A014081 -id:A014081 - cp's OEIS frontend

A033265 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.

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

Offset: 1

Clark Kimberling

			The base-2 representation of n=4 is 100 with d(0)=0, d(1)=0, d(2)=1. There are two rise-or-equal, one from d(0) to d(1) and one from d(1) to d(2), so a(4)=2. - _R. J. Mathar_, Oct 16 2015

Antti Karttunen, Table of n, a(n) for n = 1..65537
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Index entries for sequences related to binary expansion of n

Cf. A014081, A023416, A037800, A037809, A005940, A156552, A297113, A364567 [= 2^a(n)].

A033265 := 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

A033265(n) = { my(i=0); while(n>1, if((n%4)!=1, i++); n >>= 1); (i); }; \\ Antti Karttunen, Aug 06 2023

From Ralf Stephan, Oct 05 2003: (Start)
a(0) = 0, a(2n) = a(n) + 1, a(2n+1) = a(n) + [n odd].
a(n) = A014081(n) + A023416(n).
G.f.: 1/(1-x) * Sum_{k>=0} (t^2 + t^3 + t^4)/((1+t)*(1+t^2)), t=x^2^k. (End)
a(n) = -1 + A297113(A005940(1+n)). - Antti Karttunen, Dec 30 2017

Sign in Name corrected by R. J. Mathar, Oct 16 2015

A245196 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=wt(k-r-1), or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=a(j)*wt(k-r-1) (where wt(i) = A000120(i)).

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane, Jul 25 2014

Other sequences defined by a recurrence of this class (see the Formula and Maple sections) include A245180, A245195, A048896, A245536, A038374.

			May be arranged into blocks of lengths 1,2,4,8,...:
0,
0, 1,
0, 0, 1, 1,
0, 0, 0, 0, 1, 0, 1, 2,
0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 2, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 2, 0, 1, 2,
...

Cf. A000120, A160239, A245180, A245195, A014081, A048896, A038374, A245536, A245537, A245538, A245539, A048896, A245547.

Maple code for this sequence:
wt := proc(n) local w, m, i; w := 0; m := n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end:
G:=[seq(wt(n),n=0..30)];
m:=1;
f:=proc(n) option remember; global m,G; local k,r,j,np;
   k:=1+floor(log[2](n)); np:=2^k-n;
   if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
   if j=0 then G[k-r]; else m*G[k-r]*f(j); fi;
end;
[seq(f(n),n=1..120)];
# Maple code for the general recurrence:
G:=[seq(wt(n),n=0..30)]; # replace this by a list G=[G(0), G(1), G(2), ...], remembering that you have to tell Maple G[1] to get G(0), G[2] to get G(1), etc.
m:=1; # replace this by the correct multiplier
f:=proc(n) option remember; global m,G; local k,r,j,np;
   k:=1+floor(log[2](n)); np:=2^k-n;
   if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
   if j=0 then G[k-r-1+1]; else m*G[k-r-1+1]*f(j); fi;
end;
[seq(f(n),n=1..120)];
# If G(n) = wt(n) and m=1 we get the present sequence
# If G(n) = A083424(n) and m=1 we get A245537
# If G(n) = A083424(n) and m=2 we get A245538
# If G(n) = A083424(n) and m=4 we get A245539
# If G(n) = A083424(n) and m=8 we get A245180 (and presumably A160239)
# If G(n) = n (n>=0) and m=1 we get A245536
# If G(n) = n+1 (n>=0) and m=1 we get A038374
# If G(n) = (n+1)(n+2)/2 (n>=0) and m=1 we get A245541
# If G(n) = (n+1)(n+2)/2 (n>=0) and m=2 we get A245547
# If G(n) = 2^n (n>=0) and m=1 we get A245195 (= 2^A014081)
# If G(n) = 2^n (n>=0) and m=2 we get A048896

This is an example of a class of sequences defined by the following recurrence.
We first choose a sequence G = [G(0), G(1), G(2), G(3), ...], which are the terms that will appear at the ends of the blocks: a(2^k-1) = G(k-1), and we also choose a parameter m (the "multiplier"). Then the recurrence (this defines a(1), a(2), a(3), ...) is:
a(2^k-2^r)=G(k-r-1) if 0 <= r <= k-1, a(2^k-2^r+j)=m*a(j)*G(k-r-1) if 2 <= r <= k-1, 1 <= j < 2^r-1.
To help apply the recurrence, here are the values of k,r,j for the first few values of n (if n=2^k-2^r we set j=0, although it is not used):
n:  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15
k:  1  2  2  3  3  3  3  4  4  4  4  4  4  4  4
r:  0  1  0  2  2  1  0  3  3  3  3  2  2  1  0
j:  0  0  0  0  1  0  0  0  1  2  3  0  1  0  0
--------------------------------------------------
n: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
k:  5  5  5  5  5  5  5  5  5  5  5  5  5  5  5  5
r:  4  4  4  4  4  4  4  4  3  3  3  3  2  2  1  0
j:  0  1  2  3  4  5  6  7  0  1  2  3  0  1  0  0
--------------------------------------------------
In the present example G(n) = wt(n) and m=1.

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

A384715 a(n) = Sum_{k=0..n} (binomial(n, k) mod 4).

Original entry on oeis.org

1, 2, 4, 8, 4, 8, 12, 16, 4, 8, 12, 24, 12, 24, 24, 32, 4, 8, 12, 24, 12, 24, 32, 48, 12, 24, 32, 48, 24, 48, 48, 64, 4, 8, 12, 24, 12, 24, 32, 48, 12, 24, 32, 64, 32, 64, 64, 96, 12, 24, 32, 48, 32, 64, 64, 96, 24, 48, 64, 96, 48, 96, 96, 128, 4, 8, 12, 24

Offset: 0

David Radcliffe, Jun 23 2025

This is a 2-automatic sequence.

			Let b(n) be the binary expansion of n. Then a(n) = (1 + p10 + p11) * 2^c, where c is the number of set bits in b(n), p10 is the number of '10' patterns in b(n), and p11 is 1 or 0 depending on whether the pattern '11' is occurring in b(n) or not. This formula is used by _Chai Wah Wu_ in the Python function below. For instance:
  n = 25 -> b(n) = 11001 -> a(n) = (1+1+1) * 2^3 = 24.
  n = 26 -> b(n) = 11010 -> a(n) = (1+2+1) * 2^3 = 32.
  n = 27 -> b(n) = 11011 -> a(n) = (1+1+1) * 2^4 = 48.
- _Peter Luschny_, Jun 25 2025

David Radcliffe, Table of n, a(n) for n = 0..10000
Kenneth S. Davis and William A. Webb, Pascal's triangle modulo 4, Fib. Quart., 29 (1991), 79-83.

Cf. A001316, A014081, A033264, A051638 (mod 3 analog), A085357. Row sums of triangle A034931.

A384715[n_] := 2^DigitSum[n, 2]*(StringCount[IntegerString[n, 2], "10"] - Boole[BitAnd[n,2*n] == 0] + 2);
Array[A384715, 100, 0] (* Paolo Xausa, Jun 26 2025 *)

a(n) = sum(k=0, n, binomial(n,k)%4); \\ Michel Marcus, Jun 25 2025

def A001316(n): return (1 + (n % 2)) * A001316(n // 2) if n else 1
def A033264(n): return (n % 4 == 2) + A033264(n // 2) if n else 0
def A085357(n): return int(n & (n<<1) == 0)
def A384715(n): return A001316(n) * (A033264(n) - A085357(n) + 2)

def A384715(n): return (((n>>1)&~n).bit_count()+bool(n&(n<<1))+1)<Chai Wah Wu, Jun 25 2025

def a(n: int) -> int:  # after Chai Wah Wu
    b = bin(n)[2:]; p = b.count("10"); q = b.count("11")
    return (p + (2 if q else 1)) * 2**n.bit_count()  # Peter Luschny, Jun 25 2025

a(n) = A001316(n) * (A033264(n) - A085357(n) + 2) for n > 0.
Recurrences:
a(4n) = a(2n),
a(4n+1) = 2a(2n),
a(8n+2) = a(4n+2) + 2a(2n) - a(2n+1),
a(8n+3) = a(4n+3) + 4a(2n) - 4a(n),
a(8n+6) = a(4n+3) + 2a(4n+2) - 2a(2n+1),
a(8n+7) = 2a(4n+3).

A328607 Numbers whose reversed binary expansion, without the most significant digit, is a necklace.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 24, 26, 28, 30, 31, 32, 48, 52, 56, 58, 60, 62, 63, 64, 96, 100, 104, 106, 108, 112, 116, 118, 120, 122, 124, 126, 127, 128, 192, 200, 208, 212, 216, 220, 224, 228, 232, 234, 236, 240, 244, 246, 248, 250, 252, 254, 255

Offset: 0

Gus Wiseman, Oct 30 2019

nonn

Offset is 0 to be consistent with A257250.

A necklace is a finite sequence that is lexicographically minimal among all of its cyclic rotations.

			The sequence of terms together with their binary expansions and binary indices begins:
    0:        0 ~ {}
    1:        1 ~ {1}
    2:       10 ~ {2}
    3:       11 ~ {1,2}
    4:      100 ~ {3}
    6:      110 ~ {2,3}
    7:      111 ~ {1,2,3}
    8:     1000 ~ {4}
   12:     1100 ~ {3,4}
   14:     1110 ~ {2,3,4}
   15:     1111 ~ {1,2,3,4}
   16:    10000 ~ {5}
   24:    11000 ~ {4,5}
   26:    11010 ~ {2,4,5}
   28:    11100 ~ {3,4,5}
   30:    11110 ~ {2,3,4,5}
   31:    11111 ~ {1,2,3,4,5}
   32:   100000 ~ {6}
   48:   110000 ~ {5,6}
   52:   110100 ~ {3,5,6}

The dual non-reversed version is A257250.
The dual non-reversed version involving all digits is A065609.
The version involving all digits is A328595.
The non-reversed version is A328668.
Binary necklaces are A000031.
Cf. A000120, A001037, A003714, A008965, A014081, A121016, A164707, A275692, A328594, A328596.

neckQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And];
Select[Range[0,100],#<=1||neckQ[Reverse[Rest[IntegerDigits[#,2]]]]&]

A328668 Numbers whose binary expansion, without the most significant digit, is a necklace.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 8, 9, 11, 15, 16, 17, 19, 21, 23, 31, 32, 33, 35, 37, 39, 43, 47, 63, 64, 65, 67, 69, 71, 73, 75, 77, 79, 85, 87, 91, 95, 127, 128, 129, 131, 133, 135, 137, 139, 141, 143, 147, 149, 151, 155, 157, 159, 171, 175, 183, 191, 255, 256, 257

Offset: 0

Gus Wiseman, Oct 26 2019

nonn

Offset is 0 to be consistent with A257250.

A necklace is a finite sequence that is lexicographically minimal among all of its cyclic rotations.

			The sequence of terms together with their binary expansions and binary indices begins:
   0:       0 ~ {}
   1:       1 ~ {1}
   2:      10 ~ {2}
   3:      11 ~ {1,2}
   4:     100 ~ {3}
   5:     101 ~ {1,3}
   7:     111 ~ {1,2,3}
   8:    1000 ~ {4}
   9:    1001 ~ {1,4}
  11:    1011 ~ {1,2,4}
  15:    1111 ~ {1,2,3,4}
  16:   10000 ~ {5}
  17:   10001 ~ {1,5}
  19:   10011 ~ {1,2,5}
  21:   10101 ~ {1,3,5}
  23:   10111 ~ {1,2,3,5}
  31:   11111 ~ {1,2,3,4,5}
  32:  100000 ~ {6}
  33:  100001 ~ {1,6}
  35:  100011 ~ {1,2,6}

The dual version is A257250.
The version involving all digits, taken in reverse, is A328595.
The reversed version is A328607.
Binary necklaces are A000031.
Necklace compositions are A008965.
Cf. A000120, A001037, A003714, A014081, A065609, A121016, A275692, A328594, A328596.

neckQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And];
Select[Range[0,100],#<=1||neckQ[Rest[IntegerDigits[#,2]]]&]

A056973 Number of blocks of {0,0} in the binary expansion of n.

Original entry on oeis.org

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

Offset: 1

Eric W. Weisstein

Reinhard Zumkeller, Table of n, a(n) for n = 1..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. A003754, A014081, A023416, A033264, A037800.

Cf. A107782.

a056973 = f 0 where
   f y x = if x == 0 then y else f (y + 0 ^ (mod x 4)) $ div x 2
-- Reinhard Zumkeller, Mar 31 2015

f:= proc(n) option remember;
     if n mod 4 = 0 then 1 + procname(n/2)
     else procname(floor(n/2))
     fi
end proc:
f(1):= 0:
map(f, [$1..200]); # Robert Israel, Sep 02 2015

f[n_] := Count[Partition[IntegerDigits[n, 2], 2, 1], {0, 0}]; Table[f@ n, {n, 0, 102}] (* Michael De Vlieger, Sep 01 2015, after Robert G. Wilson v at A014081 *)
SequenceCount[#,{0,0},Overlaps->True]&/@(IntegerDigits[#,2]&/@Range[0,120]) (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 24 2018 *)

a(n) = { my(x = bitor(n, n>>1));
         if (x == 0, 0, 1 + logint(x, 2) - hammingweight(x)) }
vector(102, i, a(i))  \\ Gheorghe Coserea, Sep 01 2015

a(2n) = a(n) + [n is even], a(2n+1) = a(n).
G.f.: 1/(1-x) * Sum_{k>=0} t^4/((1+t)*(1+t^2)) where t=x^(2^k). - Ralf Stephan, Sep 10 2003
a(n) = A023416(n) - A033264(n). - Ralf Stephan, Sep 10 2003
Sum_{n>=1} a(n)/(n*(n+1)) = 2 - 3*log(2)/2 - Pi/4 (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021

A239906 Let cn(n,k) denote the number of times 11..1 (k 1's) appears in the binary representation of n; a(n) = n - cn(n,1) + cn(n,2) - cn(n,3).

Original entry on oeis.org

0, 0, 1, 2, 3, 3, 5, 5, 7, 7, 8, 9, 11, 11, 12, 12, 15, 15, 16, 17, 18, 18, 20, 20, 23, 23, 24, 25, 26, 26, 27, 27, 31, 31, 32, 33, 34, 34, 36, 36, 38, 38, 39, 40, 42, 42, 43, 43, 47, 47, 48, 49, 50, 50, 52, 52, 54, 54, 55, 56, 57, 57, 58, 58, 63, 63, 64, 65, 66, 66, 68, 68, 70, 70, 71, 72, 74, 74

Offset: 0

N. J. A. Sloane, Apr 07 2014

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A000120, A012081, A014082, A239904, A239907.

# From A014081:
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;
[seq(n-cn(n,1)+cn(n,2)-cn(n,3), n=0..100)];

cn[n_, k_] := Count[Partition[IntegerDigits[n, 2], k, 1], Table[1, {k}]]; Table[n - Sum[cn[n, i], {i, 1, 3, 2}] + cn[n, 2], {n, 0, 77}] (* Michael De Vlieger, Sep 18 2015 *)

a(n) = {
    my(x = bitand(n, n>>1), wt = k->hammingweight(k));
    n - wt(n) + wt(x) - wt(bitand(x, n>>2));
};
vector(78, i, a(i-1))  \\ Gheorghe Coserea, Sep 24 2015

A239907 Let cn(n,k) denote the number of times 11..1 (k 1's) appears in the binary representation of n; a(n) = n - cn(n,1) + cn(n,2) - cn(n,3) + cn(n,4) - ... .

Original entry on oeis.org

0, 0, 1, 2, 3, 3, 5, 5, 7, 7, 8, 9, 11, 11, 12, 13, 15, 15, 16, 17, 18, 18, 20, 20, 23, 23, 24, 25, 26, 26, 28, 28, 31, 31, 32, 33, 34, 34, 36, 36, 38, 38, 39, 40, 42, 42, 43, 44, 47, 47, 48, 49, 50, 50, 52, 52, 54, 54, 55, 56, 58, 58, 59, 60, 63, 63, 64, 65, 66, 66, 68, 68, 70, 70, 71, 72, 74, 74, 75

Offset: 0

N. J. A. Sloane, Apr 07 2014

Gheorghe Coserea, Table of n, a(n) for n = 0..10000
Jon Maiga, Computer-generated formulas for A239907, Sequence Machine.

Cf. A000120, A012081, A014082, A239904, A239906.

# From A014081:
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;
g:=n->add((-1)^i*cn(n,i),i=1..10); # assumes n < 1023
[seq(n+g(n), n=0..100)];

cn[n_, k_] := Count[Partition[IntegerDigits[n, 2], k, 1], Table[1, {k}]]; Table[n - Sum[cn[n, i], {i, 1, IntegerLength[n, 2], 2}] + Sum[cn[n, i], {i, 2, IntegerLength[n, 2], 2}], {n, 0, 78}] (* Michael De Vlieger, Sep 18 2015 *)

binruns(n) = {
  if (n == 0, return([1, 0]));
  my(bag = List(), v=0);
  while(n != 0,
        v = valuation(n,2); listput(bag, v); n >>= v; n++;
        v = valuation(n,2); listput(bag, v); n >>= v; n--);
  return(Vec(bag));
};
a(n) = {
  my(v = binruns(n));
  n - sum(i = 1, #v, if (i%2 == 0, (v[i] + 1)\2, 0))
};
vector(79, i, a(i-1))  \\ Gheorghe Coserea, Sep 18 2015

Conjecture: a(n) = n - A329320(n) for n >= 0 (noticed by Sequence Machine). - Mikhail Kurkov, Oct 13 2021

A126306 a(n) = number of double-rises (UU-subsequences) in the n-th Dyck path encoded by A014486(n).

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jan 02 2007

nonn

			A014486(20) = 228 (11100100 in binary), encodes the following Dyck path:
    /\
   /..\/\
  /......\
and there is one rising (left-hand side) slope with length 3 and one with length 1, so in the first slope, consisting of 3 U-steps, there are two cases with two consecutive U-steps (overlapping is allowed), thus a(20)=2.

Cf. A014081, A014486, A000120, A048735, A125976, A057514.

def ok(n):
    if n==0: return True
    B=bin(n)[2:] if n!=0 else '0'
    s=0
    for b in B:
        s+=1 if b=='1' else -1
        if s<0: return False
    return s==0
def a014081(n): return sum(((n>>i)&3==3) for i in range(len(bin(n)[2:]) - 1))
print([a014081(n) for n in range(4001) if ok(n)]) # Indranil Ghosh, Jun 13 2017

a(n) = A014081(A014486(n)).
a(n) = A000120(A048735(A014486(n))).
a(A125976(n)) = A057514(n)-1, for all n >= 1.

cp's OEIS Frontend

A033265 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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A245196 Write n>=1 as either n=2^k-2^r with 0 <= r <= k-1, in which case a(2^k-2^r)=wt(k-r-1), or as n=2^k-2^r+j with 2 <= r <= k-1, 1 <= j < 2^r-1, in which case a(2^k-2^r+j)=a(j)*wt(k-r-1) (where wt(i) = A000120(i)).

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

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A384715 a(n) = Sum_{k=0..n} (binomial(n, k) mod 4).

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A328607 Numbers whose reversed binary expansion, without the most significant digit, is a necklace.

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A328668 Numbers whose binary expansion, without the most significant digit, is a necklace.

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A056973 Number of blocks of {0,0} in the binary expansion of n.

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