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.

Previous Showing 11-20 of 118 results. Next

A005811 Number of runs in binary expansion of n (n>0); number of 1's in Gray code for n.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

Starting with a(1) = 0 mirror all initial 2^k segments and increase by one.
a(n) gives the net rotation (measured in right angles) after taking n steps along a dragon curve. - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
This sequence generates A082410: (0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...) and A014577; identical to the latter except starting 1, 1, 0, ...; by writing a "1" if a(n+1) > a(n); if not, write "0". E.g., A014577(2) = 0, since a(3) < a(2), or 1 < 2. - Gary W. Adamson, Sep 20 2003
Starting with 1 = partial sums of A034947: (1, 1, -1, 1, 1, -1, -1, 1, 1, 1, ...). - Gary W. Adamson, Jul 23 2008
The composer Per Nørgård's name is also written in the OEIS as Per Noergaard.
Can be used as a binomial transform operator: Let a(n) = the n-th term in any S(n); then extract 2^k strings, adding the terms. This results in the binomial transform of S(n). Say S(n) = 1, 3, 5, ...; then we obtain the strings: (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), ...; = the binomial transform of (1, 3, 5, ...) = (1, 4, 12, 32, 80, ...). Example: the 8-bit string has a sum of 32 with a distribution of (1, 3, 3, 1) or one 1, three 3's, three 5's, and one 7; as expected. - Gary W. Adamson, Jun 21 2012
Considers all positive odd numbers as nodes of a graph. Two nodes are connected if and only if the sum of the two corresponding odd numbers is a power of 2. Then a(n) is the distance between 2n + 1 and 1. - Jianing Song, Apr 20 2019

Examples

			Considered as a triangle with 2^k terms per row, the first few rows are:
  1
  2, 1
  2, 3, 2, 1
  2, 3, 4, 3, 2, 3, 2, 1
  ...
The n-th row becomes right half of next row; left half is mirrored terms of n-th row increased by one. - _Gary W. Adamson_, Jun 20 2012
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A037834 (-1), A088748 (+1), A246960 (mod 4), A034947 (first differences), A000975 (indices of record highs), A173318 (partial sums).
Partial sums of A112347. Recursion depth of A035327.

Programs

  • Haskell
    import Data.List (group)
    a005811 0 = 0
    a005811 n = length $ group $ a030308_row n
    a005811_list = 0 : f [1] where
       f (x:xs) = x : f (xs ++ [x + x `mod` 2, x + 1 - x `mod` 2])
    -- Reinhard Zumkeller, Feb 16 2013, Mar 07 2011
    
  • Maple
    A005811 := proc(n)
        local i, b, ans;
        if n = 0 then
            return 0 ;
        end if;
        ans := 1;
        b := convert(n, base, 2);
        for i from nops(b)-1 to 1 by -1 do
            if b[ i+1 ]<>b[ i ] then
                ans := ans+1
            fi
        od;
        return ans ;
    end proc:
    seq(A005811(i), i=1..50) ;
    # second Maple program:
    a:= n-> add(i, i=Bits[Split](Bits[Xor](n, iquo(n, 2)))):
    seq(a(n), n=0..100);  # Alois P. Heinz, Feb 01 2023
  • Mathematica
    Table[ Length[ Length/@Split[ IntegerDigits[ n, 2 ] ] ], {n, 1, 255} ]
    a[n_] := DigitCount[BitXor[n, Floor[n/2]], 2, 1]; Array[a, 100, 0] (* Amiram Eldar, Jul 11 2024 *)
  • PARI
    a(n)=sum(k=1,n,(-1)^((k/2^valuation(k,2)-1)/2))
    
  • PARI
    a(n)=if(n<1,0,a(n\2)+(a(n\2)+n)%2) \\ Benoit Cloitre, Jan 20 2014
    
  • PARI
    a(n) = hammingweight(bitxor(n, n>>1));  \\ Gheorghe Coserea, Sep 03 2015
    
  • Python
    def a(n): return bin(n^(n>>1))[2:].count("1") # Indranil Ghosh, Apr 29 2017

Formula

a(2^k + i) = a(2^k - i + 1) + 1 for k >= 0 and 0 < i <= 2^k. - Reinhard Zumkeller, Aug 14 2001
a(2n+1) = 2a(n) - a(2n) + 1, a(4n) = a(2n), a(4n+2) = 1 + a(2n+1).
a(j+1) = a(j) + (-1)^A014707(j). - Christopher Hendrie (hendrie(AT)acm.org), Sep 11 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^2^k/(1+x^2^(k+1)). - Ralf Stephan, May 02 2003
Delete the 0, make subsets of 2^n terms; and reverse the terms in each subset to generate A088696. - Gary W. Adamson, Oct 19 2003
a(0) = 0, a(2n) = a(n) + [n odd], a(2n+1) = a(n) + [n even]. - Ralf Stephan, Oct 20 2003
a(n) = Sum_{k=1..n} (-1)^((k/2^A007814(k)-1)/2) = Sum_{k=1..n} (-1)^A025480(k-1). - Ralf Stephan, Oct 29 2003
a(n) = A069010(n) + A033264(n). - Ralf Stephan, Oct 29 2003
a(0) = 0 then a(n) = a(floor(n/2)) + (a(floor(n/2)) + n) mod 2. - Benoit Cloitre, Jan 20 2014
a(n) = A037834(n) + 1.
a(n) = A000120(A003188(n)). - Amiram Eldar, Jul 11 2024

Extensions

Additional description from Wouter Meeussen

A005187 a(n) = a(floor(n/2)) + n; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1's in binary expansion of 2n.

Original entry on oeis.org

0, 1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, 22, 23, 25, 26, 31, 32, 34, 35, 38, 39, 41, 42, 46, 47, 49, 50, 53, 54, 56, 57, 63, 64, 66, 67, 70, 71, 73, 74, 78, 79, 81, 82, 85, 86, 88, 89, 94, 95, 97, 98, 101, 102, 104, 105, 109, 110, 112, 113, 116, 117, 119, 120, 127, 128
Offset: 0

Views

Author

N. J. A. Sloane, May 20 1991; Allan Wilks, Dec 11 1999

Keywords

Comments

Also exponent of the largest power of 2 dividing (2n)! (A010050) and (2n)!! (A000165).
Write n in binary: 1ab..yz, then a(n) = 1ab..yz + ... + 1ab + 1a + 1. - Ralf Stephan, Aug 27 2003
Also numbers having a partition into distinct Mersenne numbers > 0; A079559(a(n))=1; complement of A055938. - Reinhard Zumkeller, Mar 18 2009
Wikipedia's article on the "Whitney Immersion theorem" mentions that the a(n)-dimensional sphere arises in the Immersion Conjecture proved by Ralph Cohen in 1985. - Jonathan Vos Post, Jan 25 2010
For n > 0, denominators for consecutive pairs of integral numerator polynomials L(n+1,x) for the Legendre polynomials with o.g.f. 1 / sqrt(1-tx+x^2). - Tom Copeland, Feb 04 2016
a(n) is the total number of pointers in the first n elements of a perfect skip list. - Alois P. Heinz, Dec 14 2017
a(n) is the position of the n-th a (indexing from 0) in the fixed point of the morphism a -> aab, b -> b. - Jeffrey Shallit, Dec 24 2020
Numbers that can be expressed as the sum of distinct numbers of the form 2^k - 1 (lenient Mersenne numbers, A000225). This follows from the 2N - Hamming weight definition. A corollary is that these are the numbers with no 2 in their skew-binary representation (cf. A169683). - Allan C. Wechsler, Feb 25 2025

Examples

			G.f. = x + 3*x^2 + 4*x^3 + 7*x^4 + 8*x^5 + 10*x^6 + 11*x^7 + 15*x^8 + ...
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A001511 (first differences), A122247 (partial sums), A055938 (complement).

Programs

  • Haskell
    a005187 n = a005187_list !! n
    a005187_list = 0 : zipWith (+) [1..] (map (a005187 . (`div` 2)) [1..])
    -- Reinhard Zumkeller, Nov 07 2011, Oct 05 2011
    
  • Magma
    [n + Valuation(Factorial(n), 2): n in [0..70]]; // Vincenzo Librandi, Jun 11 2019
    
  • Maple
    A005187 := n -> 2*n - add(i, i=convert(n, base, 2)):
    seq(A005187(n), n=0..65); # Peter Luschny, Apr 08 2014
  • Mathematica
    a[0] = 0; a[n_] := a[n] = a[Floor[n/2]] + n; Table[ a[n], {n, 0, 50}] (* or *)
    Table[IntegerExponent[(2n)!, 2], {n, 0, 65}] (* Robert G. Wilson v, Apr 19 2006 *)
    Table[2n-DigitCount[2n,2,1],{n,0,70}] (* Harvey P. Dale, May 24 2014 *)
  • PARI
    {a(n) = if( n<0, 0, valuation((2*n)!, 2))}; /* Michael Somos, Oct 24 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, sum(k=1, n, (2*n)\2^k))}; /* Michael Somos, Oct 24 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, 2*n - subst( Pol( binary( n ) ), x, 1) ) }; /* Michael Somos, Aug 28 2007 */
    
  • PARI
    a(n)=my(s=n);while(n>>=1,s+=n);s \\ Charles R Greathouse IV, Apr 07 2012
    
  • PARI
    a(n)=2*n-hammingweight(n) \\ Charles R Greathouse IV, Jan 07 2013
    
  • Python
    def A005187(n): return 2*n-bin(n).count('1') # Chai Wah Wu, Jun 03 2021
  • Sage
    @CachedFunction
    def A005187(n): return A005187(n//2) + n if n > 0 else 0
    [A005187(n) for n in range(66)]  # Peter Luschny, Dec 13 2012
    

Formula

a(n) = A011371(2n+1) = A011371(n) + n, n >= 0.
A046161(n) = 2^a(n).
For m>0, let q = floor(log_2(m)); a(2m+1) = 2^q + 3m + Sum_{k>=1} floor((m-2^q)/2^k); a(2m) = a(2m+1) - 1. - Len Smiley
a(n) = Sum_{k >= 0} floor(n/2^k) = n + A011371(n). - Henry Bottomley, Jul 03 2001
G.f.: A(x) = Sum_{k>=0} x^(2^k)/((1-x)*(1-x^(2^k))). - Ralf Stephan, Dec 24 2002
a(n) = Sum_{k=1..n} A001511(k), sum of binary Hamming distances between consecutive integers up to n. - Gary W. Adamson, Jun 15 2003
Conjecture: a(n) = 2n + O(log(n)). - Benoit Cloitre, Oct 07 2003 [true as a(n) = 2*n - hamming_weight(2*n). Joerg Arndt, Jun 10 2019]
Sum_{n=2^k..2^(k+1)-1} a(n) = 3*4^k - (k+4)*2^(k-1) = A085354(k). - Philippe Deléham, Feb 19 2004
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence: a(n) = n + a(floor(n/2)); a(2n) = 2n + a(n); a(n*2^m) = 2*n*(2^m-1) + a(n).
a(2^m) = 2^(m+1) - 1, m >= 0.
Asymptotic behavior: a(n) = 2n + O(log(n)), a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= 2n-1; equality holds for powers of 2.
a(n) >= 2n-1-floor(log_2(n)); equality holds for n = 2^m-1, m > 0.
lim inf (2n - a(n)) = 1, for n-->oo.
lim sup (2n - log_2(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_2(n)) = 1, for n-->oo. (End)
a(n) = 2n - A000120(n). - Paul Barry, Oct 26 2007
PURRS demo results: Bounds for a(n) = n + a(n/2) with initial conditions a(1) = 1: a(n) >= -2 + 2*n - log(n)*log(2)^(-1), a(n) <= 1 + 2*n for each n >= 1. - Alexander R. Povolotsky, Apr 06 2008
If n = 2^a_1 + 2^a_2 + ... + 2^a_k, then a(n) = n-k. This can be used to prove that 2^n maximally divides (2n!)/n!. - Jon Perry, Jul 16 2009
a(n) = Sum_{k>=0} A030308(n,k)*A000225(k+1). - Philippe Deléham, Oct 16 2011
a(n) = log_2(denominator(binomial(-1/2,n))). - Peter Luschny, Nov 25 2011
a(2n+1) = a(2n) + 1. - M. F. Hasler, Jan 24 2015
a(n) = A004134(n) - n. - Cyril Damamme, Aug 04 2015
G.f.: (1/(1 - x))*Sum_{k>=0} (2^(k+1) - 1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017

A005836 Numbers whose base-3 representation contains no 2.

Original entry on oeis.org

0, 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39, 40, 81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121, 243, 244, 246, 247, 252, 253, 255, 256, 270, 271, 273, 274, 279, 280, 282, 283, 324, 325, 327, 328, 333, 334, 336, 337, 351, 352
Offset: 1

Views

Author

Keywords

Comments

3 does not divide binomial(2s, s) if and only if s is a member of this sequence, where binomial(2s, s) = A000984(s) are the central binomial coefficients.
This is the lexicographically earliest increasing sequence of nonnegative numbers that contains no arithmetic progression of length 3. - Robert Craigen (craigenr(AT)cc.umanitoba.ca), Jan 29 2001
In the notation of A185256 this is the Stanley Sequence S(0,1). - N. J. A. Sloane, Mar 19 2010
Complement of A074940. - Reinhard Zumkeller, Mar 23 2003
Sums of distinct powers of 3. - Ralf Stephan, Apr 27 2003
Numbers n such that central trinomial coefficient A002426(n) == 1 (mod 3). - Emeric Deutsch and Bruce E. Sagan, Dec 04 2003
A039966(a(n)+1) = 1; A104406(n) = number of terms <= n.
Subsequence of A125292; A125291(a(n)) = 1 for n>1. - Reinhard Zumkeller, Nov 26 2006
Also final value of n - 1 written in base 2 and then read in base 3 and with finally the result translated in base 10. - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n) modulo 2 is the Thue-Morse sequence A010060. - Dennis Tseng, Jul 16 2009
Also numbers such that the balanced ternary representation is the same as the base 3 representation. - Alonso del Arte, Feb 25 2011
Fixed point of the morphism: 0 -> 01; 1 -> 34; 2 -> 67; ...; n -> (3n)(3n+1), starting from a(1) = 0. - Philippe Deléham, Oct 22 2011
It appears that this sequence lists the values of n which satisfy the condition sum(binomial(n, k)^(2*j), k = 0..n) mod 3 <> 0, for any j, with offset 0. See Maple code. - Gary Detlefs, Nov 28 2011
Also, it follows from the above comment by Philippe Lallouet that the sequence must be generated by the rules: a(1) = 0, and if m is in the sequence then so are 3*m and 3*m + 1. - L. Edson Jeffery, Nov 20 2015
Add 1 to each term and we get A003278. - N. J. A. Sloane, Dec 01 2019

Examples

			12 is a term because 12 = 110_3.
This sequence regarded as a triangle with rows of lengths 1, 1, 2, 4, 8, 16, ...:
   0
   1
   3,  4
   9, 10, 12, 13
  27, 28, 30, 31, 36, 37, 39, 40
  81, 82, 84, 85, 90, 91, 93, 94, 108, 109, 111, 112, 117, 118, 120, 121
... - _Philippe Deléham_, Jun 06 2015
		

References

  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E10, pp. 317-323.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A039966 (characteristic function).
For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
Row 3 of array A104257.
Summary of increasing sequences avoiding arithmetic progressions of specified lengths (the second of each pair is obtained by adding 1 to the first):
3-term AP: A005836 (>=0), A003278 (>0);
4-term AP: A005839 (>=0), A005837 (>0);
5-term AP: A020654 (>=0), A020655 (>0);
6-term AP: A020656 (>=0), A005838 (>0);
7-term AP: A020657 (>=0), A020658 (>0);
8-term AP: A020659 (>=0), A020660 (>0);
9-term AP: A020661 (>=0), A020662 (>0);
10-term AP: A020663 (>=0), A020664 (>0).
See also A000452.

Programs

  • Haskell
    a005836 n = a005836_list !! (n-1)
    a005836_list = filter ((== 1) . a039966) [0..]
    -- Reinhard Zumkeller, Jun 09 2012, Sep 29 2011
    
  • Julia
    function a(n)
        m, r, b = n, 0, 1
        while m > 0
            m, q = divrem(m, 2)
            r += b * q
            b *= 3
        end
    r end; [a(n) for n in 0:57] |> println # Peter Luschny, Jan 03 2021
  • Maple
    t := (j, n) -> add(binomial(n,k)^j, k=0..n):
    for i from 1 to 400 do
        if(t(4,i) mod 3 <>0) then print(i) fi
    od; # Gary Detlefs, Nov 28 2011
    # alternative Maple program:
    a:= proc(n) option remember: local k, m:
    if n=1 then 0 elif n=2 then 1 elif n>2 then k:=floor(log[2](n-1)): m:=n-2^k: procname(m)+3^k: fi: end proc:
    seq(a(n), n=1.. 20); # Paul Weisenhorn, Mar 22 2020
    # third Maple program:
    a:= n-> `if`(n=1, 0, irem(n-1, 2, 'q')+3*a(q+1)):
    seq(a(n), n=1..100);  # Alois P. Heinz, Jan 26 2022
  • Mathematica
    Table[FromDigits[IntegerDigits[k, 2], 3], {k, 60}]
    Select[Range[0, 400], DigitCount[#, 3, 2] == 0 &] (* Harvey P. Dale, Jan 04 2012 *)
    Join[{0}, Accumulate[Table[(3^IntegerExponent[n, 2] + 1)/2, {n, 57}]]] (* IWABUCHI Yu(u)ki, Aug 01 2012 *)
    FromDigits[#,3]&/@Tuples[{0,1},7] (* Harvey P. Dale, May 10 2019 *)
  • PARI
    A=vector(100);for(n=2,#A,A[n]=if(n%2,3*A[n\2+1],A[n-1]+1));A \\ Charles R Greathouse IV, Jul 24 2012
    
  • PARI
    is(n)=while(n,if(n%3>1,return(0));n\=3);1 \\ Charles R Greathouse IV, Mar 07 2013
    
  • PARI
    a(n) = fromdigits(binary(n-1),3);  \\ Gheorghe Coserea, Jun 15 2018
    
  • Python
    def A005836(n):
        return int(format(n-1,'b'),3) # Chai Wah Wu, Jan 04 2015
    

Formula

a(n) = A005823(n)/2 = A003278(n)-1 = A033159(n)-2 = A033162(n)-3.
Numbers n such that the coefficient of x^n is > 0 in prod (k >= 0, 1 + x^(3^k)). - Benoit Cloitre, Jul 29 2003
a(n+1) = Sum_{k=0..m} b(k)* 3^k and n = Sum( b(k)* 2^k ).
a(2n+1) = 3a(n+1), a(2n+2) = a(2n+1) + 1, a(0) = 0.
a(n+1) = 3*a(floor(n/2)) + n - 2*floor(n/2). - Ralf Stephan, Apr 27 2003
G.f.: (x/(1-x)) * Sum_{k>=0} 3^k*x^2^k/(1+x^2^k). - Ralf Stephan, Apr 27 2003
a(n) = Sum_{k = 1..n-1} (1 + 3^A007814(k)) / 2. - Philippe Deléham, Jul 09 2005
From Reinhard Zumkeller, Mar 02 2008: (Start)
A081603(a(n)) = 0.
If the offset were changed to zero, then: a(0) = 0, a(n+1) = f(a(n)+1, a(n)+1) where f(x, y) = if x < 3 and x <> 2 then y else if x mod 3 = 2 then f(y+1, y+1) else f(floor(x/3), y). (End)
With offset a(0) = 0: a(n) = Sum_{k>=0} A030308(n,k)*3^k. - Philippe Deléham, Oct 15 2011
a(2^n) = A003462(n). - Philippe Deléham, Jun 06 2015
We have liminf_{n->infinity} a(n)/n^(log(3)/log(2)) = 1/2 and limsup_{n->infinity} a(n)/n^(log(3)/log(2)) = 1. - Gheorghe Coserea, Sep 13 2015
a(2^k+m) = a(m) + 3^k with 1 <= m <= 2^k and 1 <= k, a(1)=0, a(2)=1. - Paul Weisenhorn, Mar 22 2020
Sum_{n>=2} 1/a(n) = 2.682853110966175430853916904584699374821677091415714815171756609672281184705... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022
A065361(a(n)) = n-1. - Rémy Sigrist, Feb 06 2023
a(n) ≍ n^k, where k = log 3/log 2 = 1.5849625007. (I believe the constant varies from 1/2 to 1.) - Charles R Greathouse IV, Mar 29 2024

Extensions

Offset corrected by N. J. A. Sloane, Mar 02 2008
Edited by the Associate Editors of the OEIS, Apr 07 2009

A003188 Decimal equivalent of Gray code for n.

Original entry on oeis.org

0, 1, 3, 2, 6, 7, 5, 4, 12, 13, 15, 14, 10, 11, 9, 8, 24, 25, 27, 26, 30, 31, 29, 28, 20, 21, 23, 22, 18, 19, 17, 16, 48, 49, 51, 50, 54, 55, 53, 52, 60, 61, 63, 62, 58, 59, 57, 56, 40, 41, 43, 42, 46, 47, 45, 44, 36, 37, 39, 38, 34, 35, 33, 32, 96, 97, 99, 98, 102, 103, 101
Offset: 0

Views

Author

Keywords

Comments

Inverse of sequence A006068 considered as a permutation of the nonnegative integers, i.e., A006068(A003188(n)) = n = A003188(A006068(n)). - Howard A. Landman, Sep 25 2001
Restricts to a permutation of each {2^(i - 1) .. 2^i - 1}. - Jason Kimberley, Apr 02 2012
a(n) mod 2 = floor(((n + 1) mod 4) / 2), see also A021913. - Reinhard Zumkeller, Apr 28 2012
Invented by Emile Baudot (1845-1903), originally called a "cyclic-permuted" code. Gray codes are named after Frank Gray, who patented their use for shaft encoders in 1953. [F. Gray, "Pulse Code Communication", U.S. Patent 2,632,058, March 17, 1953.] - Robert G. Wilson v, Jun 22 2014
For n >= 2, let G_n be the graph whose vertices are labeled as 0,1,...,2^n-1, and two vertices are adjacent if and only if their binary expansions differ in exactly one bit, then a(0),a(1),...,a(2^n-1),a(0) is a Hamilton cycle in G_n. - Jianing Song, Jun 01 2022

Examples

			For n = 13, the binary reflected Gray code representation of n is '1011' and 1011_2 = 11_10. So, a(13) = 11. - _Indranil Ghosh_, Jan 23 2017
		

References

  • M. Gardner, Mathematical Games, Sci. Amer. Vol. 227 (No. 2, Feb. 1972), p. 107.
  • M. Gardner, Knotted Doughnuts and Other Mathematical Entertainments. Freeman, NY, 1986, p. 15.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(2*A003714(n)) = 3*A003714(n) for all n. - Antti Karttunen, Apr 26 1999
Cf. A014550 (in binary), A055975 (first differences), A048724 (even bisection), A065621 (odd bisection).

Programs

  • C
    int a(int n) { return n ^ (n>>1); }
    
  • Haskell
    import Data.Bits (xor, shiftR)
    a003188 n = n `xor` (shiftR n 1) :: Integer
    -- Reinhard Zumkeller, May 26 2013, Apr 28 2012
    
  • Magma
    // A recursive algorithm
    N := 10; s := [[]];
    for n in [1..N] do
    for j in [#s..1 by -1] do
       Append(~s,Append(s[j],1));
       Append(~s[j],0);
    end for;
    end for;
    [SequenceToInteger(b,2):b in s]; // Jason Kimberley, Apr 02 2012
    
  • Magma
    // A direct algorithm
    I2B := func< i | [b eq 1: b in IntegerToSequence(i,2)]>;
    B2I := func< s | SequenceToInteger([b select 1 else 0:b in s],2)>;
    [B2I(Xor(I2B(i),I2B(i div 2)cat[false])):i in [1..127]]; //Jason Kimberley, Apr 02 2012
    
  • Maple
    with(combinat); graycode(6); # to produce first 64 terms
    printf(cat(` %.6d`$64), op(map(convert, graycode(6), binary))); lprint(); # to produce binary strings
    # alternative:
    read("transforms"):
    A003188 := proc(n)
        XORnos(n,floor(n/2)) ;
    end proc: # R. J. Mathar, Mar 09 2015
    # another Maple program:
    a:= n-> Bits[Xor](n, iquo(n, 2)):
    seq(a(n), n=0..70);  # Alois P. Heinz, Aug 16 2020
  • Mathematica
    f[n_] := BitXor[n, Floor[n/2]]; Array[f, 70, 0] (* Robert G. Wilson v, Jun 09 2010 *)
  • PARI
    a(n)=bitxor(n,n>>1);
    
  • PARI
    a(n)=sum(k=1,n,(-1)^((k/2^valuation(k,2)-1)/2)*2^valuation(k,2))
    
  • Python
    def A003188(n):
        return int(bin(n^(n//2))[2:],2) # Indranil Ghosh, Jan 23 2017
    
  • Python
    def A003188(n): return n^ n>>1 # Chai Wah Wu, Jun 29 2022
    
  • R
    maxn <- 63 # by choice
    a <- 1
    for(n in 1:maxn){ a[2*n  ] <- 2*a[n] + (n%%2 != 0)
                      a[2*n+1] <- 2*a[n] + (n%%2 == 0)}
    (a <- c(0,a))
    # Yosu Yurramendi, Apr 10 2020
    (C#)
    static uint a(this uint n) => (n >> 1) ^ n; // Frank Hollstein, Mar 12 2021

Formula

a(n) = 2*a(floor(n/2)) + A021913(n - 1). - Henry Bottomley, Apr 05 2001
a(n) = n XOR floor(n/2), where XOR is the binary exclusive OR operator. - Paul D. Hanna, Jun 04 2002
G.f.: (1/(1-x)) * Sum_{k>=0} 2^k*x^2^k/(1 + x^2^(k+1)). - Ralf Stephan, May 06 2003
a(0) = 0, a(2n) = 2a(n) + [n odd], a(2n + 1) = 2a(n) + [n even]. - Ralf Stephan, Oct 20 2003
a(0) = 0, a(n) = 2 a(floor(n/2)) + mod(floor((n + 1)/2), 2).
a(n) = Sum_{k=1..n} 2^A007814(k) * (-1)^((k/2^A007814(k) - 1)/2). - Ralf Stephan, Oct 29 2003
a(0) = 0, a(n + 1) = a(n) XOR 2^A007814(n) - Jaume Simon Gispert (jaume(AT)nuem.com), Sep 11 2004
Inverse of sequence A006068. - Philippe Deléham, Apr 29 2005
a(n) = a(n-1) XOR A006519(n). - Franklin T. Adams-Watters, Jul 18 2011
From Mikhail Kurkov, Sep 27 2023: (Start)
a(2^m + k) = a(2^m - k - 1) + 2^m for 0 <= k < 2^m, m >= 0.
a(n) = a(A053645(A054429(n))) + A053644(n) for n > 0.
a(n) = A063946(a(A053645(n)) + A053644(n)) for n > 0. (End)

A054429 Simple self-inverse permutation of natural numbers: List each block of 2^n numbers (from 2^n to 2^(n+1) - 1) in reverse order.

Original entry on oeis.org

1, 3, 2, 7, 6, 5, 4, 15, 14, 13, 12, 11, 10, 9, 8, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 127, 126, 125, 124, 123, 122, 121
Offset: 1

Views

Author

Keywords

Comments

a(n) gives the position of the inverse of the n-th term in the full Stern-Brocot tree: A007305(a(n)+2) = A047679(n) and A047679(a(n)) = A007305(n+2). - Reinhard Zumkeller, Dec 22 2008
From Gary W. Adamson, Jun 21 2012: (Start)
The mapping and conversion rules are as follows:
By rows, we have ...
1;
3, 2;
7, 6, 5, 4;
15, 14, 13, 12, 11, 10, 9, 8;
... onto which we are to map one-half of the Stern-Brocot infinite Farey Tree:
1/2
1/3, 2/3
1/4, 2/5, 3/5, 3/4
1/5, 2/7, 3/8, 3/7, 4/7, 5/8, 5/7, 4/5
...
The conversion rules are: Convert the decimal to binary, adding a duplicate of the rightmost binary term to its right. For example, 10 = 1010, which becomes 10100. Then, from the left, record the number of runs = [1,1,1,2], the continued fraction representation of 5/8. Check: 10 decimal corresponds to 5/8 as shown in the overlaid mapping. Take decimal 9 = 1001 which becomes 10011, with a continued fraction representation of [1,2,2] = 5/7. Check: 9 decimal corresponds to 5/7 in the Farey Tree map. (End)
From Indranil Ghosh, Jan 19 2017: (Start)
a(n) is the value generated when n is converted into its Elias gamma code, the 1's and 0's are interchanged and the resultant is converted back to its decimal value for all values of n > 1. For n = 1, A054429(n) = 1 but after converting 1 to Elias gamma code, interchanging the 1's and 0's and converting it back to decimal, the result produced is 0.
For example, let n = 10. The Elias gamma code for 10 is '1110010'. After interchanging the 1's and 0's it becomes "0001101" and 1101_2 = 13_10. So a(10) = 13. (End)
From Yosu Yurramendi, Mar 09 2017 (similar to Zumkeller's comment): (Start)
A002487(a(n)) = A002487(n+1), A002487(a(n)+1) = A002487(n), n > 0.
A162909(a(n)) = A162910(n), A162910(a(n)) = A162909(n), n > 0.
A162911(a(n)) = A162912(n), A162912(a(n)) = A162911(n), n > 0.
A071766(a(n)) = A245326(n), A245326(a(n)) = A071766(n), n > 0.
A229742(a(n)) = A245325(n), A245325(a(n)) = A229742(n), n > 0.
A020651(a(n)) = A245327(n), A245327(a(n)) = A020651(n), n > 0.
A020650(a(n)) = A245328(n), A245328(a(n)) = A020650(n), n > 0. (End)
From Yosu Yurramendi, Mar 29 2017: (Start)
A063946(a(n)) = a(A063946(n)) = A117120(n), n > 0.
A065190(a(n)) = a(A065190(n)) = A092569(n), n > 0.
A258746(a(n)) = a(A258746(n)) = A165199(n), n > 0.
A258996(a(n)) = a(A258996(n)), n > 0.
A117120(a(n)) = a(A117120(n)), n > 0.
A092569(a(n)) = a(A092569(n)), n > 0. (End)

Crossrefs

See also A054424, A054430.
{A000027, A054429, A059893, A059894} form a 4-group.
This is Guy Steele's sequence GS(6, 5) (see A135416).

Programs

  • Haskell
    a054429 n = a054429_list !! (n-1)
    a054429_list = f [1..] where
       f xs@(x:_) = reverse us ++ f vs where (us, vs) = splitAt x xs
    -- Reinhard Zumkeller, Jun 01 2015, Feb 21 2014
    
  • Maple
    A054429 := n -> 3*2^ilog2(n) - n - 1:
    seq(A054429(n), n = 1..70); # [Updated by Peter Luschny, Apr 24 2024]
  • Mathematica
    Flatten[Table[Range[2^(n+1)-1,2^n,-1],{n,0,6}]] (* Harvey P. Dale, Dec 17 2013 *)
  • PARI
    A054429(n)= 3<<#binary(n\2)-n-1 \\ M. F. Hasler, Aug 18 2014
    
  • Python
    from itertools import count, islice
    def A054429_gen(): # generator of terms
        return (m for n in count(0) for m in range((1<A054429_list = list(islice(A054429_gen(),30)) # Chai Wah Wu, Jul 27 2023
  • R
    maxblock <- 10 # by choice
    a <- NULL
    for(m in 0:maxblock) a <- c(a, rev(2^m:(2^(m+1)-1)))
    a
    # Yosu Yurramendi, Mar 10 2017
    

Formula

a(n) = ReflectBinTreePermutation(n).
a(n) = if n=1 then 1 else 2*a(floor(n/2)) + 1 - n mod 2. - Reinhard Zumkeller, Feb 18 2003
G.f.: 1/(1-x) * ((x-2x^2)/(1-x) + Sum_{k>=0} 3*2^k*x^2^k). - Ralf Stephan, Sep 15 2003
A000120(a(n)) = A000120(A059894(n)) = A023416(n) + 1. - Ralf Stephan, Oct 05 2003
A115310(n, 1) = a(n). - Reinhard Zumkeller, Jan 20 2006
a(1) = 1, a(2^(m+1) + k) = a(2^m+k) + 2^(m+1),
a(2^(m+1) + 2^m+k) = a(2^m+k) + 2^m, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Apr 06 2017
a(n) = A117120(A063946(n)) = A063946(A117120(n)) = A092569(A065190(n)) = A065190(A092569(n)), n > 0. - Yosu Yurramendi, Apr 10 2017
a(n) = 3*A053644(n) - n - 1. - Alan Michael Gómez Calderón, Feb 28 2025

A016116 a(n) = 2^floor(n/2).

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64, 64, 128, 128, 256, 256, 512, 512, 1024, 1024, 2048, 2048, 4096, 4096, 8192, 8192, 16384, 16384, 32768, 32768, 65536, 65536, 131072, 131072, 262144, 262144, 524288, 524288, 1048576, 1048576, 2097152
Offset: 0

Views

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Powers of 2 doubled up. The usual OEIS policy is to omit the duplicates in such cases (when this would become A000079). This is an exception.
Number of symmetric compositions of n: e.g., 5 = 2+1+2 = 1+3+1 = 1+1+1+1+1 so a(5) = 4; 6 = 3+3 = 2+2+2 = 1+4+1 = 2+1+1+2 = 1+2+2+1 = 1+1+2+1+1 = 1+1+1+1+1+1 so a(6) = 8. - Henry Bottomley, Dec 10 2001
This sequence is the number of digits of each term of A061519. - Dmitry Kamenetsky, Jan 17 2009
Starting with offset 1 = binomial transform of [1, 1, -1, 3, -7, 17, -41, ...]; where A001333 = (1, 1, 3, 7, 17, 41, ...). - Gary W. Adamson, Mar 25 2009
a(n+1) is the number of symmetric subsets of [n]={1,2,...,n}. A subset S of [n] is symmetric if k is an element of S implies (n-k+1) is an element of S. - Dennis P. Walsh, Oct 27 2009
INVERT and inverse INVERT transforms give A006138, A039834(n-1).
The Kn21 sums, see A180662, of triangle A065941 equal the terms of this sequence. - Johannes W. Meijer, Aug 15 2011
First differences of A027383. - Jason Kimberley, Nov 01 2011
Run lengths in A079944. - Jeremy Gardiner, Nov 21 2011
Number of binary palindromes (A006995) between 2^(n-1) and 2^n (for n>1). - Hieronymus Fischer, Feb 17 2012
Pisano period lengths: 1, 1, 4, 1, 8, 4, 6, 1, 12, 8, 20, 4, 24, 6, 8, 1, 16, 12, 36, 8, ... . - R. J. Mathar, Aug 10 2012
Range of row n of the Circular Pascal array of order 4. - Shaun V. Ault, May 30 2014
a(n) is the number of permutations of length n avoiding both 213 and 312 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
Also, the decimal representation of the diagonal from the origin to the corner (and from the corner to the origin except for the initial term) of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 190", based on the 5-celled von Neumann neighborhood when initialized with a single black (ON) cell at stage zero. - Robert Price, May 10 2017
a(n + 1) + n - 1, n > 0, is the number of maximal subsemigroups of the monoid of partial order-preserving or -reversing mappings on a set with n elements. See the East et al. link. - James Mitchell and Wilf A. Wilson, Jul 21 2017
Number of symmetric stairs with n cells. A stair is a snake polyomino allowing only two directions for adjacent cells: east and north. See A005418. - Christian Barrientos, May 11 2018
For n >= 4, a(n) is the exponent of the group of the Gaussian integers in a reduced system modulo (1+i)^(n+2). See A302254. - Jianing Song, Jun 27 2018
a(n) is the number of length-(n+1) binary sequences, denoted , with s(1)=1 and with s(i+1)=s(i) for odd i. - Dennis P. Walsh, Sep 06 2018
a(n+1) is the number of subsets of {1,2,..,n} in which all differences between successive elements of subsets are even. For example, for n = 7, a(6) = 8 and the 8 subsets are {7}, {1,7}, {3,7}, {5,7}, {1,3,7}, {1,5,7}, {3,5,7}, {1,3,5,7}. For odd differences between elements see Comment in A000045 (Fibonacci numbers). - Enrique Navarrete, Jul 01 2020
Also, the number of walks of length n on the graph x--y--z, starting at x. - Sean A. Irvine, May 30 2025

Examples

			For n=5 the a(5)=4 symmetric subsets of [4] are {1,4}, {2,3}, {1,2,3,4} and the empty set. - _Dennis P. Walsh_, Oct 27 2009
For n=5 the a(5)=4 length-6 binary sequences are <1,1,0,0,0,0>, <1,1,0,0,1,1>, <1,1,1,1,0,0> and <1,1,1,1,1,1>. - _Dennis P. Walsh_, Sep 06 2018
		

Crossrefs

a(n) = A094718(3, n).
Cf. A001333.
See A052955 for partial sums (without the initial term).
A000079 gives the odd-indexed terms of a(n).
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Programs

Formula

a(n) = a(n-1)*a(n-2)/a(n-3) = 2*a(n-2) = 2^A004526(n).
G.f.: (1+x)/(1-2*x^2).
a(n) = (1/2 + sqrt(1/8))*sqrt(2)^n + (1/2 - sqrt(1/8))*(-sqrt(2))^n. - Ralf Stephan, Mar 11 2003
E.g.f.: cosh(sqrt(2)*x) + sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, Jul 16 2003
The signed sequence (-1)^n*2^floor(n/2) has a(n) = (sqrt(2))^n(1/2 - sqrt(2)/4) + (-sqrt(2))^n(1/2 + sqrt(2)/4). It is the inverse binomial transform of A000129(n-1). - Paul Barry, Apr 21 2004
Diagonal sums of A046854. a(n) = Sum_{k=0..n} binomial(floor(n/2), k). - Paul Barry, Jul 07 2004
a(n) = a(n-2) + 2^floor((n-2)/2). - Paul Barry, Jul 14 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(floor(n/2), floor(k/2)). - Paul Barry, Jul 15 2004
E.g.f.: cosh(asinh(1) + sqrt(2)*x)/sqrt(2). - Michael Somos, Feb 28 2005
a(n) = Sum_{k=0..n} A103633(n,k). - Philippe Deléham, Dec 03 2006
a(n) = 2^(n/2)*((1 + (-1)^n)/2 + (1-(-1)^n)/(2*sqrt(2))). - Paul Barry, Nov 12 2009
a(n) = 2^((2*n - 1 + (-1)^n)/4). - Luce ETIENNE, Sep 20 2014

A003602 Kimberling's paraphrases: if n = (2k-1)*2^m then a(n) = k.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 4, 1, 5, 3, 6, 2, 7, 4, 8, 1, 9, 5, 10, 3, 11, 6, 12, 2, 13, 7, 14, 4, 15, 8, 16, 1, 17, 9, 18, 5, 19, 10, 20, 3, 21, 11, 22, 6, 23, 12, 24, 2, 25, 13, 26, 7, 27, 14, 28, 4, 29, 15, 30, 8, 31, 16, 32, 1, 33, 17, 34, 9, 35, 18, 36, 5, 37, 19, 38, 10, 39, 20, 40, 3, 41, 21, 42
Offset: 1

Views

Author

Keywords

Comments

Fractal sequence obtained from powers of 2.
k occurs at (2*k-1)*A000079(m), m >= 0. - Robert G. Wilson v, May 23 2006
Sequence is T^(oo)(1) where T is acting on a word w = w(1)w(2)..w(m) as follows: T(w) = "1"w(1)"2"w(2)"3"(...)"m"w(m)"m+1". For instance T(ab) = 1a2b3. Thus T(1) = 112, T(T(1)) = 1121324, T(T(T(1))) = 112132415362748. - Benoit Cloitre, Mar 02 2009
Note that iterating the post-numbering operator U(w) = w(1) 1 w(2) 2 w(3) 3... produces the same limit sequence except with an additional "1" prepended, i.e., 1,1,1,2,1,3,2,4,... - Glen Whitney, Aug 30 2023
In the binary expansion of n, first swallow all zeros from the right, then add 1, and swallow the now-appearing 0 bit as well. - Ralf Stephan, Aug 22 2013
Although A264646 and this sequence initially agree in their digit-streams, they differ after 48 digits. - N. J. A. Sloane, Nov 20 2015
"[This is a] fractal because we get the same sequence after we delete from it the first appearance of all positive integers" - see Cobeli and Zaharescu link. - Robert G. Wilson v, Jun 03 2018
From Peter Munn, Jun 16 2022: (Start)
The sequence is the list of positive integers interleaved with the sequence itself. Provided the offset is suitable (which is the case here) a term of such a self-interleaved sequence is determined by the odd part of its index. Putting some of the formulas given here into words, a(n) is the position of the odd part of n in the list of odd numbers.
Applying the interleaving transform again, we get A110963.
(End)
Omitting all 1's leaves A131987 + 1. - David James Sycamore, Jul 26 2022
a(n) is also the smallest positive number not among the terms between a(a(n-1)) and a(n-1) inclusive (with a(0)=1 prepended). - Neal Gersh Tolunsky, Mar 07 2023

Examples

			From _Peter Munn_, Jun 14 2022: (Start)
Start of table showing the interleaving with the positive integers:
   n  a(n)  (n+1)/2  a(n/2)
   1    1      1
   2    1               1
   3    2      2
   4    1               1
   5    3      3
   6    2               2
   7    4      4
   8    1               1
   9    5      5
  10    3               3
  11    6      6
  12    2               2
(End)
		

References

  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) is the index of the column in A135764 where n appears (see also A054582).
Cf. A000079, A000265, A001511, A003603, A003961, A014577 (with offset 1, reduction mod 2), A025480, A035528, A048673, A101279, A110963, A117303, A126760, A181988, A220466, A249745, A253887, A337821 (2-adic valuation).
Cf. also A349134 (Dirichlet inverse), A349135 (sum with it), A349136 (Möbius transform), A349431, A349371 (inverse Möbius transform).
Cf. A264646.

Programs

  • Haskell
    a003602 = (`div` 2) . (+ 1) . a000265
    -- Reinhard Zumkeller, Feb 16 2012, Oct 14 2010
    
  • Haskell
    import Data.List (transpose)
    a003602 = flip div 2 . (+ 1) . a000265
    a003602_list = concat $ transpose [[1..], a003602_list]
    -- Reinhard Zumkeller, Aug 09 2013, May 23 2013
    
  • Maple
    A003602:=proc(n) options remember: if n mod 2 = 1 then RETURN((n+1)/2) else RETURN(procname(n/2)) fi: end proc:
    seq(A003602(n), n=1..83); # Pab Ter
    nmax := 83: for m from 0 to ceil(simplify(log[2](nmax))) do for k from 1 to ceil(nmax/(m+2)) do a((2*k-1)*2^m) := k od: od: seq(a(k), k=1..nmax); # Johannes W. Meijer, Feb 04 2013
    A003602 := proc(n)
        a := 1;
        for p in ifactors(n)[2] do
            if op(1,p) > 2 then
                a := a*op(1,p)^op(2,p) ;
            end if;
        end do  :
        (a+1)/2 ;
    end proc: # R. J. Mathar, May 19 2016
  • Mathematica
    a[n_] := Block[{m = n}, While[ EvenQ@m, m /= 2]; (m + 1)/2]; Array[a, 84] (* or *)
    a[1] = 1; a[n_] := a[n] = If[OddQ@n, (n + 1)/2, a[n/2]]; Array[a, 84] (* Robert G. Wilson v, May 23 2006 *)
    a[n_] := Ceiling[NestWhile[Floor[#/2] &, n, EvenQ]/2]; Array[a, 84] (* Birkas Gyorgy, Apr 05 2011 *)
    a003602 = {1}; max = 7; Do[b = {}; Do[AppendTo[b, {k, a003602[[k]]}], {k, Length[a003602]}]; a003602 = Flatten[b], {n, 2, max}]; a003602 (* L. Edson Jeffery, Nov 21 2015 *)
  • PARI
    A003602(n)=(n/2^valuation(n,2)+1)/2; /* Joerg Arndt, Apr 06 2011 */
    
  • Python
    import math
    def a(n): return (n/2**int(math.log(n - (n & n - 1), 2)) + 1)/2 # Indranil Ghosh, Apr 24 2017
    
  • Python
    def A003602(n): return (n>>(n&-n).bit_length())+1 # Chai Wah Wu, Jul 08 2022
  • Scheme
    (define (A003602 n) (let loop ((n n)) (if (even? n) (loop (/ n 2)) (/ (+ 1 n) 2)))) ;; Antti Karttunen, Feb 04 2015
    

Formula

a(n) = (A000265(n) + 1)/2.
a((2*k-1)*2^m) = k, for m >= 0 and k >= 1. - Robert G. Wilson v, May 23 2006
Inverse Weigh transform of A035528. - Christian G. Bower
G.f.: 1/x * Sum_{k>=0} x^2^k/(1-2*x^2^(k+1) + x^2^(k+2)). - Ralf Stephan, Jul 24 2003
a(2*n-1) = n and a(2*n) = a(n). - Pab Ter (pabrlos2(AT)yahoo.com), Oct 25 2005
a(A118413(n,k)) = A002024(n,k); = a(A118416(n,k)) = A002260(n,k); a(A014480(n)) = A001511(A014480(n)). - Reinhard Zumkeller, Apr 27 2006
Ordinal transform of A001511. - Franklin T. Adams-Watters, Aug 28 2006
a(n) = A249745(A126760(A003961(n))) = A249745(A253887(A048673(n))). That is, this sequence plays the same role for the numbers in array A135764 as A126760 does for the odd numbers in array A135765. - Antti Karttunen, Feb 04 2015 & Jan 19 2016
G.f. satisfies g(x) = g(x^2) + x/(1-x^2)^2. - Robert Israel, Apr 24 2015
a(n) = A181988(n)/A001511(n). - L. Edson Jeffery, Nov 21 2015
a(n) = A025480(n-1) + 1. - R. J. Mathar, May 19 2016
a(n) = A110963(2n-1) = A349135(4*n). - Antti Karttunen, Apr 18 2022
a(n) = (1 + n)/2, for n odd; a(n) = a(n/2), for n even. - David James Sycamore, Jul 28 2022
a(n) = n/2^A001511(n) + 1/2. - Alan Michael Gómez Calderón, Oct 06 2023
a(n) = A123390(A118319(n)). - Flávio V. Fernandes, Mar 02 2025

Extensions

More terms from Pab Ter (pabrlos2(AT)yahoo.com), Oct 25 2005

A011371 a(n) = n minus (number of 1's in binary expansion of n). Also highest power of 2 dividing n!.

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 4, 4, 7, 7, 8, 8, 10, 10, 11, 11, 15, 15, 16, 16, 18, 18, 19, 19, 22, 22, 23, 23, 25, 25, 26, 26, 31, 31, 32, 32, 34, 34, 35, 35, 38, 38, 39, 39, 41, 41, 42, 42, 46, 46, 47, 47, 49, 49, 50, 50, 53, 53, 54, 54, 56, 56, 57, 57, 63, 63, 64, 64, 66, 66, 67, 67, 70
Offset: 0

Views

Author

Keywords

Comments

Terms of A005187 repeated. - Lekraj Beedassy, Jul 06 2004
This sequence shows why in binary 0 and 1 are the only two numbers n such that n equals the sum of its digits raised to the consecutive powers (equivalent to the base-10 sequence A032799). 1 raised to any consecutive power is still 1 and thus any sum of digits raised to consecutive powers for any n > 1 falls short of equaling the value of n by the n-th term of this sequence. - Alonso del Arte, Jul 27 2004
Also the number of trailing zeros in the base-2 representation of n!. - Hieronymus Fischer, Jun 18 2007
Partial sums of A007814. - Philippe Deléham, Jun 21 2012
If n is in A089633 and n > 0, then a(n) = n - floor(log_2(n+1)). - Douglas Latimer, Jul 25 2012
For n > 1, denominators of integral numerator polynomials L(n,x) for the Legendre polynomials with o.g.f. 1/sqrt(1 - t*x + x^2). - Tom Copeland, Feb 04 2016
The definition of this sequence explains why, for n > 1, the highest power of 2 dividing n! added to the number of 1's in the binary expansion of n is equal to n. This result is due to the French mathematician Adrien Legendre (1752-1833) [see the Honsberger reference]. - Bernard Schott, Apr 07 2017
a(n) is the total number of 2's in the prime factorizations over the first n positive integers. The expected number of 2's in the factorization of an integer n is 1 (as n->infinity). Generally, the expected number of p's (for a prime p) is 1/(p-1). - Geoffrey Critzer, Jun 05 2017

Examples

			a(3) = 1 because 3 in binary is 11 (two 1's) and 3 - 2 = 1.
a(4) = 3 because 4 in binary is 100 (one 1 and two 0's) and 4 - 1 = 3.
a(5) = 3 because 5 in binary is 101 (a zero between two 1's) and 5 - 2 = 3.
a(100) = 97.
a(10^3) = 994.
a(10^4) = 9995.
a(10^5) = 99994.
a(10^6) = 999993.
a(10^7) = 9999992.
a(10^8) = 99999988.
a(10^9) = 999999987.
G.f. = x^2 + x^3 + 3*x^4 + 3*x^5 + 4*x^6 + 4*x^7 + 7*x^8 + 7*x^9 + 8*x^10 + ...
		

References

  • K. Atanassov, On Some of Smarandache's Problems, section 7, on the 61st problem, page 42, American Research Press, 1999, 16-21.
  • G. Bachman, Introduction to p-Adic Numbers and Valuation Theory, Academic Press, 1964; see Lemma 3.1.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 305.
  • H. Davenport, The Higher Arithmetic, 7th ed. 1999, Cambridge University Press, p. 216, exercise 1.07.
  • R. Honsberger, Mathematical Gems II, Dolciani Mathematical Expositions, 1976, pp. 1-6.

Crossrefs

a(n) = Sum_{k=1..n} A007814(k), n >= 1, a(0) = 0.

Programs

  • Haskell
    a011371 n = n - a000120 n  -- Reinhard Zumkeller, Jan 24 2014
    
  • Magma
    [Valuation(Factorial(n), 2): n in [0..80]]; // Bruno Berselli, Aug 05 2013
    
  • Maple
    A011371(n) = RETURN(((2^(l))-1)+sum('(j*floor((n-(2^l)+2^j)/(2^(j+1))))','j'=1..l)); # after K. Atanassov. Here l is [ log2(n) ].
    A011371 := n -> n - add(i,i=convert(n,base,2)): # Peter Luschny, May 02 2009
    read("transforms") : A011371 := proc(n) n-wt(n) ; end proc: # R. J. Mathar, May 15 2013
  • Mathematica
    -1 + Length[ Last[ Split[ IntegerDigits[ 2(n!), 2 ] ] ] ], FoldList[ Plus, 0, Fold[ Flatten[ {#1, #2, #1} ]&, 0, Range[ 6 ] ] ]
    Table[IntegerExponent[n!, 2], {n, 0, 127}]
    Table[n - DigitCount[n, 2, 1], {n, 0, 127}]
    Table[t = 0; p = 2; While[s = Floor[n/p]; t = t + s; s > 0, p *= 2]; t, {n, 0, 100} ]
  • PARI
    {a(n) = if( n<0, 0, valuation(n!, 2))}; /* Michael Somos, Oct 24 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, sum(k=1, n, n\2^k))}; /* Michael Somos, Oct 24 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, n - subst( Pol( binary( n ) ), x, 1))}; /* Michael Somos, Aug 28 2007 */
    
  • PARI
    a(n)=sum(k=1,log(n+1)\log(2),n>>k) \\ Charles R Greathouse IV, Oct 03 2012
    
  • PARI
    a(n)=my(s);while(n>>=1,s+=n);s \\ Charles R Greathouse IV, Aug 09 2013
    
  • PARI
    a(n) = n - hammingweight(n); \\ Michel Marcus, Jun 05 2014
    
  • Python
    [n - bin(n)[2:].count("1") for n in range(101)] # Indranil Ghosh, Apr 09 2017
    
  • Python
    # 3.10+
    def A011371(n): return n-n.bit_count() # Chai Wah Wu, Jul 09 2022

Formula

a(n) = a(floor(n/2)) + floor(n/2) = floor(n/2) + floor(n/4) + floor(n/8) + floor(n/16) + ... - Henry Bottomley, Apr 24 2001
G.f.: A(x) = (1/(1 - x))*Sum_{k>=1} x^(2^k)/(1 - x^(2^k)). - Ralf Stephan, Apr 11 2002
a(n) = n - A000120(n). - Lekraj Beedassy, Sep 01 2003
a(n) = A005187(n) - n, n >= 0.
a(n) = A007814(A000142(n)). - Reinhard Zumkeller, Apr 09 2004
From Hieronymus Fischer, Jun 25 and Aug 13 2007: (Start)
a(n) = Sum_{k=2..n} Sum_{j|k, j >= 2} (floor(log_2(j)) - floor(log_2(j - 1))).
The g.f. can be expressed in terms of a Lambert series, in that 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 is a power of 2, otherwise b(k) = 0.
G.f.: g(x) = (1/(1-x))*Sum_{k>0} c(k)*x^k, where c(k) = Sum_{j>1, j|k} (floor(log_2(j)) - floor(log_2(j-1))).
Recurrence:
a(n) = floor(n/2) + a(floor(n/2));
a(2*n) = n + a(n);
a(n*2^m) = n*(2^m - 1) + a(n).
a(2^m) = 2^m - 1, m >= 0.
Asymptotic behavior:
a(n) = n + O(log(n)),
a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
a(n) <= n - 1; equality holds for powers of 2.
a(n) >= n - 1 - floor(log_2(n)); equality holds for n = 2^m - 1, m > 0.
lim inf (n - a(n)) = 1, for n->oo.
lim sup (n - log_2(n) - a(n)) = 0, for n->oo.
lim sup (a(n+1) - a(n) - log_2(n)) = 0, for n->oo. (End)
a(n) = Sum_{k >= 0} A030308(n, k)*A000225(k). - Philippe Deléham, Oct 16 2011
a(n) = Sum_{k=0..floor(log_2(n+1))} f^(k+1)(n), where f(n) = (n - (n mod 2))/2 and f^(k+1) is the (k+1)-th composition of f. - Joseph Wheat, Mar 01 2018
a(n) = Sum_{k=1..floor(n/2)} floor(log_2(n/k)). - Ammar Khatab, Feb 01 2025

Extensions

Examples added by Hieronymus Fischer, Jun 06 2012

A006257 Josephus problem: a(2*n) = 2*a(n)-1, a(2*n+1) = 2*a(n)+1.

Original entry on oeis.org

0, 1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29
Offset: 0

Views

Author

Keywords

Comments

Write the numbers 1 through n in a circle, start at 1 and cross off every other number until only one number is left.
A version of the children's game "One potato, two potato, ...".
a(n)/A062383(n) = (0, 0.1, 0.01, 0.11, 0.001, ...) enumerates all binary fractions in the unit interval [0, 1). - Fredrik Johansson, Aug 14 2006
Iterating a(n), a(a(n)), ... eventually leads to 2^A000120(n) - 1. - Franklin T. Adams-Watters, Apr 09 2010
By inspection, the solution to the Josephus Problem is a sequence of odd numbers (from 1) starting at each power of 2. This yields a direct closed form expression (see formula below). - Gregory Pat Scandalis, Oct 15 2013
Also zero together with a triangle read by rows in which row n lists the first 2^(n-1) odd numbers (see A005408), n >= 1. Row lengths give A011782. Right border gives A000225. Row sums give A000302, n >= 1. See example. - Omar E. Pol, Oct 16 2013
For n > 0: a(n) = n + 1 - A080079(n). - Reinhard Zumkeller, Apr 14 2014
In binary, a(n) = ROL(n), where ROL = rotate left = remove the leftmost digit and append it to the right. For example, n = 41 = 101001_2 => a(n) = (0)10011_2 = 19. This also explains FTAW's comment above. - M. F. Hasler, Nov 02 2016
In the under-down Australian card deck separation: top card on bottom of a deck of n cards, next card separated on the table, etc., until one card is left. The position a(n), for n >= 1, from top will be the left over card. See, e.g., the Behrends reference, pp. 156-164. For the down-under case see 2*A053645(n), for n >= 3, n not a power of 2. If n >= 2 is a power of 2 the botton card survives. - Wolfdieter Lang, Jul 28 2020

Examples

			From _Omar E. Pol_, Jun 09 2009: (Start)
Written as an irregular triangle the sequence begins:
  0;
  1;
  1,3;
  1,3,5,7;
  1,3,5,7,9,11,13,15;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31;
  1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,
   43,45,47,49,51,53,55,57,59,61,63;
...
(End)
From _Omar E. Pol_, Nov 03 2018: (Start)
An illustration of initial terms, where a(n) is the area (or number of cells) in the n-th region of the structure:
   n   a(n)       Diagram
   0    0     _
   1    1    |_|_ _
   2    1      |_| |
   3    3      |_ _|_ _ _ _
   4    1          |_| | | |
   5    3          |_ _| | |
   6    5          |_ _ _| |
   7    7          |_ _ _ _|
(End)
		

References

  • Erhard Behrends, Der mathematische Zauberstab, Rowolth Taschenbuch Verlag, rororo 62902, 4. Auflage, 2019, pp. 156-164. [English version: The Math Behind the Magic, AMS, 2019.]
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 10.
  • M. S. Petković, "Josephus problem", Famous Puzzles of Great Mathematicians, page 179, Amer. Math. Soc. (AMS), 2009.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Paul Weisenhorn, Josephus und seine Folgen, MNU, 59(2006), pp. 18-19.

Crossrefs

Second column, and main diagonal, of triangle A032434.
Cf. A181281 (with s=5), A054995 (with s=3).
Column k=2 of A360099.

Programs

  • Coq
    Require Import ZArith.
    Fixpoint a (n : positive) : Z :=
    match n with
    | xH => 1
    | xI n' => (2*(a n') + 1)%Z
    | xO n' => (2*(a n') - 1)%Z
    end.
    (* Stefan Haan, Aug 27 2023 *)
  • Haskell
    a006257 n = a006257_list !! n
    a006257_list =
       0 : 1 : (map (+ 1) $ zipWith mod (map (+ 1) $ tail a006257_list) [2..])
    -- Reinhard Zumkeller, Oct 06 2011
    
  • Magma
    [0] cat [2*(n-2^Floor(Log(2,n)))+1: n in [1..100]]; // Vincenzo Librandi, Jan 14 2016
    
  • Maple
    a(0):=0: for n from 1 to 100 do a(n):=(a(n-1)+1) mod n +1: end do:
    seq(a(i),i=0..100); # Paul Weisenhorn, Oct 10 2010; corrected by Robert Israel, Jan 13 2016
    A006257 := proc(n)
        convert(n,base,2) ;
        ListTools[Rotate](%,-1) ;
        add( op(i,%)*2^(i-1),i=1..nops(%)) ;
    end proc: # R. J. Mathar, May 20 2016
    A006257 := n -> 2*n  - Bits:-Iff(n, n):
    seq(A006257(n), n=0..78); # Peter Luschny, Sep 24 2019
  • Mathematica
    Table[ FromDigits[ RotateLeft[ IntegerDigits[n, 2]], 2], {n, 0, 80}] (* Robert G. Wilson v, Sep 21 2003 *)
    Flatten@Table[Range[1, 2^n - 1, 2], {n, 0, 5}] (* Birkas Gyorgy, Feb 07 2011 *)
    m = 5; Range[2^m - 1] + 1 - Flatten@Table[Reverse@Range[2^n], {n, 0, m - 1}] (* Birkas Gyorgy, Feb 07 2011 *)
  • PARI
    a(n)=sum(k=1,n,if(bitxor(n,k)Paul D. Hanna
    
  • PARI
    a(n)=if(n, 2*n-2^logint(2*n,2)+1, 0) \\ Charles R Greathouse IV, Oct 29 2016
    
  • Python
    import math
    def A006257(n):
         return 0 if n==0 else 2*(n-2**int(math.log(n,2)))+1 # Indranil Ghosh, Jan 11 2017
    
  • Python
    def A006257(n): return bool(n&(m:=1<Chai Wah Wu, Jan 22 2023
    (C#)
    static long cs_A006257(this long n) => n == 0 ? 0 : 1 + (1 + (n - 1).cs_A006257()) % n; // Frank Hollstein, Feb 24 2021
    

Formula

To get a(n), write n in binary, rotate left 1 place.
a(n) = 2*A053645(n) + 1 = 2(n-msb(n))+1. - Marc LeBrun, Jul 11 2001. [Here "msb" = "most significant bit", A053644.]
G.f.: 1 + 2/(1-x) * ((3*x-1)/(2-2*x) - Sum_{k>=1} 2^(k-1)*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = number of positive integers k < n such that n XOR k < n. a(n) = n - A035327(n). - Paul D. Hanna, Jan 21 2006
a(n) = n for n = 2^k - 1. - Zak Seidov, Dec 14 2006
a(n) = n - A035327(n). - K. Spage, Oct 22 2009
a(2^m+k) = 1+2*k; with 0 <= m and 0 <= k < 2^m; n = 2^m+k; m = floor(log_2(n)); k = n-2^m; a(n) = ((a(n-1)+1) mod n) + 1; a(1) = 1. E.g., n=27; m=4; k=11; a(27) = 1 + 2*11 = 23. - Paul Weisenhorn, Oct 10 2010
a(n) = 2*(n - 2^floor(log_2(n))) + 1 (see comment above). - Gregory Pat Scandalis, Oct 15 2013
a(n) = 0 if n = 0 and a(n) = 2*a(floor(n/2)) - (-1)^(n mod 2) if n > 0. - Marek A. Suchenek, Mar 31 2016
G.f. A(x) satisfies: A(x) = 2*A(x^2)*(1 + x) + x/(1 + x). - Ilya Gutkovskiy, Aug 31 2019
For n > 0: a(n) = 2 * A062050(n) - 1. - Frank Hollstein, Oct 25 2021

Extensions

More terms from Robert G. Wilson v, Sep 21 2003

A053644 Most significant bit of n, msb(n); largest power of 2 less than or equal to n; write n in binary and change all but the first digit to zero.

Original entry on oeis.org

0, 1, 2, 2, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64, 64
Offset: 0

Views

Author

Henry Bottomley, Mar 22 2000

Keywords

Comments

Except for the initial term, 2^n appears 2^n times. - Lekraj Beedassy, May 26 2005
a(n) is the smallest k such that row k in triangle A265705 contains n. - Reinhard Zumkeller, Dec 17 2015
a(n) is the sum of totient function over powers of 2 <= n. - Anthony Browne, Jun 17 2016
Given positive n, reverse the bits of n and divide by 2^floor(log_2 n). Numerators are in A030101. Ignoring the initial 0, denominators are in this sequence. - Alonso del Arte, Feb 11 2020

Crossrefs

See A000035 for least significant bit(n).
MASKTRANS transform of A055975 (prepended with 0), MASKTRANSi transform of A048678.
Bisection of A065267, A065279, A065291, A072376.
First differences of A063915. Cf. A076877, A073121.
This is Guy Steele's sequence GS(5, 5) (see A135416).
Equals for n >= 1 the first right hand column of A160464. - Johannes W. Meijer, May 24 2009
Diagonal of A088370. - Alois P. Heinz, Oct 28 2011

Programs

  • Haskell
    a053644 n = if n <= 1 then n else 2 * a053644 (div n 2)
    -- Reinhard Zumkeller, Aug 28 2014
    a053644_list = 0 : concat (iterate (\zs -> map (* 2) (zs ++ zs)) [1])
    -- Reinhard Zumkeller, Dec 08 2012, Oct 21 2011, Oct 17 2010
    
  • Magma
    [0] cat [2^Ilog2(n): n in [1..90]]; // Vincenzo Librandi, Dec 11 2018
    
  • Maple
    a:= n-> 2^ilog2(n):
    seq(a(n), n=0..80);  # Alois P. Heinz, Dec 20 2016
  • Mathematica
    A053644[n_] := 2^(Length[ IntegerDigits[n, 2]] - 1); A053644[0] = 0; Table[A053644[n], {n, 0, 74}] (* Jean-François Alcover, Dec 01 2011 *)
    nv[n_] := Module[{c = 2^n}, Table[c, {c}]]; Join[{0}, Flatten[Array[nv, 7, 0]]] (* Harvey P. Dale, Jul 17 2012 *)
  • PARI
    a(n)=my(k=1);while(k<=n,k<<=1);k>>1 \\ Charles R Greathouse IV, May 27 2011
    
  • PARI
    a(n) = if(!n, 0, 2^exponent(n)) \\ Iain Fox, Dec 10 2018
    
  • Python
    def a(n): return 0 if n==0 else 2**(len(bin(n)[2:]) - 1) # Indranil Ghosh, May 25 2017
    
  • Python
    def A053644(n): return 1<Chai Wah Wu, Jul 27 2022
  • Scala
    (0 to 127).map(Integer.highestOneBit()) // _Alonso del Arte, Feb 26 2020
    

Formula

a(n) = a(floor(n / 2)) * 2.
a(n) = 2^A000523(n).
From n >= 1 onward, A053644(n) = A062383(n)/2.
a(0) = 0, a(1) = 1 and a(n+1) = a(n)*floor(n/a(n)). - Benoit Cloitre, Aug 17 2002
G.f.: 1/(1 - x) * (x + Sum_{k >= 1} 2^(k - 1)*x^2^k). - Ralf Stephan, Apr 18 2003
a(n) = (A003817(n) + 1)/2 = A091940(n) + 1. - Reinhard Zumkeller, Feb 15 2004
a(n) = Sum_{k = 1..n} (floor(2^k/k) - floor((2^k - 1)/k))*A000010(k). - Anthony Browne, Jun 17 2016
a(2^m+k) = 2^m, m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Aug 07 2016
Previous Showing 11-20 of 118 results. Next