A065359 -id:A065359 - cp's OEIS frontend

A020988 a(n) = (2/3)*(4^n-1).

0, 2, 10, 42, 170, 682, 2730, 10922, 43690, 174762, 699050, 2796202, 11184810, 44739242, 178956970, 715827882, 2863311530, 11453246122, 45812984490, 183251937962, 733007751850, 2932031007402, 11728124029610, 46912496118442, 187649984473770, 750599937895082

Offset: 0

N. J. A. Sloane

Numbers whose binary representation is 10, n times (see A163662(n) for n >= 1). - Alexandre Wajnberg, May 31 2005
Numbers whose base-4 representation consists entirely of 2's; twice base-4 repunits. - Franklin T. Adams-Watters, Mar 29 2006
Expected time to finish a random Tower of Hanoi problem with 2n disks using optimal moves, so (since 2n is even and A010684(2n) = 1) a(n) = A060590(2n). - Henry Bottomley, Apr 05 2001
a(n) is the number of derangements of [2n + 3] with runs consisting of consecutive integers. E.g., a(1) = 10 because the derangements of {1, 2, 3, 4, 5} with runs consisting of consecutive integers are 5|1234, 45|123, 345|12, 2345|1, 5|4|123, 5|34|12, 45|23|1, 345|2|1, 5|4|23|1, 5|34|2|1 (the bars delimit the runs). - Emeric Deutsch, May 26 2003
For n > 0, also smallest numbers having in binary representation exactly n + 1 maximal groups of consecutive zeros: A087120(n) = a(n-1), see A087116. - Reinhard Zumkeller, Aug 14 2003
Number of walks of length 2n + 3 between any two diametrically opposite vertices of the cycle graph C_6. Example: a(0) = 2 because in the cycle ABCDEF we have two walks of length 3 between A and D: ABCD and AFED. - Emeric Deutsch, Apr 01 2004
From Paul Barry, May 18 2003: (Start)
Row sums of triangle using cumulative sums of odd-indexed rows of Pascal's triangle (start with zeros for completeness):
            0  0
            1  1
         1  4  4  1
      1  6 14 14  6  1
   1  8 27 49 49 27  8  1 (End)
a(n) gives the position of the n-th zero in A173732, i.e., A173732(a(n)) = 0 for all n and this gives all the zeros in A173732. - Howard A. Landman, Mar 14 2010
Smallest number having alternating bit sum -n. Cf. A065359. For n = 0, 1, ..., the last digit of a(n) is 0, 2, 0, 2, ... . - Washington Bomfim, Jan 22 2011
Number of toothpicks minus 1 in the toothpick structure of A139250 after 2^n stages. - Omar E. Pol, Mar 15 2012
For n > 0 also partial sums of the odd powers of 2 (A004171). - K. G. Stier, Nov 04 2013
Values of m such that binomial(4*m + 2, m) is odd. Cf. A002450. - Peter Bala, Oct 06 2015
For a(n) > 2, values of m such that m is two steps away from a power of 2 under the Collatz iteration. - Roderick MacPhee, Nov 10 2016
a(n) is the position of the first occurrence of 2^(n+1)-1 in A020986. See the Brillhart and Morton link, pp. 856-857. - John Keith, Jan 12 2021
a(n) is the number of monotone paths in the n-dimensional cross-polytope for a generic linear orientation. See the Black and De Loera link. - Alexander E. Black, Feb 15 2023

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..170 from Vincenzo Librandi)
Andrei Asinowski, Cyril Banderier, and Benjamin Hackl, On extremal cases of pop-stack sorting, Permutation Patterns (Zürich, Switzerland, 2019) [link is not very stable].
Andrei Asinowski, Cyril Banderier, and Benjamin Hackl, Flip-sort and combinatorial aspects of pop-stack sorting, arXiv:2003.04912 [math.CO], 2020.
Peter Bala, A characterization of A002450, A020988 and A080674.
Alexander E. Black, Monotone Paths on Polytopes: Combinatorics and Optimization, Ph. D. Dissertation, Univ. Calif. Davis (2024). See p. 59.
Alexander Black and Jesús De Loera, Monotone paths on cross-polytopes, arXiv:2102.01237 [math.CO], Feb 2021
John Brillhart and Peter Morton, A case study in mathematical research: the Golay-Rudin-Shapiro sequence, Amer. Math. Monthly, 103 (1996) 854-869.
Nobushige Kurokawa, Zeta functions over F_1, Proc. Japan Acad., 81, Ser. A (2005), 180-184. See Theorem 3 (3).
Jonathan L. Merzel, Ján Minač, Tung T. Nguyen, and Nguyên Duy Tân, On divisibility relation graphs, 2025. See p. 14.
Andrei K. Svinin, Tuenter polynomials and a Catalan triangle, arXiv:1603.05748 [math.CO], 2016. See p.3.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A020989, A047849, A108019, A295521, A020522.

[(2/3)*(4^n-1): n in [0..40] ]; // Vincenzo Librandi, Apr 28 2011

A020988 := proc(n)
    2*(4^n-1)/3 ;
end proc: # R. J. Mathar, Feb 19 2015

(2(4^Range[0, 30] - 1))/3 (* or *) LinearRecurrence[{5, -4}, {0, 2}, 30] (* Harvey P. Dale, Sep 25 2013 *)

vector(100, n, n--; (2/3)*(4^n-1)) \\ Altug Alkan, Oct 06 2015

Vec(2*z/((1-z)*(1-4*z)) + O(z^30)) \\ Altug Alkan, Oct 11 2015

def A020988(n): return (2 * ((1 << (2 * n)) - 1)) // 3 # John Reimer Morales, Aug 05 2025

(((List.fill(20)(4: BigInt)).scanLeft(1: BigInt)( * )).map(2 * )).scanLeft(0: BigInt)( + ) // _Alonso del Arte, Sep 12 2019

a(n) = 4*a(n-1) + 2, a(0) = 0.
a(n) = A026644(2*n).
a(n) = A007583(n) - 1 = A039301(n+1) - 2 = A083584(n-1) + 1.
E.g.f. : (2/3)*(exp(4*x)-exp(x)). - Paul Barry, May 18 2003
a(n) = A007583(n+1) - 1 = A039301(n+2) - 2 = A083584(n) + 1. - Ralf Stephan, Jun 14 2003
G.f.: 2*x/((1-x)*(1-4*x)). - R. J. Mathar, Sep 17 2008
a(n) = a(n-1) + 2^(2n-1), a(0) = 0. - Washington Bomfim, Jan 22 2011
a(n) = A193652(2*n). - Reinhard Zumkeller, Aug 08 2011
a(n) = 5*a(n-1) - 4*a(n-2) (n > 1), a(0) = 0, a(1) = 2. - L. Edson Jeffery, Mar 02 2012
a(n) = (2/3)*A024036(n). - Omar E. Pol, Mar 15 2012
a(n) = 2*A002450(n). - Yosu Yurramendi, Jan 24 2017
From Seiichi Manyama, Nov 24 2017: (Start)
Zeta_{GL(2)/F_1}(s) = Product_{k = 1..4} (s-k)^(-b(2,k)), where Sum b(2,k)*t^k = t*(t-1)*(t^2-1). That is Zeta_{GL(2)/F_1}(s) = (s-3)*(s-2)/((s-4)*(s-1)).
Zeta_{GL(2)/F_1}(s) = Product_{n > 0} (1 - (1/s)^n)^(-A295521(n)) = Product_{n > 0} (1 - x^n)^(-A295521(n)) = (1-3*x)*(1-2*x)/((1-4*x)*(1-x)) = 1 + Sum_{k > 0} a(k-1)*x^k (x=1/s). (End)
From Oboifeng Dira, May 29 2020: (Start)
a(n) = A078008(2n+1) (second bisection).
a(n) = Sum_{k=0..n} binomial(2n+1, ((n+2) mod 3)+3k). (End)
From John Reimer Morales, Aug 04 2025: (Start)
a(n) = A000302(n) - A047849(n).
a(n) = A020522(n) + A000079(n) - A047849(n). (End)

Edited by N. J. A. Sloane, Sep 06 2006

A055017 Difference between sums of alternate digits of n starting with the last, i.e., (sum of ultimate digit of n, antepenultimate digit of n, ...) - (sum of penultimate digit of n, preantepenultimate digit of n, ...).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, -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

Offset: 0

Henry Bottomley, May 31 2000

a(n) is a multiple of 11 iff n is divisible by 11.
Digital sum with alternating signs starting with a positive sign for the rightmost digit. - Hieronymus Fischer, Jun 18 2007
For n < 100, a(n) = (n mod 10 - floor(n/10)) = -A076313(n). - Hieronymus Fischer, Jun 18 2007

			a(123) = 3-2+1 = 2, a(9875) = 5-7+8-9 = -3.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Cf. A225693 (alternating sum of digits).
Unsigned version differs from A040114 and A040115 when n=100 and from A040997 when n=101.
Cf. A076313, A076314, A007953, A003132.
Cf. A004086.
Cf. analogous sequences for bases 2-9: A065359, A065368, A346688, A346689, A346690, A346691, A346731, A346732 and also A373605 (for primorial base).

sumodigs := proc(n) local dg; dg := convert(n,base,10) ; add(op(1+2*i,dg), i=0..floor(nops(dg)-1)/2) ; end proc:
sumedigs := proc(n) local dg; dg := convert(n,base,10) ; add(op(2+2*i,dg), i=0..floor(nops(dg)-2)/2) ; end proc:
A055017 := proc(n) sumodigs(n)-sumedigs(n) ; end proc: # R. J. Mathar, Aug 26 2011

def A055017(n): return sum((-1 if i % 2 else 1)*int(j) for i, j in enumerate(str(n)[::-1])) # Chai Wah Wu, May 11 2022

"Recursive version for general bases"
"Set base = 10 for this sequence"
altDigitalSumRight: base
| s |
base = 1 ifTrue: [^self \\ 2].
(s := self // base) > 0
  ifTrue: [^(self - (s * base) - (s altDigitalSumRight: base))]
  ifFalse: [^self]
[by Hieronymus Fischer, Mar 23 2014]

From Hieronymus Fischer, Jun 18 2007, Jun 25 2007, Mar 23 2014: (Start)
a(n) = n + 11*Sum_{k>=1} (-1)^k*floor(n/10^k).
a(10n+k) = k - a(n), 0 <= k < 10.
G.f.: Sum_{k>=1} (x^k-x^(k+10^k)+(-1)^k*11*x^(10^k))/((1-x^(10^k))*(1-x)).
a(n) = n + 11*Sum_{k=10..n} Sum_{j|k,j>=10} (-1)^floor(log_10(j))*(floor(log_10(j)) - floor(log_10(j-1))).
G.f. expressed in terms of Lambert series: g(x) = (x/(1-x)+11*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)^floor(log_10(k)) if k>1 is a power of 10, otherwise b(k)=0.
G.f.: (1/(1-x)) * Sum_{k>=1} (1+11*c(k))*x^k, where c(k) = Sum_{j>=2,j|k} (-1)^floor(log_10(j))*(floor(log_10(j))-floor(log_10(j-1))).
Formulas for general bases b > 1 (b = 10 for this sequence).
a(n) = Sum_{k>=0} (-1)^k*(floor(n/b^k) mod b).
a(n) = n + (b+1)*Sum_{k>=1} (-1)^k*floor(n/b^k). Both sums are finite with floor(log_b(n)) as the highest index.
a(n) = a(n mod b^k) + (-1)^k*a(floor(n/b^k)), for all k >= 0.
a(n) = a(n mod b) - a(floor(n/b)).
a(n) = a(n mod b^2) + a(floor(n/b^2)).
a(n) = (-1)^m*A225693(n), where m = floor(log_b(n)).
a(n) = (-1)^k*A225693(A004086(n)), where k = is the number of trailing 0's of n, formally, k = max(j | n == 0 (mod 10^j)).
a(A004086(A004086(n))) = (-1)^k*a(n), where k = is the number of trailing 0's in the decimal representation of n. (End)

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A359359 Sum of positions of zeros in the binary expansion of n, where positions are read starting with 1 from the left (big-endian).

Original entry on oeis.org

1, 0, 2, 0, 5, 2, 3, 0, 9, 5, 6, 2, 7, 3, 4, 0, 14, 9, 10, 5, 11, 6, 7, 2, 12, 7, 8, 3, 9, 4, 5, 0, 20, 14, 15, 9, 16, 10, 11, 5, 17, 11, 12, 6, 13, 7, 8, 2, 18, 12, 13, 7, 14, 8, 9, 3, 15, 9, 10, 4, 11, 5, 6, 0, 27, 20, 21, 14, 22, 15, 16, 9, 23, 16, 17, 10

Offset: 0

Gus Wiseman, Jan 03 2023

			The binary expansion of 100 is (1,1,0,0,1,0,0), with zeros at positions {3,4,6,7}, so a(100) = 20.

The number of zeros is A023416, partial sums A059015.
For positions of 1's we have A230877, reversed A029931.
The reversed version is A359400.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion.
A039004 lists the positions of zeros in A345927.
Cf. A000120, A048793, A065359, A069010, A070939, A073642, A083652, A328594, A328595, A359402, A359495.

Table[Total[Join@@Position[IntegerDigits[n,2],0]],{n,0,100}]

a(n>0) = binomial(A029837(n)+1,2) - A230877(n).

A050292 a(2n) = 2n - a(n), a(2n+1) = 2n + 1 - a(n) (for n >= 0).

Original entry on oeis.org

0, 1, 1, 2, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 16, 17, 17, 18, 19, 20, 20, 21, 21, 22, 22, 23, 24, 25, 25, 26, 26, 27, 27, 28, 29, 30, 30, 31, 32, 33, 33, 34, 35, 36, 36, 37, 37, 38, 38, 39, 40, 41, 41, 42, 43, 44, 44, 45, 46, 47, 47, 48, 48, 49, 49, 50, 51, 52, 52, 53, 54

Offset: 0

Eric W. Weisstein

Note that the first equation implies a(0)=0, so there is no need to specify an initial value.
Maximal cardinality of a double-free subset of {1, 2, ..., n}, or in other words, maximal size of a subset S of {1, 2, ..., n} with the property that if x is in S then 2x is not. a(0)=0 by convention.
Least k such that a(k)=n is equal to A003159(n).
To construct the sequence: let [a, b, c, a, a, a, b, c, a, b, c, ...] be the fixed point of the morphism a -> abc, b ->a, c -> a, starting from a(1) = a, then write the indices of a, b, c, that of a being written twice; see A092606. - Philippe Deléham, Apr 13 2004
Number of integers from {1,...,n} for which the subtraction of 1 changes the parity of the number of 1's in their binary expansion. - Vladimir Shevelev, Apr 15 2010
Number of integers from {1,...,n} the factorization of which over different terms of A050376 does not contain 2. - Vladimir Shevelev, Apr 16 2010
a(n) modulo 2 is the Prouhet-Thue-Morse sequence A010060. Each number n appears A026465(n+1) times. - Philippe Deléham, Oct 19 2011
Another way of stating the last two comments from Philippe Deléham: the sequence can be obtained by replacing each term of the Thue-Morse sequence A010060 by the run number that term is in. - N. J. A. Sloane, Dec 31 2013

			Examples for n = 1 through 8: {1}, {1}, {1,3}, {1,3,4}, {1,3,4,5}, {1,3,4,5}, {1,3,4,5,7}, {1,3,4,5,7}.
Binary expansion of 5 is 101, so Sum{i>=0} b_i*(-1)^i = 2. Therefore a(5) = 10/3 + 2/3 = 4. - _Vladimir Shevelev_, Apr 15 2010

S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.26.
Wang, E. T. H. "On Double-Free Sets of Integers." Ars Combin. 28, 97-100, 1989.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Steven R. Finch, Triple-Free Sets of Integers [From Steven Finch, Apr 20 2019]
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Double-Free Set.

Bisection of A123087.

Cf. A001045, A050291-A050296, A050321, A035263, A121701, A030308, A080277, A120385, A197911.

a050292 n = a050292_list !! (n-1)
a050292_list = scanl (+) 0 a035263_list
-- Reinhard Zumkeller, Jan 21 2013

A050292:=n->add((-1)^k*floor(n/2^k), k=0..n); seq(A050292(n), n=0..100); # Wesley Ivan Hurt, Feb 14 2014

a[n_] := a[n] = If[n < 2, 1, n - a[Floor[n/2]]]; Table[ a[n], {n, 1, 75}]
Join[{0},Accumulate[Nest[Flatten[#/.{0->{1,1},1->{1,0}}]&,{0},7]]] (* Harvey P. Dale, Apr 29 2018 *)

PARI
```
a(n)=if(n<2,1,n-a(floor(n/2)))
```

from sympy.ntheory import digits
def A050292(n): return ((n<<1)+sum((0,1,-1,0)[i] for i in digits(n,4)[1:]))//3 # Chai Wah Wu, Jan 30 2025

Partial sums of A035263. Close to (2/3)*n.
a(n) = A123087(2*n) = n - A123087(n). - Max Alekseyev, Mar 05 2023
From Benoit Cloitre, Nov 24 2002: (Start)
a(1)=1, a(n) = n - a(floor(n/2));
a(n) = (2/3)*n + (1/3)*A065359(n);
more generally, for m>=0, a(2^m*n) - 2^m*a(n) = A001045(m)*A065359(n) where A001045(m) = (2^m - (-1)^m)/3 is the Jacobsthal sequence;
a(A039004(n)) = (2/3)*A039004(n);
a(2*A039004(n)) = 2*a(A039004(n));
a(A003159(n)) = n;
a(A003159(n)-1) = n-1;
a(n) mod 2 = A010060(n) the Thue-Morse sequence;
a(n+1) - a(n) = A035263(n+1);
a(n+2) - a(n) = abs(A029884(n)).
(End)
G.f.: (1/(x-1)) * Sum_{i>=0} (-1)^i*x^(2^i)/(x^(2^i)-1). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 17 2003
a(n) = Sum_{k>=0} (-1)^k*floor(n/2^k). - Benoit Cloitre, Jun 03 2003
a(A091785(n)) = 2n; a(A091855(n)) = 2n-1. - Philippe Deléham, Mar 26 2004
a(2^n) = (2^(n+1) + (-1)^n)/3. - Vladimir Shevelev, Apr 15 2010
If n = Sum_{i>=0} b_i*2^i is the binary expansion of n, then a(n) = 2n/3 + (1/3)Sum_{i>=0} b_i*(-1)^i. Thus a(n) = 2n/3 + O(log(n)). - Vladimir Shevelev, Apr 15 2010
Moreover, the equation a(3m)=2m has infinitely many solutions, e.g., a(3*2^k)=2*2^k; on the other hand, a((4^k-1)/3)=(2*(4^k-1))/9+k/3, i.e., limsup |a(n)-2n/3| = infinity. - Vladimir Shevelev, Feb 23 2011
a(n) = Sum_{k>=0} A030308(n,k)*A001045(k+1). - Philippe Deléham, Oct 19 2011
From Peter Bala, Feb 02 2013: (Start)
Product_{n >= 1} (1 + x^((2^n - (-1)^n)/3 )) = (1 + x)^2(1 + x^3)(1 + x^5)(1 + x^11)(1 + x^21)... = 1 + sum {n >= 1} x^a(n) = 1 + 2x + x^2 + x^3 + 2x^4 + 2x^5 + .... Hence this sequence lists the numbers representable as a sum of distinct Jacobsthal numbers A001045 = [1, 1', 3, 5, 11, 21, ...], where we distinguish between the two occurrences of 1 by writing them as 1 and 1'. For example, 9 occurs twice in the present sequence because 9 = 5 + 3 + 1 and 9 = 5 + 3 + 1'. Cf. A197911 and A080277. See also A120385.
(End)

Extended with formula by Christian G. Bower, Sep 15 1999
Corrected and extended by Reinhard Zumkeller, Aug 16 2006
Extended with formula by Philippe Deléham, Oct 19 2011
Entry revised to give a simpler definition by N. J. A. Sloane, Jan 03 2014

A124757 Zero-based weighted sum of compositions in standard order.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 2, 3, 0, 1, 2, 3, 3, 4, 5, 6, 0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14

Offset: 0

Franklin T. Adams-Watters, Nov 06 2006

The standard order of compositions is given by A066099.

Sum of all positions of 1's except the last in the reversed binary expansion of n. For example, the reversed binary expansion of 14 is (0,1,1,1), so a(14) = 2 + 3 = 5. Keeping the last position gives A029931. - Gus Wiseman, Jan 17 2023

			Composition number 11 is 2,1,1; 0*2+1*1+2*1 = 3, so a(11) = 3.
The table starts:
  0
  0
  0 1
  0 1 2 3

Alois P. Heinz, Rows n = 0..14, flattened

Cf. A066099, A070939, A029931, A011782 (row lengths), A001788 (row sums).
Row sums of A048793 if we delete the last part of every row.
For prime indices instead of standard comps we have A359674, rev A359677.
Positions of first appearances are A359756.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion, reverse A030308.
A230877 adds up positions of 1's in binary expansion, length A000120.
A359359 adds up positions of 0's in binary expansion, length A023416.
Cf. A059015, A065359, A069010, A073642, A083652, A359400, A359402, A359678.

Table[Total[Most[Join@@Position[Reverse[IntegerDigits[n,2]],1]]],{n,30}]

For a composition b(1),...,b(k), a(n) = Sum_{i=1..k} (i-1)*b(i).

For n>0, a(n) = A029931(n) - A070939(n).

A030300 Runs have lengths 2^n, n >= 0.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 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, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Jean-Paul Delahaye

An example of a sequence with property that the fraction of 1's in the first n terms does not converge to a limit. - N. J. A. Sloane, Sep 24 2007
Image, under the coding sending a,d,e -> 1 and b,c -> 0, of the fixed point, starting with a, of the morphism a -> ab, b -> cd, c -> ee, d -> eb, e -> cc. - Jeffrey Shallit, May 14 2016
This sequence taken as digits of a base-b fraction is g(1/b) = Sum_{n>=1} a(n)/b^n = b/(b-1) * Sum_{k>=0} (-1)^k/b^(2^k) per the generating function below.  With initial 0, it is binary expansion .01001111 = A275975.  With initial 0 and digits 2*a(n), it is ternary expansion .02002222 = A160386.  These and in general g(1/b) for any integer b>=2 are among forms which Kempner showed are transcendental. - Kevin Ryde, Sep 07 2019

Antti Karttunen, Table of n, a(n) for n = 1..65537
Aubrey J. Kempner, On Transcendental Numbers, Transactions of the American Mathematical Society, volume 17, number 4, October 1916, pages 476-482.
Kevin Ryde, Plot of A079947(n)/n, illustrating proportion of 1s in the first n terms here does not converge (but oscillates with rises and falls by hyperbolas)
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Index entries for sequences related to binary expansion of n
Index entries for characteristic functions

Cf. A030301. Partial sums give A079947.
Cf. A065359, A083905.
Characteristic function of A053738.

f0 := n->[seq(0,i=1..2^n)]; f1 := n->[seq(1,i=1..2^n)]; s := []; for i from 0 to 4 do s := [op(s), op(f1(2*i)), op(f0(2*i+1))]; od: A030300 := s;

nMax = 6; Table[1 - Mod[n, 2], {n, 0, nMax}, {2^n}] // Flatten (* Jean-François Alcover, Oct 20 2016 *)

a(n) = if(n, !(logint(n,2)%2)); /* Kevin Ryde, Aug 02 2019 */

def A030300(n): return n.bit_length()&1 # Chai Wah Wu, Jan 30 2023

a(n) = A065359(n) + A083905(n).
a(n) = (1/2)*(1+(-1)^floor(log_2(n))). - Benoit Cloitre, Feb 22 2003
G.f.: 1/(1-x) * Sum_{k>=0} (-1)^k*x^2^k. - Ralf Stephan, Jul 12 2003
a(n) = 1 - a(floor(n/2)). - Vladeta Jovovic, Aug 04 2003
a(n) = A115253(2n, n) mod 2. - Paul Barry, Jan 18 2006
a(n) = 1 - A030301(n). - Antti Karttunen, Oct 10 2017

A005351 Base -2 representation for n regarded as base 2, then evaluated.

Original entry on oeis.org

0, 1, 6, 7, 4, 5, 26, 27, 24, 25, 30, 31, 28, 29, 18, 19, 16, 17, 22, 23, 20, 21, 106, 107, 104, 105, 110, 111, 108, 109, 98, 99, 96, 97, 102, 103, 100, 101, 122, 123, 120, 121, 126, 127, 124, 125, 114, 115, 112, 113, 118, 119, 116, 117, 74, 75, 72, 73, 78, 79, 76

Offset: 0

N. J. A. Sloane

a(n) = n when n is a power of 4. This is because the even-indexed powers of 2 are the same as the even-indexed powers of -2. - Alonso del Arte, Feb 09 2012
a(n) = n if n is a sum of distinct powers of 4. - Michael Somos, Aug 27 2012
Write n = Sum_{i in b(n)} (-2)^(i - 1), which uniquely determines the set of positive integers b(n). Then a(n) = Sum_{i in b(n)} 2^(i - 1). For example, a(7) = 27 because 7 = (-2)^0 + (-2)^1 + (-2)^3 + (-2)^4 and 27 = 2^0 + 2^1 + 2^3 + 2^4. - Gus Wiseman, Jul 26 2019

			2 = 4+(-2)+0 = 110 => 6, 3 = 4+(-2)+1 = 111 => 7, ..., 6 = (16)+(-8)+0+(-2)+0 = 11010 => 26.

M. Gardner, Knotted Doughnuts and Other Mathematical Entertainments. Freeman, NY, 1986, p. 101.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..8191
Joerg Arndt, Matters Computational (The Fxtbook), p. 58-59
Eric Weisstein's World of Mathematics, Negabinary
Wikipedia, Negative base
A. Wilks, Email, May 22 1991

Cf. A039724. Complement of A005352.
Cf. A185269 (primes in this sequence).
Cf. A000120, A029931, A035327, A053985, A065359, A070939, A326032.

a005351 0 = 0
a005351 n = a005351 n' * 2 + m where
   (n', m) = if r < 0 then (q + 1, r + 2) else (q, r)
             where (q, r) = quotRem n (negate 2)
-- Reinhard Zumkeller, Jul 07 2012

a[n_] := Module[{t = 2(4^Floor[ Log[4, Abs[n] + 1] + 2] - 1)/3}, BitXor[n + t, t]]; Table[a[n], {n, 0, 60}] (* Robert G. Wilson v, Jan 24 2005 *)

a(n) = my(t=(32*4^logint(abs(n)+1,4)-2)/3); bitxor(n+t,t); \\ Ruud H.G. van Tol, Oct 18 2023

def A005351(n):
    s, q = '', n
    while q >= 2 or q < 0:
        q, r = divmod(q, -2)
        if r < 0:
            q += 1
            r += 2
        s += str(r)
    return int(str(q)+s[::-1],2) # Chai Wah Wu, Apr 10 2016

a(4n+2) = 4a(n+1)+2, a(4n+3) = 4a(n+1)+3, a(4n+4) = 4a(n+1), a(4n+5) = 4a(n+1)+1, n>-2, a(1)=1. - Ralf Stephan, Apr 06 2004

More terms from Robert G. Wilson v, Jan 24 2005

A065368 Alternating sum of ternary digits in n. Replace 3^k with (-1)^k in ternary expansion of n.

Original entry on oeis.org

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

Offset: 0

Marc LeBrun, Oct 31 2001

Notation: (3)[n](-1).

Fixed point of the morphism 0 -> 0,1,2; 1 -> -1,0,1; 2 -> -2,-1,0; ...; n -> -n,-n+1,-n+2. - Philippe Deléham, Oct 22 2011

			15 = +1(9)+2(3)+0(1) -> +1(+1)+2(-1)+0(+1) = -1 = a(15).

Cf. A030341, A053735, A065359, A065364.

from sympy.ntheory.digits import digits
def a(n):
    return sum(bi*(-1)**k for k, bi in enumerate(digits(n, 3)[1:][::-1]))
print([a(n) for n in range(104)]) # Michael S. Branicky, Jul 28 2021

from sympy.ntheory import digits
def A065368(n): return sum((0, 1, 2, -1, 0, 1, -2, -1, 0)[i] for i in digits(n,9)[1:]) # Chai Wah Wu, Jul 19 2024

a(n) = Sum_{k>=0} A030341(n,k)*(-1)^k. - Philippe Deléham, Oct 22 2011.

G.f. A(x) satisfies: A(x) = x * (1 + 2*x) / (1 - x^3) - (1 + x + x^2) * A(x^3). - Ilya Gutkovskiy, Jul 28 2021

Initial 0 added by Philippe Deléham, Oct 22 2011

A346690 Replace 6^k with (-1)^k in base-6 expansion of n.

Original entry on oeis.org

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

Offset: 0

Ilya Gutkovskiy, Jul 29 2021

If n has base-6 expansion abc..xyz with least significant digit z, a(n) = z - y + x - w + ...

			59 = 135_6, 5 - 3 + 1 = 3, so a(59) = 3.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A007092, A053827, A055017, A065359, A065368, A346688, A346689, A346691.

f:= proc(n) option remember; (n mod 6) - procname(floor(n/6)) end proc:
f(0):= 0:
map(f, [$1..100]); # Robert Israel, Nov 21 2022

nmax = 104; A[] = 0; Do[A[x] = x (1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4)/(1 - x^6) - (1 + x + x^2 + x^3 + x^4 + x^5) A[x^6] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[n + 7 Sum[(-1)^k Floor[n/6^k], {k, 1, Floor[Log[6, n]]}], {n, 0, 104}]

a(n) = subst(Pol(digits(n, 6)), 'x, -1); \\ Michel Marcus, Nov 22 2022

from sympy.ntheory.digits import digits
def a(n):
    return sum(bi*(-1)**k for k, bi in enumerate(digits(n, 6)[1:][::-1]))
print([a(n) for n in range(105)]) # Michael S. Branicky, Jul 29 2021

G.f. A(x) satisfies: A(x) = x * (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4) / (1 - x^6) - (1 + x + x^2 + x^3 + x^4 + x^5) * A(x^6).
a(n) = n + 7 * Sum_{k>=1} (-1)^k * floor(n/6^k).
a(6*n+j) = j - a(n) for 0 <= j <= 5. - Robert Israel, Nov 21 2022

cp's OEIS Frontend

A020988 a(n) = (2/3)*(4^n-1).

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A055017 Difference between sums of alternate digits of n starting with the last, i.e., (sum of ultimate digit of n, antepenultimate digit of n, ...) - (sum of penultimate digit of n, preantepenultimate digit of n, ...).

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A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

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A359359 Sum of positions of zeros in the binary expansion of n, where positions are read starting with 1 from the left (big-endian).

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A050292 a(2n) = 2n - a(n), a(2n+1) = 2n + 1 - a(n) (for n >= 0).

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A124757 Zero-based weighted sum of compositions in standard order.

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