A033264 -id:A033264 - cp's OEIS frontend

A366600 a(n) = (1 + A033264(n))*a(A053645(n)) for n > 0 with a(0) = 1.

1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 6, 2, 4, 4, 8, 1, 2, 2, 6, 2, 6, 6, 12, 2, 4, 4, 18, 4, 8, 8, 16, 1, 2, 2, 6, 2, 6, 6, 12, 2, 6, 6, 24, 6, 12, 12, 24, 2, 4, 4, 18, 4, 18, 18, 36, 4, 8, 8, 54, 8, 16, 16, 32, 1, 2, 2, 6, 2, 6, 6, 12, 2, 6, 6, 24, 6, 12, 12, 24, 2

Offset: 0

Mikhail Kurkov, Oct 14 2023

			a(6) = 4 because the binary expansion of 6 is 110 and we have [(10), 1(10)] -> [1, 1]. Increasing these values by 1 gives us 2*2 = 4.
a(18) = 6 because the binary expansion of 18 is 10010 and we have [(10), (10)0(10)] -> [1, 2]. Increasing these values by 1 gives us 2*3 = 6.
a(26) = 18 because the binary expansion of 26 is 11010 and we have [(10), (10)(10), 1(10)(10)] -> [1, 2, 2]. Increasing these values by 1 gives us 2*3*3 = 18.
For n=482, the bits of n and the resulting product for a(n) are
  n    = 482 = binary 1 1 1 1 0 0 0 1 0
  a(n) = 162 =        3*3*3*3      *2
n=3863 = binary 111100010111 is the same a(n) = 162 since its final trailing "111" has no effect.

Cf. A033264, A053645, A346422.

A033264[n_] := SequenceCount[IntegerDigits[n, 2], {1, 0}];
A053645[n_] := n - 2^Floor@Log2@n;
a[n_] := a[n] = If[n == 0, 1, (1 + A033264[n]) a[A053645[n]]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Nov 14 2023 *)

a(n) = my(A = 1, B = 1); if(n, for(i=1, logint(n, 2), if(bittest(n, i), A *= (B += !bittest(n, i-1))))); A

a(2n + 1) = a(n).
a(4n) = a(2n) with a(0) = 1.
a(4n + 2) = 2*b(n), b(2n + 1) = 2*b(n), b(2n) = 3*c(n - 1, 1) with b(0) = 1.
c(2n + 1,  k) = c(n, k), c(4n + 2, k) = (k + 2)*c(2n, k), c(4n, k) = (k + 3)*c(n - 1, k + 1) with c(0, k) = 1.
Another way to compute a(4n + 2):
a(2*(4^n - 1)/3) = (n + 1)!.
a(2^(2m)*(2k + 1) + 2*(4^m - 1)/3) = (m + 1)*a(2^(2m)*k + 2*(4^m - 1)/3).
a(2^(2m + 1)*(2k + 1) + 2*(4^(m + 1) - 1)/3) = a(2^(2m + 1)*k + 2*(4^(m + 1) - 1)/3).
Note that a(4n + 2) is completely defined by these 3 last formulas. However, it looks like that it is not so easy to identify m and k for a given n, which makes these formulas useless for computing this sequence.

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

N. J. A. Sloane, Jeffrey Shallit, Simon Plouffe

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

			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

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

T. D. Noe, Table of n, a(n) for n = 0..10000
Joerg Arndt, Matters Computational (The Fxtbook).
J.-P. Allouche, G.-N. Han and J. Shallit, On some conjectures of P. Barry, arXiv:2006.08909 [math.NT], 2020.
J.-P. Allouche and J. Shallit, The Ring of k-regular Sequences, II.
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
Danielle Cox and Karyn McLellan, A problem on generation sets containing Fibonacci numbers, Fib. Quart., 55 (No. 2, 2017), 105-113.
Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves -- I and II, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. Reprinted with addendum in Donald E. Knuth, Selected Papers on Fun and Games, 2010, pages 571-614. Equation 3.2 g(n) = a(n-1).
Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. [Cached copy, with permission]
P. Flajolet et al., Mellin Transforms And Asymptotics: Digital Sums, Theoret. Computer Sci. 23 (1994), 291-314.
P. Flajolet and Lyle Ramshaw, A note on Gray code and odd-even merge, SIAM J. Comput. 9 (1980), 142-158.
Sara Kropf and Stephan Wagner, q-Quasiadditive functions, arXiv:1605.03654 [math.CO], 2016.
Sara Kropf and S. Wagner, On q-Quasiadditive and q-Quasimultiplicative Functions, arXiv preprint arXiv:1608.03700 [math.CO], 2016.
Shuo Li, Palindromic length sequence of the ruler sequence and of the period-doubling sequence, arXiv:2007.08317 [math.CO], 2020.
Helmut Prodinger and Friedrich J. Urbanek, Infinite 0-1-Sequences Without Long Adjacent Identical Blocks, Discrete Mathematics, volume 28, issue 3, 1979, pages 277-289. Also first author's copy. Their "variation" v(k) at definition 3.4 is a(k).
Jeffrey Shallit, The mathematics of Per Noergaard's rhythmic infinity system, Fib. Q., 43 (2005), 262-268.
Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions, 2004.
Ralf Stephan, Table of generating functions.
Index entries for "core" sequences.
Index entries for sequences related to binary expansion of n.

Cf. A037834 (-1), A088748 (+1), A246960 (mod 4), A034947 (first differences), A000975 (indices of record highs), A173318 (partial sums).
Cf. A056539, A014707, A014577, A082410, A030308, A090079, A044813, A165413, A226227, A226228, A226229.
Partial sums of A112347. Recursion depth of A035327.
Cf. A000120, A003188.

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

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

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 *)

a(n)=sum(k=1,n,(-1)^((k/2^valuation(k,2)-1)/2))

a(n)=if(n<1,0,a(n\2)+(a(n\2)+n)%2) \\ Benoit Cloitre, Jan 20 2014

a(n) = hammingweight(bitxor(n, n>>1));  \\ Gheorghe Coserea, Sep 03 2015

def a(n): return bin(n^(n>>1))[2:].count("1") # Indranil Ghosh, Apr 29 2017

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

Additional description from Wouter Meeussen

A069010 Number of runs of 1's in the binary representation of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 3, 3, 2, 3, 2, 2, 1, 2, 2, 2, 2, 3, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 3, 3, 2, 3, 2, 2, 2, 3, 3, 3, 3, 4, 3, 3, 2, 3, 3, 3, 2, 3

Offset: 0

Henry Bottomley, Apr 02 2002

a(n) is also the number of distinct parts in 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 24 2017

Positions of first occurrences of k are A002450(k). - John Keith, Aug 30 2021

			a(11) = 2 since 11 is 1011 in binary with two runs of 1's.
a(12) = 1 since 12 is 1100 in binary with one run of 1's.

Tilman Piesk (terms 0..9999) & Antti Karttunen, Table of n, a(n) for n = 0..16384
Louis Marin, Counting Polyominoes in a Rectangle b X h, arXiv:2406.16413 [cs.DM], 2024. See p. 148.
Vladimir Shevelev, Two analogs of Thue-Morse sequence, arXiv:1603.04434 [math.NT], 2016.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions

Cf. A000035, A000120, A005811, A014081, A033264, A037800.
Cf. A268411 (parity of the terms), A268412 (positions of even terms), A268415 (of odd terms).
Cf. A002450 (positions of record highs).
Cf. also A227349, A246588.

f:= proc(n) option remember; if n::even then procname(n/2)
elif n mod 4 = 1 then 1 + procname((n-1)/2) else  procname((n-1)/2) fi end proc:
f(0):= 0:
map(f, [$0..1000]); # Robert Israel, Sep 06 2015

Count[Split@ IntegerDigits[#, 2], n_ /; First@ n == 1] & /@ Range[0, 120] (* Michael De Vlieger, Sep 05 2015 *)

a(n) = (1 + (hammingweight(bitxor(n, n>>1)))) >> 1;  \\ Gheorghe Coserea, Sep 05 2015

def A069010(n):
    return sum(1 for d in bin(n)[2:].split('0') if len(d)) # Chai Wah Wu, Nov 04 2016

(define (A069010 n) (/ (+ (A005811 n) (A000035 n)) 2)) ;; Antti Karttunen, Feb 05 2016

a(n) = ceiling(A005811(n)/2) = A005811(n) - A033264(n). If 2^k <= n < 3*2^(k-1) then a(n) = a(n-2^k)+1; if 3*2^(k-1) <= n < 2^(k+1) then a(n) = a(n-2^k).
a(2n) = a(n), a(2n+1) = a(n) + [n is even]. - Ralf Stephan, Aug 20 2003
G.f.: (1/(1-x)) * Sum_{k>=0} (t/(1+t))/(1+t^2), where t=x^2^k. - Ralf Stephan, Sep 07 2003
a(n) = A000120(n) - A014081(n) = A037800(n) + 1, n>0. - Ralf Stephan, Sep 10 2003

A297330 Total variation of base-10 digits of n; see Comments.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 8, 7, 6, 5, 4, 3, 2

Offset: 1

Clark Kimberling, Jan 17 2018

Suppose that a number n has base-b digits b(m), b(m-1), ..., b(0).  The base-b down-variation of n is the sum DV(n,b) of all d(i)-d(i-1) for which d(i) > d(i-1); the base-b up-variation of n is the sum UV(n,b) of all d(k-1)-d(k) for which d(k) < d(k-1).  The total base-b variation of n is the sum TV(n,b) = DV(n,b) + UV(n,b).  Guide to related sequences and partitions of the natural numbers:
***
Base b     {DV(n,b)}   {UV(n,b)}    {TV(n,b)}
         A033264     A037800      A037834
         A037853     A037844      A037835
         A037854     A037845      A037836
         A037855     A037846      A037837
         A037856     A037847      A037838
         A037857     A037848      A037839
         A037858     A037849      A037840
         A037859     A037850      A037841
        A037860     A037851      A297330
        A297231     A297232      A297233
        A297234     A297235      A297236
        A297237     A297238      A297239
        A297240     A297241      A297242
        A297243     A297244      A297245
        A297246     A297247      A297247
For each b, let u = {n : UV(n,b) < DV(n,b)}, e = {n : UV(n,b) = DV(n,b)}, and d = {n : UV(n,b) > DV(n,b)}.  The sets u,e,d partition the natural numbers.  A guide to the matching sequences for u, e, d follows:
***
Base b    Sequence u   Sequence e   Sequence d
         A005843      A005408      (none)
         A297249      A297250      A297251
         A297252      A297253      A297254
         A297255      A297256      A297257
         A297258      A297259      A297260
         A297261      A297262      A297263
         A297264      A297265      A297266
         A297267      A297268      A297269
        A297270      A297271      A297272
        A297273      A297274      A297275
        A297276      A297277      A297278
        A297279      A297280      A297281
        A297282      A297283      A297284
        A297285      A297286      A297287
        A297288      A297289      A297290
Not a duplicate of A151950: e.g., a(100)=1 but A151950(100)=11. - Robert Israel, Feb 06 2018

			13684632 has DV = 8-4 + 6-3 + 3-2 = 8 and has UV = 3-1 + 6-3 + 8-6 + 6-4 = 9, so that a(13684632) = DV + UV = 17.

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A037851, A297330, A297271, A297272.

f:= proc(n) local L,i; L:= convert(n,base,10);
add(abs(L[i+1]-L[i]),i=1..nops(L)-1) end proc:
map(f, [$1..100]); # Robert Israel, Feb 04 2018
# alternative
A297330 := proc(n)
    A037860(n)+A037851(n) ;
end proc: # R. J. Mathar, Sep 27 2021

b = 10; z = 120; t = Table[Total@Flatten@Map[Abs@Differences@# &, Partition[ IntegerDigits[n, b], 2, 1]], {n, z}] (* after Michael De Vlieger, e.g. A037834 *)

def A297330(n):
    s = str(n)
    return sum(abs(int(s[i])-int(s[i+1])) for i in range(len(s)-1)) # Chai Wah Wu, May 31 2022

A014081 a(n) is the number of occurrences of '11' in the binary expansion of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 1, 2, 3, 0, 0, 0, 1, 0, 0, 1, 2, 1, 1, 1, 2, 2, 2, 3, 4, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 3, 3, 3, 4, 5, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 1, 2, 3, 0, 0, 0, 1, 0, 0, 1, 2, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 1

Offset: 0

Simon Plouffe

a(n) takes the value k for the first time at n = 2^(k+1)-1. Cf. A000225. - Robert G. Wilson v, Apr 02 2009

a(n) = A213629(n,3) for n > 2. - Reinhard Zumkeller, Jun 17 2012

			The binary expansion of 15 is 1111, which contains three occurrences of 11, so a(15)=3.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
J.-P. Allouche, On an Inequality in a 1970 Paper of R. L. Graham, INTEGERS 21A (2021), #A2.
Jean-Paul Allouche and Jeffrey Shallit, Sums of digits and the Hurwitz zeta function, in: K. Nagasaka and E. Fouvry (eds.), Analytic Number Theory, Lecture Notes in Mathematics, Vol. 1434, Springer, Berlin, Heidelberg, 1990, pp. 19-30.
John Brillhart and L. Carlitz, Note on the Shapiro Polynomials, Proceedings of the American Mathematical Society, volume 25, number 1, May 1970, pages 114-118 (see A001782 for a scanned copy), with a(n) = exponent in theorem 4.
Helmut Prodinger, Generalizing the sum of digits function, SIAM J. Algebraic Discrete Methods, Vol. 3, No. 1 (1982), pp. 35-42. MR0644955 (83f:10009). [See B_2(11,n) on p. 35. - _N. J. A. Sloane_, Apr 06 2014]
Michel Rigo and Manon Stipulanti, Revisiting regular sequences in light of rational base numeration systems, arXiv:2103.16966 [cs.FL], 2021. Mentions this sequence.
Bartosz Sobolewski and Lukas Spiegelhofer, Block occurrences in the binary expansion, arXiv:2309.00142 [math.NT], 2023.
Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions, 2004.
Ralf Stephan, Table of generating functions.
Eric Weisstein's World of Mathematics, Digit Block.
Eric Weisstein's World of Mathematics, Rudin-Shapiro Sequence.
Index entries for sequences related to binary expansion of n

Cf. A014082, A033264, A037800, A056973, A000225, A213629, A000120, A069010, A100046.
First differences give A245194.
A245195 gives 2^a(n).

import Data.Bits ((.&.))
a014081 n = a000120 (n .&. div n 2)  -- Reinhard Zumkeller, Jan 23 2012

# To count occurrences of 11..1 (k times) in binary expansion of v:
cn := proc(v, k) local n, s, nn, i, j, som, kk;
som := 0;
kk := convert(cat(seq(1, j = 1 .. k)),string);
n := convert(v, binary);
s := convert(n, string);
nn := length(s);
for i to nn - k + 1 do
if substring(s, i .. i + k - 1) = kk then som := som + 1 fi od;
som; end; # This program no longer worked. Corrected by N. J. A. Sloane, Apr 06 2014.
[seq(cn(n,2),n=0..300)];
# Alternative:
A014081 := proc(n) option remember;
  if n mod 4 <= 1 then procname(floor(n/4))
elif n mod 4 = 2 then procname(n/2)
else 1 + procname((n-1)/2)
fi
end proc:
A014081(0):= 0:
map(A014081, [$0..1000]); # Robert Israel, Sep 04 2015

f[n_] := Count[ Partition[ IntegerDigits[n, 2], 2, 1], {1, 1}]; Table[ f@n, {n, 0, 104}] (* Robert G. Wilson v, Apr 02 2009 *)
Table[SequenceCount[IntegerDigits[n,2],{1,1},Overlaps->True],{n,0,120}] (* Harvey P. Dale, Jun 06 2022 *)

A014081(n)=sum(i=0,#binary(n)-2,bitand(n>>i,3)==3)  \\ M. F. Hasler, Jun 06 2012

a(n) = hammingweight(bitand(n, n>>1)) ;
vector(105, i, a(i-1))  \\ Gheorghe Coserea, Aug 30 2015

def a(n): return sum([((n>>i)&3==3) for i in range(len(bin(n)[2:]) - 1)]) # Indranil Ghosh, Jun 03 2017

from re import split
def A014081(n): return sum(len(d)-1 for d in split('0+', bin(n)[2:]) if d != '') # Chai Wah Wu, Feb 04 2022

a(4n) = a(4n+1) = a(n), a(4n+2) = a(2n+1), a(4n+3) = a(2n+1) + 1. - Ralf Stephan, Aug 21 2003
G.f.: (1/(1-x)) * Sum_{k>=0} t^3/((1+t)*(1+t^2)), where t = x^(2^k). - Ralf Stephan, Sep 10 2003
a(n) = A000120(n) - A069010(n). - Ralf Stephan, Sep 10 2003
Sum_{n>=1} A014081(n)/(n*(n+1)) = A100046 (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021

A037800 Number of occurrences of 01 in the binary expansion of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1

Offset: 0

Clark Kimberling

Number of i such that d(i)>d(i-1), where Sum{d(i)*2^i: i=0,1,...,m} is base 2 representation of n.

This is the base-2 up-variation sequence; see A297330. - Clark Kimberling, Jan 18 2017

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Jean-Paul Allouche and Jeffrey Shallit, Sums of digits and the Hurwitz zeta function, in: K. Nagasaka and E. Fouvry (eds.), Analytic Number Theory, Lecture Notes in Mathematics, Vol. 1434, Springer, Berlin, Heidelberg, 1990, pp. 19-30.
Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions, 2004.
Ralf Stephan, Table of generating functions.
Eric Weisstein's World of Mathematics, Digit Block.
Index entries for sequences related to binary expansion of n

Cf. A014081, A014082, A033264, A056974, A056975, A056976, A056977, A056978, A056979, A056980.

Cf. A030308, A231902.

a037800 = f 0 . a030308_row where
   f c [_]          = c
   f c (1 : 0 : bs) = f (c + 1) bs
   f c (_ : bs)     = f c bs
-- Reinhard Zumkeller, Feb 20 2014

Table[SequenceCount[IntegerDigits[n,2],{0,1}],{n,0,120}] (* Harvey P. Dale, Aug 10 2023 *)

a(n) = { if(n == 0, 0, -1 + hammingweight(bitnegimply(n, n>>1))) };  \\ Gheorghe Coserea, Aug 31 2015

a(2n) = a(n), a(2n+1) = a(n) + [n is even]. - Ralf Stephan, Aug 21 2003
G.f.: 1/(1-x) * Sum_{k>=0} t^5/(1+t)/(1+t^2) where t=x^2^k. - Ralf Stephan, Sep 10 2003
a(n) = A069010(n) - 1, n>0. - Ralf Stephan, Sep 10 2003
Sum_{n>=1} a(n)/(n*(n+1)) = log(2)/2 + Pi/4 - 1 = A231902 - 1 (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021

A278219 Filter-sequence related to base-2 run-length encoding: a(n) = A046523(A243353(n)).

Original entry on oeis.org

1, 2, 4, 2, 4, 8, 6, 2, 4, 12, 16, 8, 6, 12, 6, 2, 4, 12, 36, 12, 16, 32, 24, 8, 6, 30, 24, 12, 6, 12, 6, 2, 4, 12, 36, 12, 36, 72, 60, 12, 16, 48, 64, 32, 24, 72, 24, 8, 6, 30, 60, 30, 24, 48, 60, 12, 6, 30, 24, 12, 6, 12, 6, 2, 4, 12, 36, 12, 36, 72, 60, 12, 36, 180, 144, 72, 60, 180, 60, 12, 16, 48, 144, 48, 64, 128, 96, 32, 24, 120, 216, 72, 24, 72

Offset: 0

Antti Karttunen, Nov 16 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A003188, A046523, A075157, A243353, A278220.
Other base-2 related filter sequences: A278217, A278222.
Sequences that (seem to) partition N into same or coarser equivalence classes are at least these: A005811, A136004, A033264, A037800, A069010, A087116, A090079 and many others like A105500, A106826, A166242, A246960, A277561, A037834, A225081 although these have not been fully checked yet.

f[n_, i_, x_] := Which[n == 0, x, EvenQ@ n, f[n/2, i + 1, x], True, f[(n - 1)/2, i, x Prime@ i]]; g[n_] := If[n == 1, 1, Times @@ MapIndexed[ Prime[First@ #2]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]]];
Table[g@ f[BitXor[n, Floor[n/2]], 1, 1], {n, 0, 93}] (* Michael De Vlieger, May 09 2017 *)

from sympy import prime, factorint
import math
def A(n): return n - 2**int(math.floor(math.log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
def a005940(n): return b(n - 1)
def P(n):
    f = factorint(n)
    return sorted([f[i] for i in f])
def a046523(n):
    x=1
    while True:
        if P(n) == P(x): return x
        else: x+=1
def a003188(n): return n^int(n/2)
def a243353(n): return a005940(1 + a003188(n))
def a(n): return a046523(a243353(n)) # Indranil Ghosh, May 07 2017

(define (A278219 n) (A046523 (A243353 n)))

a(n) = A046523(A243353(n)).
a(n) = A278222(A003188(n)).
a(n) = A278220(1+A075157(n)).

A014082 Number of occurrences of '111' in binary expansion of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 3, 4, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 2, 3, 0, 0, 0, 0, 0, 0, 0, 1, 0

Offset: 0

Simon Plouffe

a(n) = A213629(n,7) for n > 6. - Reinhard Zumkeller, Jun 17 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Digit Block
Index entries for sequences related to binary expansion of n

Cf. A014081, A033264, A056974, A056975, A056976, A056977, A056978, A056979, A056980, A213629, A239906, A239907.

import Data.List (tails, isPrefixOf)
a014082 = sum . map (fromEnum . ([1,1,1] `isPrefixOf`)) .
                    tails . a030308_row
-- Reinhard Zumkeller, Jun 17 2012

See A014081.
f:= proc(n) option remember;
  if n::even then procname(n/2)
  elif n mod 8 = 7 then 1 + procname((n-1)/2)
  else procname((n-1)/2)
fi
end proc:
f(0):= 0:
map(f, [$0..1000]); # Robert Israel, Sep 11 2015

f[n_] := Count[ Partition[ IntegerDigits[n, 2], 3, 1], {1, 1, 1}]; Table[f@n, {n, 0, 104}] (* Robert G. Wilson v, Apr 02 2009 *)
a[0] = a[1] = 0; a[n_] := a[n] = If[EvenQ[n], a[n/2], a[(n - 1)/2] + Boole[Mod[(n - 1)/2, 4] == 3]]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Oct 22 2012, after Ralf Stephan *)
Table[SequenceCount[IntegerDigits[n,2],{1,1,1},Overlaps->True],{n,0,110}] (* Harvey P. Dale, Mar 05 2023 *)

a(n) = hammingweight(bitand(n, bitand(n>>1, n>>2))); \\ Gheorghe Coserea, Aug 30 2015

a(2n) = a(n), a(2n+1) = a(n) + [n congruent to 3 mod 4]. - Ralf Stephan, Aug 21 2003

G.f.: 1/(1-x) * Sum_{k>=0} t^7(1-t)/(1-t^8), where t=x^2^k. - Ralf Stephan, Sep 08 2003

A087116 Number of maximal groups of consecutive zeros in binary representation of n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 2, 2, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2

Offset: 0

Reinhard Zumkeller, Aug 14 2003

nonn

The following four statements are equivalent: a(n) = 0; n = 2^k - 1 for some k; A087117(n) = 0; A023416(n) = 0.

			G.f. = 1 + x^2 + x^4 + x^5 + x^6 + x^8 + x^9 + 2*x^10 + x^11 + x^12 + x^13 + x^14 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A087118, A087119, A087120, A023416, A007088.
Cf. A023416, A087117.
Cf. A033264, A003817, A069010.
Essentially the same as A033264.

a087116 0 = 1
a087116 n = f 0 n where
   f y 0 = y
   f y x = if r == 0 then g x' else f y x'
           where (x', r) = divMod x 2
                 g z = if r == 0 then g z' else f (y + 1) z'
                       where (z', r) = divMod z 2
-- Reinhard Zumkeller, Mar 31 2015

a[n_] := SequenceCount[IntegerDigits[n, 2], {Longest[0..]}];
Table[a[n], {n, 0, 101}] (* Jean-François Alcover, Oct 18 2021 *)

a(n) = if (n == 0, 1, hammingweight(bitxor(n, n>>1)) >> 1);
vector(102, i, a(i-1))  \\ Gheorghe Coserea, Sep 17 2015

def A087116(n):
    return sum(1 for d in bin(n)[2:].split('1') if len(d)) # Chai Wah Wu, Nov 04 2016

a(n) = A033264(n) for n > 0 since strings of 0's alternate with strings of 1's. - Jonathan Sondow, Jan 17 2016
a(n) = a(2*n + 1) = a(4*n + 2) - 1, if n > 0. - Michael Somos, Nov 04 2016
a(n) = A069010(A003817(n)-n) for n > 0. - Chai Wah Wu, Nov 04 2016

A173318 Partial sums of A005811.

Original entry on oeis.org

0, 1, 3, 4, 6, 9, 11, 12, 14, 17, 21, 24, 26, 29, 31, 32, 34, 37, 41, 44, 48, 53, 57, 60, 62, 65, 69, 72, 74, 77, 79, 80, 82, 85, 89, 92, 96, 101, 105, 108, 112, 117, 123, 128, 132, 137, 141, 144, 146, 149, 153, 156, 160, 165, 169, 172, 174, 177, 181, 184, 186, 189

Offset: 0

Jonathan Vos Post, Feb 16 2010

Partial sums of number of runs in binary expansion of n (n>0). Partial sums of number of 1's in Gray code for n. The subsequence of squares in this partial sum begins 0, 1, 4, 9, 144, 169, 256, 289, 324 (since we also have 32 and 128 I wonder about why so many powers). The subsequence of primes in this partial sum begins: 3, 11, 17, 29, 31, 37, 41, 53, 79, 89, 101, 137, 149, 181, 191, 197, 229, 271.

Note: A227744 now gives the squares, which occur at positions given by A227743. - Antti Karttunen, Jul 27 2013

			1 has 1 run in its binary representation "1".
2 has 2 runs in its binary representation "10".
3 has 1 run in its binary representation "11".
4 has 2 runs in its binary representation "100".
5 has 3 runs in its binary representation "101".
Thus a(1) = 1, a(2) = 1+2 = 3, a(3) = 1+2+1 = 4, a(4) = 1+2+1+2 = 6, a(5) = 1+2+1+2+3 = 9.

Antti Karttunen, Table of n, a(n) for n = 0..8191
Richard Blecksmith and Purushottam W. Laud, Some Exact Number Theory Computations via Probability Mechanisms, American Mathematical Monthly, volume 102, number 10, December 1995, pages 893-903, with a(n) = Sum_{j=0..n} b_j as calculated in section 2.
Hsien-Kuei Hwang, Svante Janson and Tsung-Hsi Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, volume 13, number 4, December 2017, article number 47, pages 1-43. Also first authors' copy, 2016. See example 5.5.
Kevin Ryde, Iterations of the Dragon Curve, see index "DirCumul".

Cf. A005811, A000788, A056539, A014707, A014577, A082410, A000975, A034947.
Cf. also A227737, A227741, A227742.
Cf. A227744 (squares occurring), A227743 (indices of squares).

Accumulate[Join[{0},Table[Length[Split[IntegerDigits[n,2]]],{n,110}]]] (* Harvey P. Dale, Jul 29 2013 *)

a(n) = my(v=binary(n+1),d=0,e=4); for(i=1,#v, if(v[i], v[i]=#v-i+d;d+=e;e=0, e=4)); fromdigits(v,2)>>1; \\ Kevin Ryde, Aug 27 2021

a(n) = sum(i=0..n) A005811(i) = sum(i=0..n) (A037834(i)+1) = sum(i=0..n) (A069010(i) + A033264(i)).
a(A000225(n)) = A001787(n) = A000788(A000225(n)). - Antti Karttunen, Jul 27 2013 & Aug 09 2013
a(2n) = 2*a(n) + n - 2*(ceiling(A005811(n)/2) - (n mod 2)), a(2n+1) = 2*a(n) + n + 1. - Ralf Stephan, Aug 11 2013

cp's OEIS Frontend

A366600 a(n) = (1 + A033264(n))*a(A053645(n)) for n > 0 with a(0) = 1.

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

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A069010 Number of runs of 1's in the binary representation of n.

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A297330 Total variation of base-10 digits of n; see Comments.

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A014081 a(n) is the number of occurrences of '11' in the binary expansion of n.

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A037800 Number of occurrences of 01 in the binary expansion of n.

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