar 06 2023

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jan 15 2001

A055641 Number of zero digits in n.

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Henry Bottomley, Jun 06 2000

Keywords

Examples

			a(99) = 0 because the digits of 99 are 9 and 9, a(100) = 2 because the digits of 100 are 1, 0 and 0 and there are two 0's.
		

Crossrefs

Programs

  • Haskell
    a055641 n | n < 10    = 0 ^ n
              | otherwise = a055641 n' + 0 ^ d where (n',d) = divMod n 10
    -- Reinhard Zumkeller, Apr 30 2013
    
  • Mathematica
    Array[Last@ DigitCount@ # &, 105] (* Michael De Vlieger, Jul 02 2015 *)
  • PARI
    a(n)=if(n,n=digits(n); sum(i=2,#n,n[i]==0), 1) \\ Charles R Greathouse IV, Sep 13 2015
    
  • PARI
    A055641(n)=#select(d->!d,digits(n))+!n \\ M. F. Hasler, Jun 22 2018
    
  • Python
    def a(n): return str(n).count("0")
    print([a(n) for n in range(106)]) # Michael S. Branicky, May 26 2022

Formula

From Hieronymus Fischer, Jun 06 2012: (Start)
a(n) = m + 1 - A055640(n) = Sum_{j=1..m+1} (1 + floor(n/10^j) - floor(n/10^j+0.9)), where m = floor(log_10(n)).
G.f.: g(x) = 1 + (1/(1-x))*Sum_{j>=0} (x^(10*10^j) - x^(11*10^j))/(1-x^10^(j+1)). (End)
a(n) = if n<10 then A000007(n) else a(A059995(n)) + A000007(A010879(n)). - Reinhard Zumkeller, Apr 30 2013, corrected by M. F. Hasler, Jun 22 2018

A196564 Number of odd digits in decimal representation of n.

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1
Offset: 0

Views

Author

Reinhard Zumkeller, Oct 04 2011

Keywords

Crossrefs

Programs

  • Haskell
    a196564 n = length [d | d <- show n, d `elem` "13579"]
    -- Reinhard Zumkeller, Feb 22 2012, Oct 04 2011
    
  • Maple
    A196564 := proc(n)
            if n =0 then
                    0;
            else
                    convert(n,base,10) ;
                    add(d mod 2,d=%) ;
            end if:
    end proc: # R. J. Mathar, Jul 13 2012
  • Mathematica
    Table[Total[Mod[IntegerDigits[n],2]],{n,0,100}] (* Zak Seidov, Oct 13 2015 *)
  • PARI
    a(n) = #select(x->x%2, digits(n)); \\ Michel Marcus, Oct 14 2015
    
  • Python
    def a(n): return sum(1 for d in str(n) if d in "13579")
    print([a(n) for n in range(100)]) # Michael S. Branicky, May 15 2022

Formula

a(n) = A055642(n) - A196563(n);
a(A014263(n)) = 0; a(A007957(n)) > 0.
From Hieronymus Fischer, May 30 2012: (Start)
a(n) = Sum_{j=0..m} (floor(n/(2*10^j) + (1/2)) - floor(n/(2*10^j))), where m=floor(log_10(n)).
a(10*n+k) = a(n) + a(k), 0<=k<10, n>=0.
a(n) = a(floor(n/10)) + a(n mod 10), n>=0.
a(n) = Sum_{j=0..m} a(floor(n/10^j) mod 10), n>=0.
a(A014261(n)) = floor(log_5(4*n+1)), n>0.
G.f.: g(x) = (1/(1-x))*Sum_{j>=0} x^10^j/(1+x^10^j). (End)

A196563 Number of even digits in decimal representation of n.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1
Offset: 0

Views

Author

Reinhard Zumkeller, Oct 04 2011

Keywords

Crossrefs

Programs

  • Haskell
    a196563 n = length [d | d <- show n, d `elem` "02468"]
    -- Reinhard Zumkeller, Feb 22 2012, Oct 04 2011
    
  • Maple
    A196563 := proc(n)
            if n =0 then
                    1;
            else
                    convert(n,base,10) ;
                    add(1-(d mod 2),d=%) ;
            end if:
    end proc: # R. J. Mathar, Jul 13 2012
  • Mathematica
    Table[Count[Mod[IntegerDigits[n],2],0][n],{n,0,100}] (* Zak Seidov, Oct 13 2015 *)
    Table[Count[IntegerDigits[n],?EvenQ],{n,0,120}] (* _Harvey P. Dale, Feb 22 2020 *)
  • PARI
    a(n) = #select(x->(!(x%2)), if (n, digits(n), [0])); \\ Michel Marcus, Oct 14 2015
    
  • Python
    def a(n): return sum(1 for d in str(n) if d in "02468")
    print([a(n) for n in range(100)]) # Michael S. Branicky, May 15 2022

Formula

a(n) = A055642(n) - A196564(n);
a(A014261(n)) = 0; a(A007928(n)) > 0.
From Hieronymus Fischer, May 30 2012: (Start)
a(n) = Sum_{j=0..m} (1 + floor(n/(2*10^j)) - floor(n/(2*10^j) + (1/2))), where m=floor(log_10(n)).
a(10*n+k) = a(n) + a(k), 0<=k<10, n>=1.
a(n) = a(floor(n/10))+a(n mod 10), n>=10.
a(n) = Sum_{j=0..m} a(floor(n/10^j) mod 10), n>=0.
a(A014263(n)) = 1 + floor(log_5(n-1)), n>1.
G.f.: g(x) = 1 + (1/(1-x))*Sum_{j>=0} x^(2*10^j)/(1+x^10^j). (End)

A059015 Total number of 0's in binary expansions of 0, ..., n.

Original entry on oeis.org

1, 1, 2, 2, 4, 5, 6, 6, 9, 11, 13, 14, 16, 17, 18, 18, 22, 25, 28, 30, 33, 35, 37, 38, 41, 43, 45, 46, 48, 49, 50, 50, 55, 59, 63, 66, 70, 73, 76, 78, 82, 85, 88, 90, 93, 95, 97, 98, 102, 105, 108, 110, 113, 115, 117, 118, 121, 123, 125, 126, 128, 129, 130, 130, 136, 141
Offset: 0

Views

Author

Patrick De Geest, Dec 15 2000

Keywords

Comments

Partial sums of A023416. - Reinhard Zumkeller, Jul 15 2011
The graph of this sequence is a version of the Takagi curve: see Lagarias (2012), Section 9, especially Theorem 9.1. - N. J. A. Sloane, Mar 12 2016

Crossrefs

The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015, A070939, A083652.

Programs

  • Haskell
    a059015 n = a059015_list !! n
    a059015_list = scanl1 (+) $ map a023416 [0..]
    -- Reinhard Zumkeller, Jul 15 2011
    
  • Maple
    a:= proc(n) option remember; `if`(n=0, 1, a(n-1)+add(1-i, i=Bits[Split](n))) end:
    seq(a(n), n=0..65);  # Alois P. Heinz, Nov 11 2024
  • Mathematica
    Accumulate[ Table[ Count[ IntegerDigits[n, 2], 0], {n, 0, 65}]] (* Jean-François Alcover, Oct 03 2012 *)
    Accumulate[DigitCount[Range[0,70],2,0]] (* Harvey P. Dale, Jun 24 2017 *)
  • PARI
    v=vector(100,i,1);for(i=1,#v-1,v[i+1] = v[i] + #binary(i) - hammingweight(i)); v \\ Charles R Greathouse IV, Nov 20 2012
    
  • PARI
    a(n)=if(n, my(m=logint(n,2)); 2 + (m+1)*(n+1) - 2^(m+1) + sum(j=1,m+1, my(t=floor(n/2^j + 1/2)); (n>>j)*(2*n + 2 - (1 + (n>>j))<Charles R Greathouse IV, Dec 14 2015
    
  • Python
    def A059015(n): return 2+(n+1)*(m:=(n+1).bit_length())-(1<Chai Wah Wu, Mar 01 2023
    
  • Python
    def A059015(n): return 2+(n+1)*((t:=(n+1).bit_length())-n.bit_count())-(1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1))>>1) # Chai Wah Wu, Nov 11 2024

Formula

a(n) = b(n)+1, with b(2n) = b(n)+b(n-1)+n, b(2n+1) = 2b(n)+n. - Ralf Stephan, Sep 13 2003
From Hieronymus Fischer, Jun 10 2012: (Start)
With m = floor(log_2(n)):
a(n) = 2 + (m+1)*(n+1) - 2^(m+1) + (1/2)*Sum_{j=1..m+1} (floor(n/2^j)*(2*n + 2 - (1 + floor(n/2^j))*2^j) - floor(n/2^j + 1/2)*(2*n + 2 - floor(n/2^j + 1/2)*2^j)).
a(n) = A083652(n) - (n+1)*A000120(n) + 2^(m-1) - (1/4) + (1/2)*sum_{j=1..m+1} (floor(n/2^j + 1/2)^2 - (floor(n/2^j) + 1/2)^2)*2^j.
a(2^m-1) = 2 + (m-2)*2^(m-1)
(this is the total number of zero digits occurring in all the numbers with <= m places).
G.f.: 1/(1 - x) + (1/(1 - x)^2)*Sum_{j>=0} x^(2*2^j)/(1 + x^(2^j)); corrected by Ilya Gutkovskiy, Mar 28 2018
General formulas for the number of digits <= d in the base p representations of all integers from 0 to n, where 0 <= d < p.
With m = floor(log_p(n)):
a(n) = 1 + (m+1)*(n+1) - (p^(m+1)-1)/(p-1) + (1/2)*sum_{j=1..m+1} (floor(n/p^j)*(2n + 2 - (1 + floor(n/p^j))*p^j) - floor(n/p^j + (p-d-1)/p)*(2n + 2 + ((p-2*d-2)/p - floor(n/p^j + (p-d-1)/p))*p^j)).
a(n) = H(n,p) - (n+1)*F(n,p,d+1) + (1/2)*sum_{j=1..m+1} ((floor(n/p^j + (p-d-1)/p)^2 - floor(n/p^j)^2)*p^j - (((p - 2*d-2)/p)*floor(n/p^j + (p-d-1)/p) + floor(n/p^j))*p^j), where H(n,p) = sum of number of digits in the base p representations of 0 to n and F(n,p,d) = number of digits >=d in the base p representation of n.
a(p^m-1) = 1 + (d+1)*m*p^(m-1) - (p^m-1)/(p-1).
(this is the total number of digits <= d occurring in all the numbers with <= m places in base p representation).
G.f.: 1/(1-x) + (1/(1-x)^2)*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)

A055640 Number of nonzero digits in decimal expansion of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2
Offset: 0

Views

Author

Henry Bottomley, Jun 06 2000

Keywords

Comments

Comment from Antti Karttunen, Sep 05 2004: (Start)
Also number of characters needed to write the number n in classical Greek alphabetic system, up to n=999. The Greek alphabetic system assigned values to the letters as follows:
alpha = 1, beta = 2, gamma = 3, delta = 4, epsilon = 5, digamma = 6, zeta = 7, eta = 8, theta = 9, iota = 10, kappa = 20, lambda = 30, mu = 40, nu = 50, xi = 60, omicron = 70, pi = 80, koppa = 90, rho = 100, sigma = 200, tau = 300, upsilon = 400, phi = 500, chi = 600, psi = 700, omega = 800, sampi = 900. (End)
For partial sums see A102685. - Hieronymus Fischer, Jun 06 2012

Examples

			129 is written as rho kappa theta in the old Greek system.
		

References

  • L. Threatte, The Greek Alphabet, in The World's Writing Systems, edited by Peter T. Daniels and William Bright, Oxford Univ. Press, 1996, p. 278.

Crossrefs

Differs from A098378 for the first time at position n=200 with a(200)=1, as only one nonzero Arabic digit (and only one Greek letter) is needed for two hundred, while A098378(200)=2 as two characters are needed in the Ethiopic system.

Programs

Formula

From Hieronymus Fischer, Jun 06 2012: (Start)
a(n) = Sum_{j=1..m+1} (floor(n/10^j+0.9) - floor(n/10^j)), where m = floor(log_10(n)).
a(n) = m + 1 - A055641(n).
G.f.: (1/(1-x))*Sum_{j>=0} (x^10^j - x^(10*10^j))/(1-x^10^(j+1)). (End)
a(n) = A055642(n) - A055641(n).

A117804 Natural position of n in the string 12345678910111213....

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120
Offset: 1

Views

Author

Warut Roonguthai, Jul 23 2007

Keywords

Comments

The number of digits necessary to write down all the numbers 0, 1, 2, ..., n-1. Thus, the partial sums of A055642 are given by a(n+1). - Hieronymus Fischer, Jun 08 2012
From Daniel Forgues, Mar 21 2013: (Start)
From n = 10^0 + 1 to 10^1: a(n) - a(n-1) = 1 (9 * 10^0 terms);
From n = 10^1 + 1 to 10^2: a(n) - a(n-1) = 2 (9 * 10^1 terms);
From n = 10^2 + 1 to 10^3: a(n) - a(n-1) = 3 (9 * 10^2 terms);
...
From n = 10^k + 1 to 10^(k+1): a(n) - a(n-1) = k+1 (9 * 10^k terms). (End)
By the "number of digits" definition, a(n) = 1 + A058183(n-1) for n > 1. - David Fifield, Jun 02 2019

Examples

			12 begins at the 14th place in 12345678910111213... (we are ignoring "early bird" occurrences here, cf. A116700), so a(12) = 14.
From _Daniel Forgues_, Mar 21 2013: (Start)
a(10^1) = 10. (1*10^1 - 0)
a(10^2) = 190. (2*10^2 - 10)
a(10^3) = 2890. (3*10^3 - 110)
a(10^4) = 38890. (4*10^4 - 1110)
a(10^5) = 488890. (5*10^5 - 11110)
a(10^6) = 5888890. (6*10^6 - 111110)
...
a(10^k) = k*10^k - R_k + 1, R_k := k-th repunit (cf. A002275)
(the number of digits necessary to write down the numbers 0..10^k-1). (End)
		

Crossrefs

Formula

a(n) = d*n + 1 - (10^d - 1)/9 where d is the number of decimal digits in n, i.e., d = floor(log_10(n)) + 1.
From Hieronymus Fischer, Jun 08 2012: (Start)
a(n) = Sum_{j=0..n-1} A055642(j).
a(n) = 1 + A055642(n-1)*n - (10^A055642(n-1)-1)/9.
a(n) = 1 + A055642(n)*n - (10^A055642(n)-1)/9.
a(10^n) = (9*n-1)*(10^n-1)/9 + n + 1. (This is the total number of digits necessary to write down all the numbers with <= n places.)
G.f.: g(x) = x/(1-x) + (x/(1-x)^2)*Sum_{j>=0} x^10^j; corrected by Ilya Gutkovskiy, Jan 09 2017 (End)

A102685 Partial sums of A055640.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123
Offset: 0

Views

Author

N. J. A. Sloane, Feb 03 2005

Keywords

Comments

The total number of nonzero digits occurring in all the numbers 0, 1, 2, ... n (in decimal representation). - Hieronymus Fischer, Jun 10 2012

Crossrefs

Formula

From Hieronymus Fischer, Jun 06 2012: (Start)
a(n) = (1/2)*Sum_{j=1..m+1} (floor((n/10^j)+0.9)*(2n + 2 + (0.8 - floor((n/10^j)+0.9))*10^j) - floor(n/10^j)*(2n + 2 - (floor(n/10^j)+1) * 10^j)), where m = floor(log_10(n)).
a(n) = (n+1)*A055640(n) + (1/2)*Sum_{j=1..m+1} ((8*floor((n/10^j)+0.9)/10 + floor(n/10^j))*10^j - (floor((n/10^j)+0.9)^2 - floor(n/10^j)^2)*10^j), where m = floor(log_10(n)).
a(10^m-1) = 9*m*10^(m-1). (This is the total number of nonzero digits occurring in all the numbers with <= m digits.)
G.f.: g(x) = (1/(1-x)^2) * Sum_{j>=0} (x^10^j - x^(10*10^j))/(1-x^10^(j+1)). (End)
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