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.

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A003159 Numbers whose binary representation ends in an even number of zeros.

Original entry on oeis.org

1, 3, 4, 5, 7, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 23, 25, 27, 28, 29, 31, 33, 35, 36, 37, 39, 41, 43, 44, 45, 47, 48, 49, 51, 52, 53, 55, 57, 59, 60, 61, 63, 64, 65, 67, 68, 69, 71, 73, 75, 76, 77, 79, 80, 81, 83, 84, 85, 87, 89, 91, 92, 93, 95, 97, 99, 100, 101, 103, 105
Offset: 1

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Author

Keywords

Comments

Fraenkel (2010) called these the "vile" numbers.
Minimal with respect to property that parity of number of 1's in binary expansion alternates.
Minimal with respect to property that sequence is half its complement. [Corrected by Aviezri S. Fraenkel, Jan 29 2010]
If k appears then 2k does not.
Increasing sequence of positive integers k such that A035263(k)=1 (from paper by Allouche et al.). - Emeric Deutsch, Jan 15 2003
a(n) is an odious number (see A000069) for n odd; a(n) is an evil number (see A001969) for n even. - Philippe Deléham, Mar 16 2004
Indices of odd numbers in A007913, in A001511. - Philippe Deléham, Mar 27 2004
Partial sums of A026465. - Paul Barry, Dec 09 2004
A121701(2*a(n)) = A121701(a(n)); A096268(a(n)-1) = 0. - Reinhard Zumkeller, Aug 16 2006
A different permutation of the same terms may be found in A141290 and the accompanying array. - Gary W. Adamson, Jun 14 2008
a(n) = n-th clockwise Tower of Hanoi move; counterclockwise if not in the sequence. - Gary W. Adamson, Jun 14 2008
Indices of terms of Thue-Morse sequence A010060 which are different from the previous term. - Tanya Khovanova, Jan 06 2009
The sequence has the following fractal property. Remove the odd numbers from the sequence, leaving 4,12,16,20,28,36,44,48,52,... Dividing these terms by 4 we get 1,3,4,5,7,9,11,12,..., which is the original sequence back again. - Benoit Cloitre, Apr 06 2010
From Gary W. Adamson, Mar 21 2010: (Start)
A conjectured identity relating to the partition sequence, A000041 as polcoeff p(x); A003159, and its characteristic function A035263: (1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); and A036554 indicating n-th terms with zeros in A035263: (2, 6, 8, 10, 14, 18, 22, ...).
The conjecture states that p(x) = A(x) = A(x^2) when A(x) = polcoeff A174065 = the Euler transform of A035263 = 1/((1-x)*(1-x^3)*(1-x^4)*(1-x^5)*...) = 1 + x + x^2 + 2*x^3 + 3*x^4 + 4*x^5 + ... and the aerated variant = the Euler transform of the complement of A035263: 1/((1-x^2)*(1-x^6)*(1-x^8)*...) = 1 + x^2 + x^4 + 2*x^6 + 3*x^8 + 4*x^10 + ....
(End)
The conjecture above was proved by Jean-Paul Allouche on Dec 21 2013. - Gary W. Adamson, Jan 22 2014
If the lower s-Wythoff sequence of s is s, then s=A003159. (See A184117 for the definition of lower and upper s-Wythoff sequences.) Starting with any nondecreasing sequence s of positive integers, A003159 is the limit when the lower s-Wythoff operation is iterated. For example, starting with s=(1,4,9,16,...)=(n^2), we obtain lower and upper s-Wythoff sequences
a=(1,3,4,5,6,8,9,10,11,12,14,...)=A184427;
b=(2,7,12,21,31,44,58,74,...)=A184428.
Then putting s=a and repeating the operation gives a'=(1,3,4,5,7,9,11,12,14,...), which has the same first eight terms as A003159. - Clark Kimberling, Jan 14 2011

Examples

			1=1, 3=11, 5=101 and 7=111 have no (0 = even) trailing zeros, 4=100 has 2 (= even) trailing zeros in the base-2 representation.
2=10 and 6=110 end in one (=odd number) of trailing zeros in their base-2 representation, therefore are not terms of this sequence. - _M. F. Hasler_, Oct 29 2013
		

References

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

Crossrefs

For the actual binary numbers see A280049.
Indices of even numbers in A007814.
Complement of A036554, also one-half of A036554.

Programs

  • Haskell
    import Data.List (delete)
    a003159 n = a003159_list !! (n-1)
    a003159_list = f [1..] where f (x:xs) = x : f (delete  (2*x) xs)
    -- Reinhard Zumkeller, Nov 04 2011
    
  • Maple
    filter:= n -> type(padic:-ordp(n,2),even):
    select(filter,[$1..1000]); # Robert Israel, Jul 07 2014
  • Mathematica
    f[n_Integer] := Block[{k = n, c = 0}, While[ EvenQ[k], c++; k /= 2]; c]; Select[ Range[105], EvenQ[ f[ # ]] & ]
    Select[Range[150],EvenQ[IntegerExponent[#,2]]&] (* Harvey P. Dale, Oct 19 2011 *)
  • PARI
    a(n)=if(n<2,n>0,n=a(n-1); until(valuation(n,2)%2==0,n++); n)
    
  • PARI
    is(n)=valuation(n,2)%2==0 \\ Charles R Greathouse IV, Sep 23 2012
    
  • Python
    from itertools import count, islice
    def A003159_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda n:(n&-n).bit_length()&1,count(max(startvalue,1)))
    A003159_list = list(islice(A003159_gen(),30)) # Chai Wah Wu, Jul 11 2022
    
  • Python
    def A003159(n):
        def f(x):
            c, s = n+x, bin(x)[2:]
            l = len(s)
            for i in range(l&1^1,l,2):
                c -= int(s[i])+int('0'+s[:i],2)
            return c
        m, k = n, f(n)
        while m != k: m, k = k, f(k)
        return m # Chai Wah Wu, Jan 29 2025

Formula

a(0) = 1; for n >= 0, a(n+1) = a(n) + 1 if (a(n) + 1)/2 is not already in the sequence, = a(n) + 2 otherwise.
Limit_{n->oo} a(n)/n = 3/2. - Benoit Cloitre, Jun 13 2002
More precisely, a(n) = 3*n/2 + O(log n). - Charles R Greathouse IV, Sep 23 2012
a(n) = Sum_{k = 1..n} A026465(k). - Benoit Cloitre, May 31 2003
a(n+1) = (if a(n) mod 4 = 3 then A007814(a(n) + 1) mod 2 else a(n) mod 2) + a(n) + 1; a(1) = 1. - Reinhard Zumkeller, Aug 03 2003
a(A003157(n)) is even. - Philippe Deléham, Feb 22 2004
Sequence consists of numbers of the form 4^i*(2*j + 1), i>=0, j>=0. - Jon Perry, Jun 06 2004
G.f.: (1/(1-x)) * Product_{k >= 1} (1 + x^A001045(k)). - Paul Barry, Dec 09 2004
a(1) = 1, a(2) = 3, and for n >= 2 we get a(n+1) = 4 + a(n) + a(n-1) - a(a(n)-n+1) - a(a(n-1)-n+2). - Benoit Cloitre, Apr 08 2010
If A(x) is the counting function for a(n) <= x, then A(2^n) = (2^(n+1) + (-1)^n)/3. - Vladimir Shevelev, Apr 15 2010
a(n) = A121539(n) + 1. - Reinhard Zumkeller, Mar 01 2012
A003159 = { N | A007814(N) is even }. - M. F. Hasler, Oct 29 2013

Extensions

Additional comments from Michael Somos
Edited by M. F. Hasler, Oct 29 2013
Incorrect formula removed by Peter Munn, Dec 04 2020

A035263 Trajectory of 1 under the morphism 0 -> 11, 1 -> 10; parity of 2-adic valuation of 2n: a(n) = A000035(A001511(n)).

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1
Offset: 1

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Author

Keywords

Comments

First Feigenbaum symbolic (or period-doubling) sequence, corresponding to the accumulation point of the 2^{k} cycles through successive bifurcations.
To construct the sequence: start with 1 and concatenate: 1,1, then change the last term (1->0; 0->1) gives: 1,0. Concatenate those 2 terms: 1,0,1,0, change the last term: 1,0,1,1. Concatenate those 4 terms: 1,0,1,1,1,0,1,1 change the last term: 1,0,1,1,1,0,1,0, etc. - Benoit Cloitre, Dec 17 2002
Let T denote the present sequence. Here is another way to construct T. Start with the sequence S = 1,0,1,,1,0,1,,1,0,1,,1,0,1,,... and fill in the successive holes with the successive terms of the sequence T (from paper by Allouche et al.). - Emeric Deutsch, Jan 08 2003 [Note that if we fill in the holes with the terms of S itself, we get A141260. - N. J. A. Sloane, Jan 14 2009]
From N. J. A. Sloane, Feb 27 2009: (Start)
In more detail: define S to be 1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1,0,1___...
If we fill the holes with S we get A141260:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........0.......1.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.0.1.... = A141260.
But instead, if we define T recursively by filling the holes in S with the terms of T itself, we get A035263:
1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,
........1.........0.........1.........1.........1.......0.........1.........0...
- the result is
1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.1.1.0.1.0.1..0..1.1.1..0..1.0.1.. = A035263. (End)
Characteristic function of A003159, i.e., A035263(n)=1 if n is in A003159 and A035263(n)=0 otherwise (from paper by Allouche et al.). - Emeric Deutsch, Jan 15 2003
This is the sequence of R (=1), L (=0) moves in the Towers of Hanoi puzzle: R, L, R, R, R, L, R, L, R, L, R, R, R, ... - Gary W. Adamson, Sep 21 2003
Manfred Schroeder, p. 279 states, "... the kneading sequences for unimodal maps in the binary notation, 0, 1, 0, 1, 1, 1, 0, 1..., are obtained from the Morse-Thue sequence by taking sums mod 2 of adjacent elements." On p. 278, in the chapter "Self-Similarity in the Logistic Parabola", he writes, "Is there a closer connection between the Morse-Thue sequence and the symbolic dynamics of the superstable orbits? There is indeed. To see this, let us replace R by 1 and C and L by 0." - Gary W. Adamson, Sep 21 2003
Partial sums modulo 2 of the sequence 1, a(1), a(1), a(2), a(2), a(3), a(3), a(4), a(4), a(5), a(5), a(6), a(6), ... . - Philippe Deléham, Jan 02 2004
Parity of A007913, A065882 and A065883. - Philippe Deléham, Mar 28 2004
The length of n-th run of 1's in this sequence is A080426(n). - Philippe Deléham, Apr 19 2004
Also parity of A005043, A005773, A026378, A104455, A117641. - Philippe Deléham, Apr 28 2007
Equals parity of the Towers of Hanoi, or ruler sequence (A001511), where the Towers of Hanoi sequence (1, 2, 1, 3, 1, 2, 1, 4, ...) denotes the disc moved, labeled (1, 2, 3, ...) starting from the top; and the parity of (1, 2, 1, 3, ...) denotes the direction of the move, CW or CCW. The frequency of CW moves converges to 2/3. - Gary W. Adamson, May 11 2007
A conjectured identity relating to the partition sequence, A000041: p(x) = A(x) * A(x^2) when A(x) = the Euler transform of A035263 = polcoeff A174065: (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...). - Gary W. Adamson, Mar 21 2010
a(n) is 1 if the number of trailing zeros in the binary representation of n is even. - Ralf Stephan, Aug 22 2013
From Gary W. Adamson, Mar 25 2015: (Start)
A conjectured identity relating to the partition sequence, A000041 as polcoeff p(x); A003159, and its characteristic function A035263: (1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, ...); and A036554 indicating n-th terms with zeros in A035263: (2, 6, 8, 10, 14, 18, 22, ...).
The conjecture states that p(x) = A(x) = A(x^2) when A(x) = polcoeffA174065 = the Euler transform of A035263 = 1/(1-x)*(1-x^3)*(1-x^4)*(1-x^5)*... = (1 + x + x^2 + 2x^3 + 3x^4 + 4x^5 + ...) and the aerated variant = the Euler transform of the complement of A035263: 1/(1-x^2)*(1-x^6)*(1-x^8)*... = (1 + x^2 + x^4 + 2x^6 + 3x^8 + 4x^10 + ...).
(End)
The conjecture above was proved by Jean-Paul Allouche on Dec 21 2013.
Regarded as a column vector, this sequence is the product of A047999 (Sierpinski's gasket) regarded as an infinite lower triangular matrix and A036497 (the Fredholm-Rueppel sequence) where the 1's have alternating signs, 1, -1, 0, 1, 0, 0, 0, -1, .... - Gary W. Adamson, Jun 02 2021
The numbers of 1's through n (A050292) can be determined by starting with the binary (say for 19 = 1 0 0 1 1) and writing: next term is twice current term if 0, otherwise twice plus 1. The result is 1, 2, 4, 9, 19. Take the difference row, = 1, 1, 2, 5, 10; and add the odd-indexed terms from the right: 5, 4, 3, 2, 1 = 10 + 2 + 1 = 13. The algorithm is the basis for determining the disc configurations in the tower of Hanoi game, as shown in the Jul 24 2021 comment of A060572. - Gary W. Adamson, Jul 28 2021

References

  • Karamanos, Kostas. "From symbolic dynamics to a digital approach." International Journal of Bifurcation and Chaos 11.06 (2001): 1683-1694. (Full version. See p. 1685)
  • Karamanos, K. (2000). From symbolic dynamics to a digital approach: chaos and transcendence. In Michel Planat (Ed.), Noise, Oscillators and Algebraic Randomness (Lecture Notes in Physics, pp. 357-371). Springer, Berlin, Heidelberg. (Short version. See p. 359)
  • Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W. H. Freeman, 1991
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 892, column 2, Note on p. 84, part (a).

Crossrefs

Parity of A001511. Anti-parity of A007814.
Absolute values of first differences of A010060. Apart from signs, same as A029883. Essentially the same as A056832.
Swapping 0 and 1 gives A096268.
Cf. A033485, A050292 (partial sums), A089608, A088172, A019300, A039982, A073675, A121701, A141260, A000041, A174065, A220466, A154269 (Mobius transform).
Limit of A317957(n) for large n.

Programs

  • Haskell
    import Data.Bits (xor)
    a035263 n = a035263_list !! (n-1)
    a035263_list = zipWith xor a010060_list $ tail a010060_list
    -- Reinhard Zumkeller, Mar 01 2012
    
  • Maple
    nmax:=105: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := (p+1) mod 2 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 07 2013
    A035263 := n -> 1 - padic[ordp](n, 2) mod 2:
    seq(A035263(n), n=1..105); # Peter Luschny, Oct 02 2018
  • Mathematica
    a[n_] := a[n] = If[ EvenQ[n], 1 - a[n/2], 1]; Table[ a[n], {n, 1, 105}] (* Or *)
    Rest[ CoefficientList[ Series[ Sum[ x^(2^k)/(1 + (-1)^k*x^(2^k)), {k, 0, 20}], {x, 0, 105}], x]]
    f[1] := True; f[x_] := Xor[f[x - 1], f[Floor[x/2]]]; a[x_] := Boole[f[x]] (* Ben Branman, Oct 04 2010 *)
    a[n_] := If[n == 0, 0, 1 - Mod[ IntegerExponent[n, 2], 2]]; (* Jean-François Alcover, Jul 19 2013, after Michael Somos *)
    Nest[ Flatten[# /. {0 -> {1, 1}, 1 -> {1, 0}}] &, {0}, 7] (* Robert G. Wilson v, Jul 23 2014 *)
    SubstitutionSystem[{0->{1,1},1->{1,0}},1,{7}][[1]] (* Harvey P. Dale, Jun 06 2022 *)
  • PARI
    {a(n) = if( n==0, 0, 1 - valuation(n, 2)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); subst( Pol(binary(n)) - Pol(binary(n-1)), x, 1)%2)}; /* Michael Somos, Sep 04 2006 */
    
  • PARI
    {a(n) = if( n==0, 0, n = abs(n); direuler(p=2, n, 1 / (1 - X^((p<3) + 1)))[n])}; /* Michael Somos, Sep 04 2006 */
    
  • Python
    def A035263(n): return (n&-n).bit_length()&1 # Chai Wah Wu, Jan 09 2023
  • Scheme
    (define (A035263 n) (let loop ((n n) (i 1)) (cond ((odd? n) (modulo i 2)) (else (loop (/ n 2) (+ 1 i)))))) ;; (Use mod instead of modulo in R6RS) Antti Karttunen, Sep 11 2017
    

Formula

Absolute values of first differences (A029883) of Thue-Morse sequence (A001285 or A010060). Self-similar under 10->1 and 11->0.
Series expansion: (1/x) * Sum_{i>=0} (-1)^(i+1)*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+1)/2^k)-floor(n/2^k)). - Benoit Cloitre, Jun 03 2003
Another g.f.: Sum_{k>=0} x^(2^k)/(1+(-1)^k*x^(2^k)). - Ralf Stephan, Jun 13 2003
a(2*n) = 1-a(n), a(2*n+1) = 1. - Ralf Stephan, Jun 13 2003
a(n) = parity of A033485(n). - Philippe Deléham, Aug 13 2003
Equals A088172 mod 2, where A088172 = 1, 2, 3, 7, 13, 26, 53, 106, 211, 422, 845, ... (first differences of A019300). - Gary W. Adamson, Sep 21 2003
a(n) = a(n-1) - (-1)^n*a(floor(n/2)). - Benoit Cloitre, Dec 02 2003
a(1) = 1 and a(n) = abs(a(n-1) - a(floor(n/2))). - Benoit Cloitre, Dec 02 2003
a(n) = 1 - A096268(n+1); A050292 gives partial sums. - Reinhard Zumkeller, Aug 16 2006
Multiplicative with a(2^k) = 1 - (k mod 2), a(p^k) = 1, p > 2. Dirichlet g.f.: Product_{n = 4 or an odd prime} (1/(1-1/n^s)). - Christian G. Bower, May 18 2005
a(-n) = a(n). a(0)=0. - Michael Somos, Sep 04 2006
Dirichlet g.f.: zeta(s)*2^s/(2^s+1). - Ralf Stephan, Jun 17 2007
a(n+1) = a(n) XOR a(ceiling(n/2)), a(1) = 1. - Reinhard Zumkeller, Jun 11 2009
Let D(x) be the generating function, then D(x) + D(x^2) == x/(1-x). - Joerg Arndt, May 11 2010
a(n) = A010060(n) XOR A010060(n+1); a(A079523(n)) = 0; a(A121539(n)) = 1. - Reinhard Zumkeller, Mar 01 2012
a((2*n-1)*2^p) = (p+1) mod 2, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 07 2013
a(n) = A000035(A001511(n)). - Omar E. Pol, Oct 29 2013
a(n) = 2-A056832(n) = (5-A089608(n))/4. - Antti Karttunen, Sep 11 2017, after Benoit Cloitre
For n >= 0, a(n+1) = M(2n) mod 2 where M(n) is the Motzkin number A001006 (see Deutsch and Sagan 2006 link). - David Callan, Oct 02 2018
a(n) = A038712(n) mod 3. - Kevin Ryde, Jul 11 2019
Given any n in the form (k * 2^m, k odd), extract k and m. Categorize the results into two outcomes of (k, m, even or odd). If (k, m) is (odd, even) substitute 1. If (odd, odd), denote the result 0. Example: 5 = (5 * 2^0), (odd, even, = 1). (6 = 3 * 2^1), (odd, odd, = 0). - Gary W. Adamson, Jun 23 2021

Extensions

Alternative description added to the name by Antti Karttunen, Sep 11 2017

A081706 Numbers n such that binary representation ends either in an odd number of ones followed by one zero or in an even number of ones.

Original entry on oeis.org

2, 3, 10, 11, 14, 15, 18, 19, 26, 27, 34, 35, 42, 43, 46, 47, 50, 51, 58, 59, 62, 63, 66, 67, 74, 75, 78, 79, 82, 83, 90, 91, 98, 99, 106, 107, 110, 111, 114, 115, 122, 123, 130, 131, 138, 139, 142, 143, 146, 147, 154, 155, 162, 163, 170, 171, 174, 175, 178, 179, 186
Offset: 1

Views

Author

Emeric Deutsch and Bruce E. Sagan, Apr 02 2003

Keywords

Comments

Values of k such that the Motzkin number A001006(k) is even. Values of k such that the number of restricted hexagonal polyominoes with k+1 cells (A002212) is even.
Or union of sequences {2*A079523(n)+k}, k=0,1. A generalization see in comment to A161639. - Vladimir Shevelev, Jun 15 2009
Or intersection of sequences A121539 and {A121539(n)-1}. A generalization see in comment to A161890. - Vladimir Shevelev, Jul 03 2009
Also numbers n for which A010060(n+2) = A010060(n). - Vladimir Shevelev, Jul 06 2009
The asymptotic density of this sequence is 1/3 (Rowland and Yassawi, 2015; Burns, 2016). - Amiram Eldar, Jan 30 2021
Numbers of the form 4^k*(2*n-1)-2 and 4^k*(2*n-1)-1 where n and k are positive integers. - Michael Somos, Oct 22 2021

Crossrefs

Programs

  • Mathematica
    (* m = MotzkinNumber *) m[0] = 1; m[n_] := m[n] = m[n - 1] + Sum[m[k]*m[n - 2 - k], {k, 0, n - 2}]; Select[Range[200], Mod[m[#], 2] == 0 &] (* Jean-François Alcover, Jul 10 2013 *)
    Select[Range[200], EvenQ@Hypergeometric2F1[3/2, -#, 3, 4]&] (* Vladimir Reshetnikov, Nov 02 2015 *)
  • PARI
    is(n)=valuation(bitor(n,1)+1,2)%2==0 \\ Charles R Greathouse IV, Mar 07 2013
    
  • Python
    from itertools import count, islice
    def A081706_gen(): # generator of terms
        for n in count(0):
            if (n&-n).bit_length()&1:
                m = n<<2
                yield m-2
                yield m-1
    A081706_list = list(islice(A081706_gen(),30)) # Chai Wah Wu, Jan 09 2023
    
  • Python
    def A081706(n):
        def f(x):
            c, s = (n+1>>1)+x, bin(x)[2:]
            l = len(s)
            for i in range(l&1^1,l,2):
                c -= int(s[i])+int('0'+s[:i],2)
            return c
        m, k = n+1>>1, f(n+1>>1)
        while m != k: m, k = k, f(k)
        return (m<<2)-1-(n&1) # Chai Wah Wu, Jan 29 2025

Formula

a(2n-1) = 2*A079523(n) = 4*A003159(n)-2; a(2n) = 4*A003159(n)-1.
Note that a(2n) = 1+a(2n-1).

A131323 Odd numbers whose binary expansion ends in an even number of 1's.

Original entry on oeis.org

3, 11, 15, 19, 27, 35, 43, 47, 51, 59, 63, 67, 75, 79, 83, 91, 99, 107, 111, 115, 123, 131, 139, 143, 147, 155, 163, 171, 175, 179, 187, 191, 195, 203, 207, 211, 219, 227, 235, 239, 243, 251, 255, 259, 267, 271, 275, 283, 291, 299, 303, 307, 315, 319, 323, 331
Offset: 1

Views

Author

Nadia Heninger and N. J. A. Sloane, Dec 16 2007

Keywords

Comments

Also numbers of the form (4^a)*b - 1 with positive integer a and odd integer b. The sequence has linear growth and the limit of a(n)/n is 6. - Stefan Steinerberger, Dec 18 2007
Evil and odious terms alternate. - Vladimir Shevelev, Jun 22 2009
Also odd numbers of the form m = (A079523(k)-1)/2. - Vladimir Shevelev, Jul 06 2009
As a set, this is the complement of A079523 in the odd numbers. - Michel Dekking, Feb 13 2019
From Ctibor O. Zizka, Dec 28 2024: (Start)
Numbers k >= 1 such that (k + 1)*(k + 2*r)/2 is not a square for any r >= 1.
Numbers k such that A076114(k + 1) = 0. (End)

Examples

			11 in binary is 1011, which ends with two 1's.
		

Crossrefs

Programs

  • Maple
    N:= 1000: # to get all terms up to N
    Odds:= [seq(2*i+1,i=0..floor((N-1)/2)]:
    f:= proc(n) local L,x;
       L:= convert(n,base,2);
       x:= ListTools:-Search(0,L);
       if x = 0 then type(nops(L),even) else type(x,odd) fi
    end proc:
    A131323:= select(f,Odds); # Robert Israel, Apr 02 2014
  • Mathematica
    Select[Range[500], OddQ[ # ] && EvenQ[FactorInteger[ # + 1][[1, 2]]] &] (* Stefan Steinerberger, Dec 18 2007 *)
    en1Q[n_]:=Module[{ll=Last[Split[IntegerDigits[n,2]]]},Union[ll] =={1} &&EvenQ[Length[ll]]]; Select[Range[1,501,2],en1Q] (* Harvey P. Dale, May 18 2011 *)
  • PARI
    is(n)=n%2 && valuation(n+1,2)%2==0 \\ Charles R Greathouse IV, Aug 20 2013
    
  • Python
    from itertools import count, islice
    def A131323_gen(startvalue=3): # generator of terms >= startvalue
        return map(lambda n:(n<<1)+1,filter(lambda n:(~(n+1)&n).bit_length()&1,count(max(startvalue>>1,1))))
    A131323_list = list(islice(A131323_gen(),30)) # Chai Wah Wu, Sep 11 2024
    
  • Python
    def A131323(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x):
            c, s = n+x, bin(x+1)[2:]
            l = len(s)
            for i in range(l&1,l,2):
                c -= int(s[i])+int('0'+s[:i],2)
            return c
        return bisection(f,n,n)<<1|1 # Chai Wah Wu, Jan 29 2025

Formula

a(n) = 2*A079523(n) + 1. - Michel Dekking, Feb 13 2019

Extensions

More terms from Stefan Steinerberger, Dec 18 2007

A342050 Numbers k which have an odd number of trailing zeros in their primorial base representation A049345(k).

Original entry on oeis.org

2, 4, 8, 10, 14, 16, 20, 22, 26, 28, 30, 32, 34, 38, 40, 44, 46, 50, 52, 56, 58, 60, 62, 64, 68, 70, 74, 76, 80, 82, 86, 88, 90, 92, 94, 98, 100, 104, 106, 110, 112, 116, 118, 120, 122, 124, 128, 130, 134, 136, 140, 142, 146, 148, 150, 152, 154, 158, 160, 164, 166, 170, 172, 176, 178, 180, 182, 184, 188, 190, 194, 196, 200, 202, 206, 208, 212
Offset: 1

Views

Author

Amiram Eldar, Feb 26 2021

Keywords

Comments

Numbers k such that A276084(k) is odd.
All the terms are even since odd numbers have 0 trailing zeros, and 0 is not odd.
The number of terms not exceeding A002110(m) for m>=1 is A002110(m) * Sum_{k=1..m}(-1)^k/A002110(k) = 1, 2, 11, 76, 837, 10880, 184961, ...
The asymptotic density of this sequence is Sum_{k>=1} (-1)^(k+1)/A002110(k) = 0.362306... (A132120).
Also Heinz numbers of partitions with even least gap. The least gap (mex or minimal excludant) of a partition is the least positive integer that is not a part. The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), giving a bijective correspondence between positive integers and integer partitions. - Gus Wiseman, Apr 23 2021
Numbers k such that A000720(A053669(k)) is even. Differences from the related A353531 seem to be terms that are multiples of 210, but not all of them, for example primorial 30030 (= 143*210) is in neither sequence. Consider also A038698. - Antti Karttunen, Apr 25 2022

Examples

			2 is a term since A049345(2) = 10 has 1 trailing zero.
4 is a term since A049345(2) = 20 has 1 trailing zero.
30 is a term since A049345(2) = 1000 has 3 trailing zeros.
From _Gus Wiseman_, Apr 23 2021: (Start)
The sequence of terms together with their prime indices begins:
      2: {1}             46: {1,9}             90: {1,2,2,3}
      4: {1,1}           50: {1,3,3}           92: {1,1,9}
      8: {1,1,1}         52: {1,1,6}           94: {1,15}
     10: {1,3}           56: {1,1,1,4}         98: {1,4,4}
     14: {1,4}           58: {1,10}           100: {1,1,3,3}
     16: {1,1,1,1}       60: {1,1,2,3}        104: {1,1,1,6}
     20: {1,1,3}         62: {1,11}           106: {1,16}
     22: {1,5}           64: {1,1,1,1,1,1}    110: {1,3,5}
     26: {1,6}           68: {1,1,7}          112: {1,1,1,1,4}
     28: {1,1,4}         70: {1,3,4}          116: {1,1,10}
     30: {1,2,3}         74: {1,12}           118: {1,17}
     32: {1,1,1,1,1}     76: {1,1,8}          120: {1,1,1,2,3}
     34: {1,7}           80: {1,1,1,1,3}      122: {1,18}
     38: {1,8}           82: {1,13}           124: {1,1,11}
     40: {1,1,1,3}       86: {1,14}           128: {1,1,1,1,1,1,1}
     44: {1,1,5}         88: {1,1,1,5}        130: {1,3,6}
(End)
		

Crossrefs

Complement of A342051.
A099800 is subsequence.
Analogous sequences: A001950 (Zeckendorf representation), A036554 (binary), A145204 (ternary), A217319 (base 4), A232745 (factorial base).
The version for reversed binary expansion is A079523.
Positions of even terms in A257993.
A000070 counts partitions with a selected part.
A056239 adds up prime indices, row sums of A112798.
A073491 lists numbers with gap-free prime indices.
A079067 counts gaps in prime indices.
A238709 counts partitions by sum and least difference.
A333214 lists positions of adjacent unequal prime gaps.
A339662 gives greatest gap in prime indices.
Differs from A353531 for the first time at n=77, where a(77) = 212, as this sequence misses A353531(77) = 210.

Programs

  • Mathematica
    seq[max_] := Module[{bases = Prime@Range[max, 1, -1], nmax}, nmax = Times @@ bases - 1; Select[Range[nmax], OddQ @ LengthWhile[Reverse @ IntegerDigits[#, MixedRadix[bases]], #1 == 0 &] &]]; seq[4]
    Select[Range[100],EvenQ[Min@@Complement[Range[PrimeNu[#]+1],PrimePi/@First/@FactorInteger[#]]]&] (* Gus Wiseman, Apr 23 2021 *)
  • PARI
    A353525(n) = { for(i=1,oo,if(n%prime(i),return((i+1)%2))); }
    isA342050(n) = A353525(n);
    k=0; n=0; while(k<77, n++; if(isA342050(n), k++; print1(n,", "))); \\ Antti Karttunen, Apr 25 2022

Extensions

More terms added (to differentiate from A353531) by Antti Karttunen, Apr 25 2022

A342051 Numbers k which have an even number of trailing zeros in their primorial base representation A049345(k).

Original entry on oeis.org

1, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19, 21, 23, 24, 25, 27, 29, 31, 33, 35, 36, 37, 39, 41, 42, 43, 45, 47, 48, 49, 51, 53, 54, 55, 57, 59, 61, 63, 65, 66, 67, 69, 71, 72, 73, 75, 77, 78, 79, 81, 83, 84, 85, 87, 89, 91, 93, 95, 96, 97, 99, 101, 102, 103
Offset: 1

Views

Author

Amiram Eldar, Feb 26 2021

Keywords

Comments

Numbers k such that A276084(k) is even.
The number of terms not exceeding A002110(m) for m>=1 is A002110(m) * (1 - Sum_{k=1..m}(-1)^k/A002110(k)) = 1, 4, 19, 134, 1473, 19150, 325549 ...
The asymptotic density of this sequence is Sum_{k>=0} (-1)^k/A002110(k) = 0.637693... = 1 - A132120.
Also Heinz numbers of partitions with odd least gap. The least gap (mex or minimal excludant) of a partition is the least positive integer that is not a part. The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), giving a bijective correspondence between positive integers and integer partitions. - Gus Wiseman, Apr 23 2021

Examples

			1 is a term since A049345(1) = 1 has 0 trailing zero.
6 is a term since A049345(6) = 100 has 2 trailing zeros.
From _Gus Wiseman_, Apr 23 2021: (Start)
The sequence of terms together with their prime indices begins:
     1: {}           25: {3,3}          51: {2,7}
     3: {2}          27: {2,2,2}        53: {16}
     5: {3}          29: {10}           54: {1,2,2,2}
     6: {1,2}        31: {11}           55: {3,5}
     7: {4}          33: {2,5}          57: {2,8}
     9: {2,2}        35: {3,4}          59: {17}
    11: {5}          36: {1,1,2,2}      61: {18}
    12: {1,1,2}      37: {12}           63: {2,2,4}
    13: {6}          39: {2,6}          65: {3,6}
    15: {2,3}        41: {13}           66: {1,2,5}
    17: {7}          42: {1,2,4}        67: {19}
    18: {1,2,2}      43: {14}           69: {2,9}
    19: {8}          45: {2,2,3}        71: {20}
    21: {2,4}        47: {15}           72: {1,1,1,2,2}
    23: {9}          48: {1,1,1,1,2}    73: {21}
    24: {1,1,1,2}    49: {4,4}          75: {2,3,3}
(End)
		

Crossrefs

Complement of A342050.
A099788 is subsequence.
Analogous sequences: A000201 (Zeckendorf representation), A003159 (binary), A007417 (ternary), A232744 (factorial base).
The version for reversed binary expansion is A121539.
Positions of odd terms in A257993.
A000070 counts partitions with a selected part.
A056239 adds up prime indices, row sums of A112798.
A073491 lists numbers with gap-free prime indices.
A079067 counts gaps in prime indices.
A238709 counts partitions by sum and least difference.
A339662 gives greatest gap in prime indices.

Programs

  • Mathematica
    seq[max_] := Module[{bases = Prime@Range[max, 1, -1], nmax}, nmax = Times @@ bases - 1; Select[Range[nmax], EvenQ @ LengthWhile[Reverse @ IntegerDigits[#, MixedRadix[bases]], #1 == 0 &] &]]; seq[4]
    Select[Range[100],OddQ[Min@@Complement[Range[PrimeNu[#]+1],PrimePi/@First/@FactorInteger[#]]]&] (* Gus Wiseman, Apr 23 2021 *)

A161579 Positions n such that A010060(n) = A010060(n+3).

Original entry on oeis.org

0, 1, 3, 4, 6, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 22, 24, 25, 27, 28, 30, 32, 33, 35, 36, 38, 40, 41, 43, 44, 45, 47, 48, 49, 51, 52, 54, 56, 57, 59, 60, 61, 63, 64, 65, 67, 68, 70, 72, 73, 75, 76, 77, 79, 80, 81, 83, 84, 86, 88, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104, 105, 107
Offset: 1

Views

Author

Vladimir Shevelev, Jun 14 2009

Keywords

Comments

Or: union of A131323 with the sequence of terms of the form A131323(n)-2, and with the sequence of terms of the form A036554(n)-2.
Conjecture: In every sequence of numbers n such that A010060(n)=A010060(n+k), for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. - Vladimir Shevelev, Jul 31 2009

Crossrefs

Programs

  • Mathematica
    tm[0] = 0; tm[n_?EvenQ] := tm[n] = tm[n/2]; tm[n_] := tm[n] = 1 - tm[(n-1)/2]; Reap[For[n = 0, n <= 200, n++, If[tm[n] == tm[n+3], Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 24 2013 *)
  • PARI
    is(n)=hammingweight(n)%2==hammingweight(n+3)%2 \\ Charles R Greathouse IV, Aug 20 2013

Formula

Equals {A001477} \ {A161580}.

Extensions

More terms from R. J. Mathar, Aug 17 2009

A161580 Positions n such that A010060(n) + A010060(n+3) = 1.

Original entry on oeis.org

2, 5, 7, 10, 14, 18, 21, 23, 26, 29, 31, 34, 37, 39, 42, 46, 50, 53, 55, 58, 62, 66, 69, 71, 74, 78, 82, 85, 87, 90, 93, 95, 98, 101, 103, 106, 110, 114, 117, 119, 122, 125, 127, 130, 133, 135, 138, 142, 146, 149, 151, 154, 157, 159, 162, 165, 167, 170, 174, 178, 181, 183, 186
Offset: 1

Views

Author

Vladimir Shevelev, Jun 14 2009

Keywords

Comments

Conjecture: In every sequence of numbers n such that A010060(n) + A010060(n+k) = 1, for fixed odd k, the odious (A000069) and evil (A001969) terms alternate. [From Vladimir Shevelev, Jul 31 2009]

Crossrefs

Programs

  • Mathematica
    tm[0] = 0; tm[n_?EvenQ] := tm[n] = tm[n/2]; tm[n_] := tm[n] = 1 - tm[(n-1)/2]; Reap[For[n = 0, n <= 200, n++, If[tm[n] + tm[n+3] == 1, Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 24 2013 *)
  • PARI
    is(n)=hammingweight(n)%2+hammingweight(n+3)%2==1 \\ Charles R Greathouse IV, Mar 22 2013

Formula

Extensions

More terms from R. J. Mathar, Aug 17 2009

A161627 Positions n such that A010060(n)=A010060(n+4).

Original entry on oeis.org

4, 5, 6, 7, 20, 21, 22, 23, 28, 29, 30, 31, 36, 37, 38, 39, 52, 53, 54, 55, 68, 69, 70, 71, 84, 85, 86, 87, 92, 93, 94, 95, 100, 101, 102, 103, 116, 117, 118, 119, 124, 125, 126, 127, 132, 133, 134, 135, 148, 149, 150, 151, 156, 157, 158, 159, 164, 165, 166, 167, 180, 181, 182
Offset: 1

Views

Author

Vladimir Shevelev, Jun 15 2009

Keywords

Comments

Or: union of the numbers of the form 4*A079523(n)+k, k=0, 1, 2, or 3.
Locates patterns of the form 1xxx1 or 0xxx0 in the Thue-Morse sequence.

Crossrefs

Programs

  • Mathematica
    tm[0] = 0; tm[n_?EvenQ] := tm[n] = tm[n/2]; tm[n_] := tm[n] = 1 - tm[(n-1)/2]; Reap[For[n = 0, n <= 200, n++, If[tm[n] == tm[n+4], Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 24 2013 *)
    SequencePosition[ThueMorse[Range[200]],{x_,,,_,x_}][[All,1]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Apr 16 2017 *)
  • PARI
    is(n)=hammingweight(n)%2==hammingweight(n+4)%2 \\ Charles R Greathouse IV, Aug 20 2013

Extensions

Extended by R. J. Mathar, Aug 28 2009

A161639 Positions n such that A010060(n) = A010060(n+8).

Original entry on oeis.org

8, 9, 10, 11, 12, 13, 14, 15, 40, 41, 42, 43, 44, 45, 46, 47, 56, 57, 58, 59, 60, 61, 62, 63, 72, 73, 74, 75, 76, 77, 78, 79, 104, 105, 106, 107, 108, 109, 110, 111, 136, 137, 138, 139, 140, 141, 142, 143, 168, 169, 170, 171, 172, 173, 174, 175, 184, 185, 186, 187, 188, 189
Offset: 1

Views

Author

Vladimir Shevelev, Jun 15 2009

Keywords

Comments

Locates correlations of the form 1xxxxxxx1 or 0xxxxxxx0 in the Thue-Morse sequence.
Or: union of numbers 8*A079523(n)+k, k=0, 1, 2, 3, 4, 5, 6, or 7.
Generalization: the numbers n such that A010060(n) = A010060(n+2^m) constitute the union of sequences {2^m*A079523(n)+k}, k=0,1,...,2^m-1.

Crossrefs

Programs

  • Mathematica
    tm[0] = 0; tm[n_?EvenQ] := tm[n] = tm[n/2]; tm[n_] := tm[n] = 1 - tm[(n-1)/2]; Reap[For[n = 0, n <= 200, n++, If[tm[n] == tm[n+8], Sow[n]]]][[2, 1]] (* Jean-François Alcover, Oct 24 2013 *)
    SequencePosition[ThueMorse[Range[0,200]],{x_,,,_,,,_,,x}][[All,1]]-1 (* Harvey P. Dale, Jul 23 2021 *)
  • PARI
    is(n)=hammingweight(n)%2==hammingweight(n+8)%2 \\ Charles R Greathouse IV, Aug 20 2013

Extensions

Duplicate of 174 removed by R. J. Mathar, Aug 28 2009
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