A056973 -id:A056973 - cp's OEIS frontend

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

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

A092339 Number of adjacent identical digits in the binary representation of n.

Original entry on oeis.org

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

Offset: 0

Ralf Stephan, Mar 18 2004

In binary: number of 00 blocks plus number of 11 blocks. (Note: the blocks can overlap. See the example below.)

			60 in binary is 111100, it has 4 blocks of adjacent digits, so a(60)=4.
Equally, 60's binary Gray code expansion is A003188(60)=34, 100010 in binary, which contains four zeros.

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 84.

Cf. A005811.

a(n)=local(v); v=binary(n); sum(k=1, length(v)-1, v[k]==v[k+1])

a(n)=if(n<1,0,if(n%2==0,a(n/2)+(n>0&&(n/2)%2==0),a((n-1)/2)+((n-1)/2)%2))

(define (A092339 n) (A080791 (A003188 n))) ;; Antti Karttunen, Jul 05 2013

Recurrence: a(2n) = a(n) + [n even], a(2n+1) = a(n) + [n odd].
a(n) = A014081(n) + A056973(n).
For n>0, A227185(n) = a(n)+1.
a(n) = A080791(A003188(n)) [because the sequence gives the number of nonleading zeros in binary Gray code expansion of n] - Antti Karttunen, Jul 05 2013

A107782 In binary representation of n: (number of zeros) minus (number of blocks of contiguous zeros).

Original entry on oeis.org

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

Offset: 0

Reinhard Zumkeller, May 25 2005

nonn

a(n) = A023416(n) - A087116(n); a(A003754(n)) = 0.

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

Cf. A007088, A107345.
Cf. A056973. - R. J. Mathar, Aug 24 2008
Cf. A023416, A087116, A003754.

a107782 n = a023416 n - a087116 n  -- Reinhard Zumkeller, Mar 31 2015

A037809 Number of i such that d(i) <= d(i-1), where Sum_{i=0..m} d(i)*2^i is the base-2 representation of n.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

			The base-2 representation of n=4 is 100 with d(0)=0, d(1)=0, d(2)=1. Only d(1) <= d(0) is true, so a(4)=1. - _R. J. Mathar_, Oct 16 2015

Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions

Cf. A033265.

A037809 := proc(n)
    a := 0 ;
    dgs := convert(n,base,2);
    for i from 2 to nops(dgs) do
        if op(i,dgs)<=op(i-1,dgs) then
            a := a+1 ;
        end if;
    end do:
    a ;
end proc: # R. J. Mathar, Oct 16 2015

From Ralf Stephan, Oct 05 2003: (Start)
G.f.: -x/(1-x) + 1/(1-x) * Sum_{k>=0} (t + t^3 + t^4)/(1 + t + t^2 + t^3), where t=x^2^k.
a(n) = A056973(n) + A000120(n) - 1.
a(n) = b(n) - 1, with b(0)=0, b(2n) = b(n) + [n even], b(2n+1) = b(n) + 1. (End)

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

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A092339 Number of adjacent identical digits in the binary representation of n.

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A107782 In binary representation of n: (number of zeros) minus (number of blocks of contiguous zeros).

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A037809 Number of i such that d(i) <= d(i-1), where Sum_{i=0..m} d(i)*2^i is the base-2 representation of n.

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