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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A160094 a(n) = 1 + A122840(n).

Original entry on oeis.org

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

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Author

Anonymous, May 01 2009

Keywords

Comments

a(n) is the Levenshtein distance from the decimal expansion of n - 1 to the decimal expansion of n. For example, to convert "9" to "10", substitute "0" for "9" and insert "1". Since two such operations are required, a(10) = 2. See the analogous A091090 (binary expansion) and A115777 (full definition). - Rick L. Shepherd, Mar 25 2015

Examples

			a(160) = 2 because the last nonzero digit of 160 (counting from left to right), when 160 is written in base 10, is 6, and that 6 occurs 2 digits from the right in 160.
		

Crossrefs

Programs

  • Mathematica
    IntegerExponent[Range[150]]+1 (* Harvey P. Dale, Feb 06 2015 *)

Formula

From Hieronymus Fischer, Jun 08 2012: (Start)
With m = floor(log_10(n)), frac(x) = x-floor(x):
a(n) = Sum_{j=0..m} (1 - ceiling(frac(n/10^j))).
a(n) = m + 1 + Sum_{j=1..m} (floor(-frac(n/10^j))).
a(n) = 1 + A054899(n) - A054899(n-1).
G.f.: g(x) = (x/(1-x)) + Sum_{j>0} x^10^j/(1-x^10^j). (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 10/9. - Amiram Eldar, Jul 10 2023
a(n) = A122840(10*n). - R. J. Mathar, Jun 28 2025

Extensions

Name simplified by Jon E. Schoenfield, Feb 26 2014

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

A005117 Squarefree numbers: numbers that are not divisible by a square greater than 1.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 46, 47, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 101, 102, 103, 105, 106, 107, 109, 110, 111, 113
Offset: 1

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Author

Keywords

Comments

1 together with the numbers that are products of distinct primes.
Also smallest sequence with the property that a(m)*a(k) is never a square for k != m. - Ulrich Schimke (ulrschimke(AT)aol.com), Dec 12 2001
Numbers k such that there is only one Abelian group with k elements, the cyclic group of order k (the numbers such that A000688(k) = 1). - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 25 2001
Numbers k such that A007913(k) > phi(k). - Benoit Cloitre, Apr 10 2002
a(n) is the smallest m with exactly n squarefree numbers <= m. - Amarnath Murthy, May 21 2002
k is squarefree <=> k divides prime(k)# where prime(k)# = product of first k prime numbers. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 30 2004
Numbers k such that omega(k) = Omega(k) = A072047(k). - Lekraj Beedassy, Jul 11 2006
The LCM of any finite subset is in this sequence. - Lekraj Beedassy, Jul 11 2006
This sequence and the Beatty Pi^2/6 sequence (A059535) are "incestuous": the first 20000 terms are bounded within (-9, 14). - Ed Pegg Jr, Jul 22 2008
Let us introduce a function D(n) = sigma_0(n)/2^(alpha(1) + ... + alpha(r)), sigma_0(n) number of divisors of n (A000005), prime factorization of n = p(1)^alpha(1) * ... * p(r)^alpha(r), alpha(1) + ... + alpha(r) is sequence (A001222). Function D(n) splits the set of positive integers into subsets, according to the value of D(n). Squarefree numbers (A005117) has D(n)=1, other numbers are "deviated" from the squarefree ideal and have 0 < D(n) < 1. For D(n)=1/2 we have A048109, for D(n)=3/4 we have A060687. - Ctibor O. Zizka, Sep 21 2008
Numbers k such that gcd(k,k')=1 where k' is the arithmetic derivative (A003415) of k. - Giorgio Balzarotti, Apr 23 2011
Numbers k such that A007913(k) = core(k) = k. - Franz Vrabec, Aug 27 2011
Numbers k such that sqrt(k) cannot be simplified. - Sean Loughran, Sep 04 2011
Indices m where A057918(m)=0, i.e., positive integers m for which there are no integers k in {1,2,...,m-1} such that k*m is a square. - John W. Layman, Sep 08 2011
It appears that these are numbers j such that Product_{k=1..j} (prime(k) mod j) = 0 (see Maple code). - Gary Detlefs, Dec 07 2011. - This is the same claim as Mohammed Bouayoun's Mar 30 2004 comment above. To see why it holds: Primorial numbers, A002110, a subsequence of this sequence, are never divisible by any nonsquarefree number, A013929, and on the other hand, the index of the greatest prime dividing any n is less than n. Cf. A243291. - Antti Karttunen, Jun 03 2014
Conjecture: For each n=2,3,... there are infinitely many integers b > a(n) such that Sum_{k=1..n} a(k)*b^(k-1) is prime, and the smallest such an integer b does not exceed (n+3)*(n+4). - Zhi-Wei Sun, Mar 26 2013
The probability that a random natural number belongs to the sequence is 6/Pi^2, A059956 (see Cesàro reference). - Giorgio Balzarotti, Nov 21 2013
Booker, Hiary, & Keating give a subexponential algorithm for testing membership in this sequence without factoring. - Charles R Greathouse IV, Jan 29 2014
Because in the factorizations into prime numbers these a(n) (n >= 2) have exponents which are either 0 or 1 one could call the a(n) 'numbers with a fermionic prime number decomposition'. The levels are the prime numbers prime(j), j >= 1, and the occupation numbers (exponents) e(j) are 0 or 1 (like in Pauli's exclusion principle). A 'fermionic state' is then denoted by a sequence with entries 0 or 1, where, except for the zero sequence, trailing zeros are omitted. The zero sequence stands for a(1) = 1. For example a(5) = 6 = 2^1*3^1 is denoted by the 'fermionic state' [1, 1], a(7) = 10 by [1, 0, 1]. Compare with 'fermionic partitions' counted in A000009. - Wolfdieter Lang, May 14 2014
From Vladimir Shevelev, Nov 20 2014: (Start)
The following is an Eratosthenes-type sieve for squarefree numbers. For integers > 1:
1) Remove even numbers, except for 2; the minimal non-removed number is 3.
2) Replace multiples of 3 removed in step 1, and remove multiples of 3 except for 3 itself; the minimal non-removed number is 5.
3) Replace multiples of 5 removed as a result of steps 1 and 2, and remove multiples of 5 except for 5 itself; the minimal non-removed number is 6.
4) Replace multiples of 6 removed as a result of steps 1, 2 and 3 and remove multiples of 6 except for 6 itself; the minimal non-removed number is 7.
5) Repeat using the last minimal non-removed number to sieve from the recovered multiples of previous steps.
Proof. We use induction. Suppose that as a result of the algorithm, we have found all squarefree numbers less than n and no other numbers. If n is squarefree, then the number of its proper divisors d > 1 is even (it is 2^k - 2, where k is the number of its prime divisors), and, by the algorithm, it remains in the sequence. Otherwise, n is removed, since the number of its squarefree divisors > 1 is odd (it is 2^k-1).
(End)
The lexicographically least sequence of integers > 1 such that each entry has an even number of proper divisors occurring in the sequence (that's the sieve restated). - Glen Whitney, Aug 30 2015
0 is nonsquarefree because it is divisible by any square. - Jon Perry, Nov 22 2014, edited by M. F. Hasler, Aug 13 2015
The Heinz numbers of partitions with distinct parts. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} prime(j) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] the Heinz number is 2*2*3*7*29 = 2436. The number 30 (= 2*3*5) is in the sequence because it is the Heinz number of the partition [1,2,3]. - Emeric Deutsch, May 21 2015
It is possible for 2 consecutive terms to be even; for example a(258)=422 and a(259)=426. - Thomas Ordowski, Jul 21 2015. [These form a subsequence of A077395 since their product is divisible by 4. - M. F. Hasler, Aug 13 2015]
There are never more than 3 consecutive terms. Runs of 3 terms start at 1, 5, 13, 21, 29, 33, ... (A007675). - Ivan Neretin, Nov 07 2015
a(n) = product of row n in A265668. - Reinhard Zumkeller, Dec 13 2015
Numbers without excess, i.e., numbers k such that A001221(k) = A001222(k). - Juri-Stepan Gerasimov, Sep 05 2016
Numbers k such that b^(phi(k)+1) == b (mod k) for every integer b. - Thomas Ordowski, Oct 09 2016
Boreico shows that the set of square roots of the terms of this sequence is linearly independent over the rationals. - Jason Kimberley, Nov 25 2016 (reference found by Michael Coons).
Numbers k such that A008836(k) = A008683(k). - Enrique Pérez Herrero, Apr 04 2018
The prime zeta function P(s) "has singular points along the real axis for s=1/k where k runs through all positive integers without a square factor". See Wolfram link. - Maleval Francis, Jun 23 2018
Numbers k such that A007947(k) = k. - Kyle Wyonch, Jan 15 2021
The Schnirelmann density of the squarefree numbers is 53/88 (Rogers, 1964). - Amiram Eldar, Mar 12 2021
Comment from Isaac Saffold, Dec 21 2021: (Start)
Numbers k such that all groups of order k have a trivial Frattini subgroup [Dummit and Foote].
Let the group G have order n. If n is squarefree and n > 1, then G is solvable, and thus by Hall's Theorem contains a subgroup H_p of index p for all p | n. Each H_p is maximal in G by order considerations, and the intersection of all the H_p's is trivial. Thus G's Frattini subgroup Phi(G), being the intersection of G's maximal subgroups, must be trivial. If n is not squarefree, the cyclic group of order n has a nontrivial Frattini subgroup. (End)
Numbers for which the squarefree divisors (A206778) and the unitary divisors (A077610) are the same; moreover they are also the set of divisors (A027750). - Bernard Schott, Nov 04 2022
0 = A008683(a(n)) - A008836(a(n)) = A001615(a(n)) - A000203(a(n)). - Torlach Rush, Feb 08 2023
From Robert D. Rosales, May 20 2024: (Start)
Numbers n such that mu(n) != 0, where mu(n) is the Möbius function (A008683).
Solutions to the equation Sum_{d|n} mu(d)*sigma(d) = mu(n)*n, where sigma(n) is the sum of divisors function (A000203). (End)
a(n) is the smallest root of x = 1 + Sum_{k=1..n-1} floor(sqrt(x/a(k))) greater than a(n-1). - Yifan Xie, Jul 10 2024
Number k such that A001414(k) = A008472(k). - Torlach Rush, Apr 14 2025
To elaborate on the formula from Greathouse (2018), the maximum of a(n) - floor(n*Pi^2/6 + sqrt(n)/17) is 10 at indices n = 48715, 48716, 48721, and 48760. The maximum is 11, at the same indices, if floor is taken individually for the two addends and the square root. If the value is rounded instead, the maximum is 9 at 10 indices between 48714 and 48765. - M. F. Hasler, Aug 08 2025

References

  • Jean-Marie De Koninck, Ces nombres qui nous fascinent, Entry 165, p. 53, Ellipses, Paris, 2008.
  • David S. Dummit and Richard M. Foote, Abstract algebra. Vol. 1999. Englewood Cliffs, NJ: David S.Prentice Hall, 1991.
  • Ivan M. Niven and Herbert S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 251.
  • Michael Pohst and Hans J. Zassenhaus, Algorithmic Algebraic Number Theory, Cambridge Univ. Press, page 432.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Complement of A013929. Subsequence of A072774 and A209061.
Characteristic function: A008966 (mu(n)^2, where mu = A008683).
Subsequences: A000040, A002110, A235488.
Subsequences: numbers j such that j*a(k) is squarefree where k > 1: A056911 (k = 2), A261034 (k = 3), A274546 (k = 5), A276378 (k = 6).

Programs

  • Haskell
    a005117 n = a005117_list !! (n-1)
    a005117_list = filter ((== 1) . a008966) [1..]
    -- Reinhard Zumkeller, Aug 15 2011, May 10 2011
    
  • Magma
    [ n : n in [1..1000] | IsSquarefree(n) ];
    
  • Maple
    with(numtheory); a := [ ]; for n from 1 to 200 do if issqrfree(n) then a := [ op(a), n ]; fi; od:
    t:= n-> product(ithprime(k),k=1..n): for n from 1 to 113 do if(t(n) mod n = 0) then print(n) fi od; # Gary Detlefs, Dec 07 2011
    A005117 := proc(n) option remember; if n = 1 then 1; else for a from procname(n-1)+1 do if numtheory[issqrfree](a) then return a; end if; end do: end if; end proc:  # R. J. Mathar, Jan 09 2013
  • Mathematica
    Select[ Range[ 113], SquareFreeQ] (* Robert G. Wilson v, Jan 31 2005 *)
    Select[Range[150], Max[Last /@ FactorInteger[ # ]] < 2 &] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
    NextSquareFree[n_, k_: 1] := Block[{c = 0, sgn = Sign[k]}, sf = n + sgn; While[c < Abs[k], While[ ! SquareFreeQ@ sf, If[sgn < 0, sf--, sf++]]; If[ sgn < 0, sf--, sf++]; c++]; sf + If[ sgn < 0, 1, -1]]; NestList[ NextSquareFree, 1, 70] (* Robert G. Wilson v, Apr 18 2014 *)
    Select[Range[250], MoebiusMu[#] != 0 &] (* Robert D. Rosales, May 20 2024 *)
  • PARI
    bnd = 1000; L = vector(bnd); j = 1; for (i=1,bnd, if(issquarefree(i),L[j]=i; j=j+1)); L
    
  • PARI
    {a(n)= local(m,c); if(n<=1,n==1, c=1; m=1; while( cMichael Somos, Apr 29 2005 */
    
  • PARI
    list(n)=my(v=vectorsmall(n,i,1),u,j); forprime(p=2,sqrtint(n), forstep(i=p^2, n, p^2, v[i]=0)); u=vector(sum(i=1,n,v[i])); for(i=1,n,if(v[i],u[j++]=i)); u \\ Charles R Greathouse IV, Jun 08 2012
    
  • PARI
    for(n=1, 113, if(core(n)==n, print1(n, ", "))); \\ Arkadiusz Wesolowski, Aug 02 2016
    
  • PARI
    S(n) = my(s); forsquarefree(k=1,sqrtint(n),s+=n\k[1]^2*moebius(k)); s;
    a(n) = my(min=1, max=231, k=0, sc=0); if(n >= 144, min=floor(zeta(2)*n - 5*sqrt(n)); max=ceil(zeta(2)*n + 5*sqrt(n))); while(min <= max, k=(min+max)\2; sc=S(k); if(abs(sc-n) <= sqrtint(n), break); if(sc > n, max=k-1, if(sc < n, min=k+1, break))); while(!issquarefree(k), k-=1); while(sc != n, my(j=1); if(sc > n, j = -1); k += j; sc += j; while(!issquarefree(k), k += j)); k; \\ Daniel Suteu, Jul 07 2022
    
  • PARI
    first(n)=my(v=vector(n),i); forsquarefree(k=1,if(n<268293,(33*n+30)\20,(n*Pi^2/6+0.058377*sqrt(n))\1), if(i++>n, return(v)); v[i]=k[1]); v \\ Charles R Greathouse IV, Jan 10 2023
    
  • PARI
    A5117=[1..3]; A005117(n)={if(n>#A5117, my(N=#A5117); A5117=Vec(A5117, max(n+999, N*5\4)); iferr(forsquarefree(k=A5117[N]+1, #A5117*Pi^2\6+sqrtint(#A5117)\17+11, A5117[N++]=k[1]),E,)); A5117[n]} \\ M. F. Hasler, Aug 08 2025
    
  • Python
    from sympy.ntheory.factor_ import core
    def ok(n): return core(n, 2) == n
    print(list(filter(ok, range(1, 114)))) # Michael S. Branicky, Jul 31 2021
    
  • Python
    from itertools import count, islice
    from sympy import factorint
    def A005117_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda n:all(x == 1 for x in factorint(n).values()),count(max(startvalue,1)))
    A005117_list = list(islice(A005117_gen(),20)) # Chai Wah Wu, May 09 2022
    
  • Python
    from math import isqrt
    from sympy import mobius
    def A005117(n):
        def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
        m, k = n, f(n)
        while m != k:
            m, k = k, f(k)
        return m # Chai Wah Wu, Jul 22 2024

Formula

Limit_{n->oo} a(n)/n = Pi^2/6 (see A013661). - Benoit Cloitre, May 23 2002
Equals A039956 UNION A056911. - R. J. Mathar, May 16 2008
A122840(a(n)) <= 1; A010888(a(n)) < 9. - Reinhard Zumkeller, Mar 30 2010
a(n) = A055229(A062838(n)) and a(n) > A055229(m) for m < A062838(n). - Reinhard Zumkeller, Apr 09 2010
A008477(a(n)) = 1. - Reinhard Zumkeller, Feb 17 2012
A055653(a(n)) = a(n); A055654(a(n)) = 0. - Reinhard Zumkeller, Mar 11 2012
A008966(a(n)) = 1. - Reinhard Zumkeller, May 26 2012
Sum_{n>=1} 1/a(n)^s = zeta(s)/zeta(2*s). - Enrique Pérez Herrero, Jul 07 2012
A056170(a(n)) = 0. - Reinhard Zumkeller, Dec 29 2012
A013928(a(n)+1) = n. - Antti Karttunen, Jun 03 2014
A046660(a(n)) = 0. - Reinhard Zumkeller, Nov 29 2015
Equals {1} UNION A000040 UNION A006881 UNION A007304 UNION A046386 UNION A046387 UNION A067885 UNION A123321 UNION A123322 UNION A115343 ... - R. J. Mathar, Nov 05 2016
|a(n) - n*Pi^2/6| < 0.058377*sqrt(n) for n >= 268293; this result can be derived from Cohen, Dress, & El Marraki, see links. - Charles R Greathouse IV, Jan 18 2018
From Amiram Eldar, Jul 07 2021: (Start)
Sum_{n>=1} (-1)^(a(n)+1)/a(n)^2 = 9/Pi^2.
Sum_{k=1..n} 1/a(k) ~ (6/Pi^2) * log(n).
Sum_{k=1..n} (-1)^(a(k)+1)/a(k) ~ (2/Pi^2) * log(n).
(all from Scott, 2006) (End)

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

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

A112765 Exponent of highest power of 5 dividing n. Or, 5-adic valuation of n.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, Sep 18 2005

Keywords

Comments

A027868 gives partial sums.
This is also the 5-adic valuation of Fibonacci(n). See Lengyel link. - Michel Marcus, May 06 2017

Crossrefs

Cf. A343251, A000351 (positions of records, greedy inverse).

Programs

  • Haskell
    a112765 n = fives n 0 where
       fives n e | r > 0     = e
                 | otherwise = fives n' (e + 1) where (n',r) = divMod n 5
    -- Reinhard Zumkeller, Apr 08 2011
    
  • Maple
    A112765 := proc(n)
        padic[ordp](n,5) ;
    end proc: # R. J. Mathar, Jul 12 2016
  • Mathematica
    a[n_] := IntegerExponent[n, 5]; Array[a, 105] (* Jean-François Alcover, Jan 25 2018 *)
  • PARI
    A112765(n)=valuation(n,5); /* Joerg Arndt, Apr 08 2011 */
    
  • Python
    def a(n):
        k = 0
        while n > 0 and n%5 == 0: n //= 5; k += 1
        return k
    print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Aug 06 2021

Formula

Totally additive with a(p) = 1 if p = 5, 0 otherwise.
From Hieronymus Fischer, Jun 08 2012: (Start)
With m = floor(log_5(n)), frac(x) = x-floor(x):
a(n) = Sum_{j=1..m} (1 - ceiling(frac(n/5^j))).
a(n) = m + Sum_{j=1..m} (floor(-frac(n/5^j))).
a(n) = A027868(n) - A027868(n-1).
G.f.: Sum_{j>0} x^5^j/(1-x^5^j). (End)
a(5n) = A055457(n). - R. J. Mathar, Jul 17 2012
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/4. - Amiram Eldar, Feb 14 2021
a(n) = 5*Sum_{j=1..floor(log(n)/log(5))} frac(binomial(n, 5^j)*5^(j-1)/n). - Dario T. de Castro, Jul 10 2022

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

A054899 a(n) = Sum_{k>0} floor(n/10^k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 11, 11, 11
Offset: 0

Views

Author

Henry Bottomley, May 23 2000

Keywords

Comments

The old definition of this sequence was "Highest power of 10 dividing n!", but that is wrong (see A027868). For example, the highest power of 10 dividing 5!=120 is 1; however, a(5)=0. - Hieronymus Fischer, Jun 18 2007
Highest power of 10 dividing the quotient of multifactorials Product_{k>=1} M(10^k, 10^k*floor(n/10^k)) /( Product_{k>=1} M(10^(k-1), 10^(k-1) * floor(n/10^k)) ) where M(r,s) is the r-multifactorial (r>=1) of s which is defined by M(r,s) = s*M(r,s-r) for s > 0, M(r,s) = 1 for s <= 0. This is because that quotient of multifactorials evaluates to the product 10^floor(n/10)*10^floor(n/100)*... - Hieronymus Fischer, Jun 14 2007
Partial sums of A122840. - Hieronymus Fischer, Jun 06 2012
Called the "terminating nines function" by Kennedy et al. (1989). a(n) is the number of terminating nines which occur up to n but not including n. - Amiram Eldar, Sep 06 2024

Examples

			          a(11) = 1
         a(111) = 12.
        a(1111) = 123.
       a(11111) = 1234.
      a(111111) = 12345.
     a(1111111) = 123456.
    a(11111111) = 1234567.
   a(111111111) = 12345678.
  a(1111111111) = 123456789.
		

Crossrefs

Cf. A011371 and A054861 for analogs involving powers of 2 and 3.
Different from the highest power of 10 dividing n! (see A027868 for reference).

Programs

  • Magma
    m:=10;
    function a(n) // a = A054899, m = 10
      if n eq 0 then return 0;
      else return a(Floor(n/m)) + Floor(n/m);
      end if; end function;
    [a(n): n in [0..103]]; // G. C. Greubel, Apr 28 2023
    
  • Mathematica
    Table[t=0; p=10; While[s=Floor[n/p]; t=t+s; s>0, p*=10]; t, {n,0,100}]
  • PARI
    a(n)=my(s);while(n\=10,s+=n);s \\ Charles R Greathouse IV, Jul 19 2011
    
  • SageMath
    m=10 # a = A054899
    def a(n): return 0 if (n==0) else a(n//m) + (n//m)
    [a(n) for n in range(104)] # G. C. Greubel, Apr 28 2023

Formula

a(n) = floor(n/10) + floor(n/100) + floor(n/1000) + ...
a(n) = (n - A007953(n))/9.
From Hieronymus Fischer, Jun 14 2007, Jun 25 2007, and Aug 13 2007: (Start)
a(n) = Sum_{k>0} floor(n/10^k).
a(n) = Sum_{k=10..n} Sum_{j|k, j>=10} ( floor(log_10(j)) -floor(log_10(j-1)) ).
G.f.: g(x) = ( Sum_{k>0} x^(10^k)/(1-x^(10^k)) )/(1-x).
G.f. expressed in terms of Lambert series:
g(x) = L[b(k)](x)/(1-x) where L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1-x^k) is a Lambert series with b(k)=1, if k>1 is a power of 10, else b(k)=0.
G.f.: g(x) = ( Sum_{k>0} c(k)*x^k )/(1-x), where c(k) = Sum_{j>1, j|k} (floor(log_10(j)) - floor(log_10(j-1)) ).
a(n) = Sum_{k=0..floor(log_10(n))} ds_10(floor(n/10^k))*10^k - n where ds_10(x) = digital sum of x in base 10.
a(n) = Sum_{k=0..floor(log_10(n))} A007953(floor(n/10^k))*10^k - n.
Recurrence:
a(n) = floor(n/10) + a(floor(n/10)).
a(10*n) = n + a(n).
a(n*10^m) = n*(10^m-1)/9 + a(n).
a(k*10^m) = k*(10^m-1)/9, for 0 <= k < 10, m >= 0.
Asymptotic behavior:
a(n) = n/9 + O(log(n)),
a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
a(n) <= (n - 1)/9; equality holds for powers of 10.
a(n) >= n/9 - 1 - floor(log_10(n)); equality holds for n=10^m-1, m>0.
lim inf (n/9 - a(n)) = 1/9, for n --> oo.
lim sup (n/9 - log_10(n) - a(n)) = 0, for n --> oo.
lim sup (a(n+1) - a(n) - log_10(n)) = 0, for n --> oo. (End)

Extensions

An incorrect g.f. was deleted by N. J. A. Sloane, Sep 13 2009
Examples added by Hieronymus Fischer, Jun 06 2012

A122841 Greatest k such that 6^k divides n.

Original entry on oeis.org

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

Views

Author

Reinhard Zumkeller, Sep 13 2006

Keywords

Comments

See A054895 for the partial sums. - Hieronymus Fischer, Jun 08 2012

Crossrefs

Programs

  • Haskell
    a122841 = f 0 where
       f y x = if r > 0 then y else f (y + 1) x'
               where (x', r) = divMod x 6
    -- Reinhard Zumkeller, Nov 10 2013
    
  • Mathematica
    Table[IntegerExponent[n, 6], {n, 1, 100}] (* Amiram Eldar, Sep 14 2020 *)
  • PARI
    a(n) = valuation(n, 6); \\ Michel Marcus, Jan 17 2022

Formula

From Hieronymus Fischer, Jun 03 2012: (Start)
With m = floor(log_6(n)), frac(x) = x-floor(x):
a(n) = Sum_{j=1..m} (1 - ceiling(frac(n/6^j))).
a(n) = m + Sum_{j=1..m} (floor(-frac(n/6^j))).
a(n) = A054895(n) - A054895(n-1).
G.f.: Sum_{j>0} x^6^j/(1-x^6^j). (End)
a(A047253(n)) = 0; a(A008588(n)) > 0; a(A044102(n)) > 1. - Reinhard Zumkeller, Nov 10 2013
6^a(n) = A234959(n), n >= 1. - Wolfdieter Lang, Jun 30 2014
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/5. - Amiram Eldar, Jan 17 2022
a(n) = min(A007814(n), A007949(n)). - Jianing Song, Jul 23 2022
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