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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A339566 Primes p such that A007088(p) == 1 (mod p).

Original entry on oeis.org

5, 137, 3967, 25087, 242899421
Offset: 1

Views

Author

Robert Israel, Dec 09 2020

Keywords

Examples

			a(3) = 3967 is in the sequence because 3967 = 111101111111_2 and 111101111111 == 1 (mod 3967).
		

Crossrefs

Primes in A339567.

Programs

  • Maple
    p:= 1: R:= NULL:
    while p < 3*10^8 do
    p:= nextprime(p);
    if convert(p,binary) mod p = 1 then R:= R, p fi
    od:
    R;
  • Python
    from sympy import nextprime
    A339566_list, p = [], 2
    while p < 10**10:
        if int(bin(p)[2:]) % p == 1:
            A339566_list.append(p)
        p = nextprime(p) # Chai Wah Wu, Dec 14 2020

A354838 For n >= 1, a(n) = A007088(n)/gcd(n, A007088(n)).

Original entry on oeis.org

1, 5, 11, 25, 101, 55, 111, 125, 1001, 101, 1011, 275, 1101, 555, 1111, 625, 10001, 5005, 10011, 505, 481, 5055, 10111, 1375, 11001, 5505, 11011, 2775, 11101, 1111, 11111, 3125, 9091, 50005, 100011, 25025, 100101, 50055, 100111, 2525, 101001, 2405, 101011, 25275
Offset: 1

Views

Author

Ctibor O. Zizka, Jun 08 2022

Keywords

Examples

			a(6) = A007088(6)/gcd(6, A007088(6)) = 110/gcd(6,110) = 110/2 = 55.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := (b = FromDigits[IntegerDigits[n, 2]])/GCD[n, b]; Array[a, 50] (* Amiram Eldar, Jun 08 2022 *)
  • PARI
    a(n) = numerator(fromdigits(binary(n))/n); \\ Kevin Ryde, Jun 09 2022

A339545 Primes p such that A007088(p) == A151799(p) (mod p).

Original entry on oeis.org

3, 19, 29, 691
Offset: 1

Views

Author

J. M. Bergot and Robert Israel, Dec 08 2020

Keywords

Comments

Primes p such that the binary representation of p, considered as a decimal number, is congruent mod p to the prime previous to p.
No other terms < 10^11. - Max Alekseyev, Feb 04 2024

Examples

			a(3) = 29 is a member because 29 = 11101_2, 11101 == 23 (mod 29), and 23 is the prime previous to 29.
		

Crossrefs

Programs

  • Maple
    select(t -> isprime(t) and convert(t,binary) mod t = prevprime(t), [seq(i,i=3..1000,2)]);

A355297 a(n) = A007088(n) mod n.

Original entry on oeis.org

0, 0, 2, 0, 1, 2, 6, 0, 2, 0, 10, 8, 9, 4, 1, 0, 5, 2, 17, 0, 0, 12, 14, 8, 1, 12, 22, 12, 23, 10, 13, 0, 11, 16, 16, 20, 16, 18, 37, 0, 18, 0, 4, 32, 31, 2, 14, 32, 45, 10, 4, 16, 20, 4, 1, 8, 22, 56, 32, 40, 20, 6, 42, 0, 41, 44, 36, 24, 15, 20, 5, 56, 25, 12, 61, 28, 24, 58, 23, 0
Offset: 1

Views

Author

Ctibor O. Zizka, Jun 27 2022

Keywords

Comments

a(n) = 0 see A032533, a(n) = 1 see A339567.

Examples

			n = 7; a(7) = A007088(7) mod 7 = 111 mod 7 = 6.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Mod[FromDigits[IntegerDigits[n, 2]], n]; Array[a, 100] (* Amiram Eldar, Jun 27 2022 *)

Formula

a(n) = A007088(n) mod n. a(n) = A007088(n) - n * floor(A007088(n)/n).

A355300 a(0) = 0; for n >= 1, a(n) = a(A007088(n) mod n) + 1.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 1, 2, 2, 3, 2, 2, 1, 3, 2, 4, 1, 1, 3, 3, 2, 2, 3, 4, 3, 4, 2, 4, 1, 3, 2, 2, 2, 2, 3, 3, 1, 3, 1, 2, 2, 5, 2, 3, 2, 6, 2, 2, 2, 2, 2, 2, 2, 4, 3, 2, 2, 2, 3, 2, 1, 4, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 4, 3, 4, 4, 1, 3, 2, 4, 1, 3, 2, 4, 3, 3, 2, 2, 2
Offset: 0

Views

Author

Ctibor O. Zizka, Jun 27 2022

Keywords

Comments

Number of steps needed to reach zero when starting from k = n and repeatedly applying the map that replaces k by A007088(k) mod k. For a(n) = 1 see A032533.
a(95980631) = 26. - Charles R Greathouse IV, Jun 30 2022

Examples

			n = 12, a(12) = a(1100 mod 12) + 1 = a(8) + 1 = a(1000 mod 8) + 2 = a(0) + 2 = 2.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = a[Mod[FromDigits[IntegerDigits[n, 2]], n]] + 1; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Jun 27 2022 *)
  • PARI
    f(n) = fromdigits(binary(n), 10); \\ A007088
    a(n) = if (n, a(f(n) % n)+1, 0); \\ Michel Marcus, Jun 27 2022

A360372 Numbers k >= 1 such that k divides Sum_{i=1..k} A007088(i).

Original entry on oeis.org

1, 11, 21, 23, 37, 461, 94101, 14958901, 16364133, 134375017, 192594821, 416095237, 4074380993, 82482257999
Offset: 1

Views

Author

Ctibor O. Zizka, Feb 04 2023

Keywords

Comments

The arithmetic mean of the first k binary numbers, taken as decimal numbers, is an integer.

Examples

			k = 11: (1 + 10 + 11 + 100 + 101 + 110 + 111 + 1000 + 1001 + 1010 + 1011) / 11 = 406, thus 11 is a term.
		

Crossrefs

Programs

  • Mathematica
    s = 0; seq = {}; Do[s += FromDigits[IntegerDigits[k, 2]]; If[Divisible[s, k], AppendTo[seq, k]], {k, 1, 1000}]; seq (* Amiram Eldar, Feb 04 2023 *)

Extensions

a(7)-a(13) from Amiram Eldar, Feb 04 2023
a(14) from Bert Dobbelaere, Feb 14 2023

A361312 Smallest prime p such that the decimal expansion of p remains prime through exactly n iterations of base-10 to base-2 conversion (A007088).

Original entry on oeis.org

2, 3, 5, 3893257, 9632552297
Offset: 0

Views

Author

Ya-Ping Lu, Mar 08 2023

Keywords

Comments

Prime numbers that remain primes after 1, 2, 3, and 4 iterations are A065720, A123266, A256621 and A256622, respectively.

Examples

			a(0) = 2 because prime number 2 in base 2 is 10, and 10 in base 10 is not a prime.
a(1) = 3 because 3 = 11_2 and 11_10 is a prime. In the second iteration, however, 11_10 = 1011_2 and 1011_10 is not a prime.
a(2) = 5 because 5 = 101_2 and 101_10 = 1100101_2. Both 101 and 1100101 are primes in base 10. In the third iteration, 1100101_10 = 100001100100101000101_2 and 100001100100101000101_10 is not a prime.
		

Crossrefs

Programs

  • Python
    from sympy import isprime, nextprime
    p = 1; mx = 5; I = [*range(mx)]; R = [*range(mx)]
    while I:
        p = nextprime(p); ct = 0; q = p
        while isprime(int(bin(q)[2:])): ct += 1; q = int(bin(q)[2:])
        if ct in I: R[ct] = p; I.remove(ct)
    print(*R, sep = ", ")

A379779 Numbers k >= 1 such that A007088(k) / A007088(A007953(k)) is an integer.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 24, 27, 30, 36, 40, 45, 48, 54, 60, 63, 70, 72, 80, 90, 100, 102, 108, 110, 111, 112, 114, 120, 126, 132, 133, 144, 150, 152, 153, 156, 180, 190, 192, 195, 200, 201, 204, 210, 216, 220, 222, 224, 228, 240, 252, 264, 266, 288, 300, 306, 312, 315, 320
Offset: 1

Views

Author

Ctibor O. Zizka, Jan 15 2025

Keywords

Examples

			k = 18: A007088(18)/A007088(A007953(18)) = 10010/1001 = 10, thus k = 18 is a member of the sequence.
		

Crossrefs

Programs

  • Mathematica
    q[k_] := Divisible @@ (FromDigits[IntegerDigits[#, 2]] & /@ {k, DigitSum[k]}); Select[Range[320], q] (* Amiram Eldar, Jan 15 2025 *)
  • PARI
    bin(n) = fromdigits(binary(n), 10); \\ A007088
    isok(k) = denominator(bin(k)/bin(sumdigits(k))) == 1; \\ Michel Marcus, Jan 15 2025

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

Views

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

A070939 Length of binary representation of n.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7
Offset: 0

Views

Author

N. J. A. Sloane, May 18 2002

Keywords

Comments

Zero is assumed to be represented as 0.
For n>1, n appears 2^(n-1) times. - Lekraj Beedassy, Apr 12 2006
a(n) is the permanent of the n X n 0-1 matrix whose (i,j) entry is 1 iff i=1 or i=j or i=2*j. For example, a(4)=3 is per([[1, 1, 1, 1], [1, 1, 0, 0], [0, 0, 1, 0], [0, 1, 0, 1]]). - David Callan, Jun 07 2006
a(n) is the number of different contiguous palindromic bit patterns in the binary representation of n; for examples, for 5=101_2 the bit patterns are 0, 1, 101; for 7=111_2 the corresponding patterns are 1, 11, 111; for 13=1101_2 the patterns are 0, 1, 11, 101. - Hieronymus Fischer, Mar 13 2012
A103586(n) = a(n + a(n)); a(A214489(n)) = A103586(A214489(n)). - Reinhard Zumkeller, Jul 21 2012
Number of divisors of 2^n that are <= n. - Clark Kimberling, Apr 21 2019

Examples

			8 = 1000 in binary has length 4.
		

References

  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • L. Levine, Fractal sequences and restricted Nim, Ars Combin. 80 (2006), 113-127.

Crossrefs

A029837(n+1) gives the length of binary representation of n without the leading zeros (i.e., when zero is represented as the empty sequence). For n > 0 this is equal to a(n).
This is Guy Steele's sequence GS(4, 4) (see A135416).
Cf. A083652 (partial sums).

Programs

  • Haskell
    a070939 n = if n < 2 then 1 else a070939 (n `div` 2) + 1
    a070939_list = 1 : 1 : l [1] where
       l bs = bs' ++ l bs' where bs' = map (+ 1) (bs ++ bs)
    -- Reinhard Zumkeller, Jul 19 2012, Jun 07 2011
    
  • Magma
    A070939:=func< n | n eq 0 select 1 else #Intseq(n, 2) >; [ A070939(n): n in [0..104] ]; // Klaus Brockhaus, Jan 13 2011
    
  • Maple
    A070939 := n -> `if`(n=0, 1, ilog2(2*n)):
    seq(A070939(n), n=0..104); # revised by Peter Luschny, Aug 10 2017
  • Mathematica
    Table[Length[IntegerDigits[n, 2]], {n, 0, 50}] (* Stefan Steinerberger, Apr 01 2006 *)
    Join[{1},IntegerLength[Range[110],2]] (* Harvey P. Dale, Aug 18 2013 *)
    a[ n_] := If[ n < 1, Boole[n == 0], BitLength[n]]; (* Michael Somos, Jul 10 2018 *)
  • PARI
    {a(n) = if( n<1, n==0, #binary(n))} /* Michael Somos, Aug 31 2012 */
    
  • PARI
    apply( {A070939(n)=exponent(n+!n)+1}, [0..99]) \\ works for negative n and is much faster than the above. - M. F. Hasler, Jan 04 2014, updated Feb 29 2020
    
  • Python
    def a(n): return len(bin(n)[2:])
    print([a(n) for n in range(105)]) # Michael S. Branicky, Jan 01 2021
    
  • Python
    def A070939(n): return 1 if n == 0 else n.bit_length() # Chai Wah Wu, May 12 2022
  • Sage
    def A070939(n) : return (2*n).exact_log(2) if n != 0 else 1
    [A070939(n) for n in range(100)] # Peter Luschny, Aug 08 2012
    

Formula

a(0) = 1; for n >= 1, a(n) = 1 + floor(log_2(n)) = 1 + A000523(n).
G.f.: 1 + 1/(1-x) * Sum(k>=0, x^2^k). - Ralf Stephan, Apr 12 2002
a(0)=1, a(1)=1 and a(n) = 1+a(floor(n/2)). - Benoit Cloitre, Dec 02 2003
a(n) = A000120(n) + A023416(n). - Lekraj Beedassy, Apr 12 2006
a(2^m + k) = m + 1, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Mar 14 2017
a(n) = A113473(n) if n>0.

Extensions

a(4) corrected by Antti Karttunen, Feb 28 2003
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