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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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

Views

Author

Karamanos Konstantinos, N. J. A. Sloane

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).

Links

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.
Cf. A060572

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

A065359 Alternating bit sum for n: replace 2^k with (-1)^k in binary expansion of n.

Original entry on oeis.org

0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2, -1, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2, -1, 0, 1, -1, 0, -2, -1, -3, -2, -1, 0, -2, -1, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2, -1, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 3, 4, 2, 3, 1, 2, 0, 1, 2, 3, 1, 2, 0, 1, -1, 0, 1, 2, 0, 1, -1, 0, -2
Offset: 0

Views

Author

Marc LeBrun, Oct 31 2001

Keywords

Comments

Notation: (2)[n](-1)
From David W. Wilson and Ralf Stephan, Jan 09 2007: (Start)
a(n) is even iff n in A001969; a(n) is odd iff n in A000069.
a(n) == 0 (mod 3) iff n == 0 (mod 3).
a(n) == 0 (mod 6) iff (n == 0 (mod 3) and n/3 not in A036556).
a(n) == 3 (mod 6) iff (n == 0 (mod 3) and n/3 in A036556). (End)
a(n) = A030300(n) - A083905(n). - Ralf Stephan, Jul 12 2003
From Robert G. Wilson v, Feb 15 2011: (Start)
First occurrence of k and -k: 0, 1, 2, 5, 10, 21, 42, 85, ..., (A000975); i.e., first 0 occurs for 0, first 1 occurs for 1, first -1 occurs at 2, first 2 occurs for 5, etc.;
a(n)=-3 only if n mod 3 = 0,
a(n)=-2 only if n mod 3 = 1,
a(n)=-1 only if n mod 3 = 2,
a(n)= 0 only if n mod 3 = 0,
a(n)= 1 only if n mod 3 = 1,
a(n)= 2 only if n mod 3 = 2,
a(n)= 3 only if n mod 3 = 0, ..., . (End)
a(n) modulo 2 is the Prouhet-Thue-Morse sequence A010060. - Philippe Deléham, Oct 20 2011
In the Koch curve, number the segments starting with n=0 for the first segment. The net direction (i.e., the sum of the preceding turns) of segment n is a(n)*60 degrees. This is since in the curve each base-4 digit 0,1,2,3 of n is a sub-curve directed respectively 0, +60, -60, 0 degrees, which is the net 0,+1,-1,0 of two bits in the sum here. - Kevin Ryde, Jan 24 2020

Examples

			Alternating bit sum for 11 = 1011 in binary is 1 - 1 + 0 - 1 = -1, so a(11) = -1.

Links

Crossrefs

Cf. A005536 (partial sums), A056832 (abs first differences), A010060 (mod 2), A039004 (indices of 0's).
Cf. A000120, A065360, A065081, A019565, A195017.
Cf. also A004718.
Cf. analogous sequences for bases 3-10: A065368, A346688, A346689, A346690, A346691, A346731, A346732, A055017 and also A373605 (for primorial base).

Programs

  • Haskell
    a065359 0 = 0
a065359 n = - a065359 n' + m where (n', m) = divMod n 2
-- Reinhard Zumkeller, Mar 20 2015
  • Maple
    A065359 := proc(n) local dgs ; dgs := convert(n,base,2) ; add( -op(i,dgs)*(-1)^i,i=1..nops(dgs)) ; end proc: # R. J. Mathar, Feb 04 2011
  • Mathematica
    f[0]=0; f[n_] := Plus @@ (-(-1)^Range[ Floor[ Log2@ n + 1]] Reverse@ IntegerDigits[n, 2]); Array[ f, 107, 0]
  • PARI
    a(n) = my(s=0, u=1); for(k=0,#binary(n)-1,s+=bittest(n,k)*u;u=-u);s /* Washington Bomfim, Jan 18 2011 */
  • PARI
    a(n) = my(b=binary(n)); b*[(-1)^k|k<-[-#b+1..0]]~; \\ Ruud H.G. van Tol, Oct 16 2023
  • PARI
    a(n) = if(n==0, 0, 2*hammingweight(bitand(n, ((4<<(2*logint(n,4)))-1)/3)) - hammingweight(n)) \\ Andrew Howroyd, Dec 14 2024
  • Python
    def a(n):
    return sum((-1)**k for k, bi in enumerate(bin(n)[2:][::-1]) if bi=='1')
print([a(n) for n in range(107)]) # Michael S. Branicky, Jul 13 2021
  • Python
    from sympy.ntheory import digits
def A065359(n): return sum((0,1,-1,0)[i] for i in digits(n,4)[1:]) # Chai Wah Wu, Jul 19 2024

Formula

G.f.: (1/(1-x)) * Sum_{k>=0} (-1)^k*x^2^k/(1+x^2^k). - Ralf Stephan, Mar 07 2003
a(0) = 0, a(2n) = -a(n), a(2n+1) = 1-a(n). - Ralf Stephan, Mar 07 2003
a(n) = Sum_{k>=0} A030308(n,k)*(-1)^k. - Philippe Deléham, Oct 20 2011
a(n) = -a(floor(n/2)) + n mod 2. - Reinhard Zumkeller, Mar 20 2015
a(n) = A139351(n) - A139352(n). - Kevin Ryde, Jan 24 2020
G.f. A(x) satisfies: A(x) = x / (1 - x^2) - (1 + x) * A(x^2). - Ilya Gutkovskiy, Jul 28 2021
a(n) = A195017(A019565(n)). - Antti Karttunen, Jun 19 2024

Extensions

More terms from Ralf Stephan, Jul 12 2003

A096268 Period-doubling sequence (or period-doubling word): fixed point of the morphism 0 -> 01, 1 -> 00.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Jun 22 2004

Keywords

Comments

Take highest power of 2 dividing n (A007814(n+1)), read modulo 2.
For the scale-invariance properties see Hendriks et al., 2012.
This is the sequence that results from the ternary Thue-Morse sequence (A036577) if all twos in that sequence are replaced by zeros. - Nathan Fox, Mar 12 2013
This sequence can be used to draw the Von Koch snowflake with a suitable walk in the plane. Start from the origin then the n-th step is "turn +Pi/3 if a(n)=0 and turn -2*Pi/3 if a(n)=1" (see link for a plot of the first 200000 steps). - Benoit Cloitre, Nov 10 2013
1 iff the number of trailing zeros in the binary representation of n+1 is odd. - Ralf Stephan, Nov 11 2013
Equivalently, with offset 1, the characteristic function of A036554 and an indicator for the A003159/A036554 classification of positive integers. - Peter Munn, Jun 02 2020

Examples

			Start: 0
Rules:
  0 --> 01
  1 --> 00
-------------
0:   (#=1)
  0
1:   (#=2)
  01
2:   (#=4)
  0100
3:   (#=8)
  01000101
4:   (#=16)
  0100010101000100
5:   (#=32)
  01000101010001000100010101000101
6:   (#=64)
  0100010101000100010001010100010101000101010001000100010101000100
7:   (#=128)
  010001010100010001000101010001010100010101000100010001010100010001000101010...
[_Joerg Arndt_, Jul 06 2011]

References

  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Links

Crossrefs

Not the same as A073059!
Swapping 0 and 1 gives A035263.
Cf. A096269, A096270, A071858, A220466, A036577.
Cf. A056832, A123087 (partial sums).
With offset 1, classification indicator for A003159/A036554.
Also with offset 1: A007814 mod 2 (cf. A096271 for mod 3), A048675 mod 2 (cf. A332813 for mod 3), A059975 mod 2.

Programs

  • Haskell
    a096268 = (subtract 1) . a056832 . (+ 1)
-- Reinhard Zumkeller, Jul 29 2014
  • Magma
    [Valuation(n+1, 2) mod 2: n in [0..100]]; // Vincenzo Librandi, Jul 20 2016
  • Maple
    nmax:=104: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 0 to ceil(nmax/(p+2))+1 do a((2*n+1)*2^p-1) := p mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Feb 02 2013
# second Maple program:
a:= proc(n) a(n):= `if`(n::even, 0, 1-a((n-1)/2)) end:
seq(a(n), n=0..125);  # Alois P. Heinz, Mar 20 2019
  • Mathematica
    Nest[ Flatten[ # /. {0 -> {1, 0}, 1 -> {0, 0}}] &, {1}, 7] (* Robert G. Wilson v, Mar 05 2005 *)
{{0}}~Join~SubstitutionSystem[{0 -> {0, 1}, 1 -> {0, 0}}, {1}, 6] // Flatten (* Michael De Vlieger, Aug 15 2016 *)
  • PARI
    a(n)=valuation(n+1,2)%2 \\ Ralf Stephan, Nov 11 2013
  • Python
    def A096268(n): return (~(n+1)&n).bit_length()&1 # Chai Wah Wu, Jan 09 2023

Formula

Recurrence: a(2*n) = 0, a(4*n+1) = 1, a(4*n+3) = a(n). - Ralf Stephan, Dec 11 2004
The recurrence may be extended backwards, with a(-1) = 1. - S. I. Ben-Abraham, Apr 01 2013
a(n) = 1 - A035263(n-1). - Reinhard Zumkeller, Aug 16 2006
Dirichlet g.f.: zeta(s)/(1+2^s). - Ralf Stephan, Jun 17 2007
Let T(x) be the g.f., then T(x) + T(x^2) = x^2/(1-x^2). - Joerg Arndt, May 11 2010
Let 2^k||n+1. Then a(n)=1 if k is odd, a(n)=0 if k is even. - Vladimir Shevelev, Aug 25 2010
a(n) = A007814(n+1) mod 2. - Robert G. Wilson v, Jan 18 2012
a((2*n+1)*2^p-1) = p mod 2, p >= 0 and n >= 0. - Johannes W. Meijer, Feb 02 2013
a(n) = A056832(n+1) - 1. - Reinhard Zumkeller, Jul 29 2014
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/3. = Amiram Eldar, Sep 18 2022

Extensions

Corrected by Jeremy Gardiner, Dec 12 2004
More terms from Robert G. Wilson v, Feb 26 2005

A197911 Representable by A001045 (Jacobsthal sequence). Complement of A003158.

Original entry on oeis.org

0, 1, 3, 4, 5, 6, 8, 9, 11, 12, 14, 15, 16, 17, 19, 20, 21, 22, 24, 25, 26, 27, 29, 30, 32, 33, 35, 36, 37, 38, 40, 41, 43, 44, 46, 47, 48, 49, 51, 52, 54, 55, 57, 58, 59, 60, 62, 63, 64, 65, 67, 68, 69, 70, 72, 73, 75, 76, 78, 79, 80, 81, 83, 84, 85, 86, 88
Offset: 0

Views

Author

Philippe Deléham, Oct 19 2011

Keywords

Comments

a(n) modulo 2 is the Prouhet-Thue-Morse sequence A010060.
The sequence of Jacobsthal numbers A001045 begins [1, 1, 3, 5, 11, 21, ...] with two occurrences of the term 1. Allowing for this, we find that the numbers representable as a sum of distinct Jacobsthal numbers form A050292. - Peter Bala, Feb 02 2013
Partial sums of A056832. - Reinhard Zumkeller, Jul 29 2014

Links

Crossrefs

Cf. A001045, A003158, A010060. A050292.
Cf. A056832.

Programs

  • Haskell
    a197911 n = a197911_list !! n
a197911_list = scanl (+) 0 a056832_list
-- Reinhard Zumkeller, Jul 29 2014
  • Python
    def A197911(n): return n+sum((~(i+1)&i).bit_length()&1 for i in range(n)) # Chai Wah Wu, Jan 09 2023

Formula

a(n) = Sum_{k>=0} A030308(n,k)*A001045(k+2).

A214682 Remove 2's that do not contribute to a factor of 4 from the prime factorization of n.

Original entry on oeis.org

1, 1, 3, 4, 5, 3, 7, 4, 9, 5, 11, 12, 13, 7, 15, 16, 17, 9, 19, 20, 21, 11, 23, 12, 25, 13, 27, 28, 29, 15, 31, 16, 33, 17, 35, 36, 37, 19, 39, 20, 41, 21, 43, 44, 45, 23, 47, 48, 49, 25, 51, 52, 53, 27, 55, 28, 57, 29
Offset: 1

Views

Author

Tyler Ball, Jul 25 2012

Keywords

Comments

In this sequence, the number 4 exhibits some characteristics of a prime number since all extraneous 2's have been removed from the prime factorizations of all other numbers.

Examples

			For n=8, v_4(8)=1, v_2(8)=3, so a(8)=(8*4^1)/(2^3)=4.
For n=12, v_4(12)=1, v_2(12)=2, so a(12)=(12*4^1)/(2^2)=12.

Links

Crossrefs

Cf. A214681, A214685.
Range of values: A003159.
Missing values: A036554.
A056832, A059895, A073675 are used in a formula defining this sequence.
A059897 is used to express relationship between terms of this sequence.
Cf. A007814 (v_2(n)), A235127 (v_4(n)).

Programs

  • Mathematica
    a[n_] := n/(2^Mod[IntegerExponent[n, 2], 2]); Array[a, 100] (* Amiram Eldar, Dec 09 2020 *)
  • PARI
    a(n)=n>>(valuation(n,2)%2) \\ Charles R Greathouse IV, Jul 26 2012
  • Python
    def A214682(n): return n>>1 if (~n&n-1).bit_length()&1 else n # Chai Wah Wu, Jan 09 2023
  • SageMath
    C = []
for i in [1..n]:
    C.append(i*4^(Integer(i).valuation(4))/2^(Integer(i).valuation(2)))

Formula

a(n) = (n*4^(v_4(n)))/(2^(v_2(n))) where v_k(n) is the k-adic valuation of n. That is, v_k(n) is the largest power of k, a, such that k^a divides n.
For n odd, a(n)=n since n has no factors of 2 (or 4).
From Peter Munn, Nov 29 2020: (Start)
a(A003159(n)) = n.
a(A036554(n)) = n/2.
a(n) = n/A056832(n) = n/A059895(n, 2) = min(n, A073675(n)).
a(A059897(n, k)) = A059897(a(n), a(k)). (End)
Multiplicative with a(2^e) = 2^(2*floor(e/2)), and a(p^e) = p^e for odd primes p. - Amiram Eldar, Dec 09 2020
Sum_{k=1..n} a(k) ~ (5/12) * n^2. - Amiram Eldar, Nov 10 2022
Dirichlet g.f.: zeta(s-1)*(2^s+1)/(2^s+2). - Amiram Eldar, Dec 30 2022

A339245 Partrich numbers: positive integers whose square part and squarefree part are divisible by 2 and an odd prime.

Original entry on oeis.org

216, 360, 504, 600, 792, 864, 936, 1000, 1080, 1176, 1224, 1368, 1400, 1440, 1512, 1656, 1944, 1960, 2016, 2088, 2200, 2232, 2376, 2400, 2520, 2600, 2664, 2744, 2808, 2904, 2952, 3000, 3096, 3168, 3240, 3384, 3400, 3456, 3672, 3744, 3800, 3816, 3960, 4000, 4056, 4104, 4200
Offset: 1

Views

Author

Peter Munn, Nov 28 2020

Keywords

Comments

Not named after anyone, partrich numbers have the square part of their odd part, the square part of their even part (A234957), the squarefree part of their odd part and the squarefree part of their even part (A056832) all greater than 1.
Numbers whose odd part and even part are nonsquare and nonsquarefree.
All terms are divisible by 8. If m is present, 2m is absent and 4m is present.
Closed under multiplication by any square and under application of A059896: for n, k >= 1, A059896(a(n), k) is in the sequence.
From Peter Munn, Apr 07 2021: (Start)
The first deficient partrich number is 39304 = 2^3 * 17^3. (ascertained by Amiram Eldar)
The first 7 terms generate Carmichael numbers using the method of Erdős described in A287840.
(End)

Examples

			A positive integer is present if and only if it factorizes as 2 times an odd squarefree number > 1, an even square that is a power of 4 and an odd square > 1. This factorization of the initial terms is shown below.
   n  a(n)
   1   216 = 2 *  3 *  4 *  9,
   2   360 = 2 *  5 *  4 *  9,
   3   504 = 2 *  7 *  4 *  9,
   4   600 = 2 *  3 *  4 * 25,
   5   792 = 2 * 11 *  4 *  9,
   6   864 = 2 *  3 * 16 *  9,
   7   936 = 2 * 13 *  4 *  9,
   8  1000 = 2 *  5 *  4 * 25,
   9  1080 = 2 * 15 *  4 *  9,
  10  1176 = 2 *  3 *  4 * 49,
  ...

Links

Crossrefs

Subsequence of A008586, A028983, A036554, A036785, A038838, A190892.
Subsequences: A017139, A017643.
Cf. A056832, A059896, A234957, A287840.

Programs

  • Mathematica
    q[n_] := Module[{ie = IntegerExponent[n, 2], odd}, ie > 2 && OddQ[ie] && !SquareFreeQ[(odd = n/2^ie)] && !IntegerQ @ Sqrt[odd]]; Select[Range[4200], q] (* Amiram Eldar, Dec 04 2020 *)

Formula

A008586 INTERSECT A028983 INTERSECT A036554 INTERSECT A038838.
Asymptotic density is 1/12 - 2/(3 * Pi^2) = 0.01578587757... . (Formula due to Amiram Eldar.)

A089608 a(n) = ((-1)^(n+1)*A002425(n)) modulo 6.

Original entry on oeis.org

1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 5, 1, 5, 1, 1, 1, 5, 1, 5, 1
Offset: 1

Views

Author

Benoit Cloitre, Dec 30 2003

Keywords

Comments

Let S(1)={1} and S(n+1)=S(n)S'(n) where S'(n) is obtained from S(n) by changing last term using the cyclic permutation 1->5->1, then sequence is S(infinity).

Links

Crossrefs

Cf. A002425, A035263, A056832.

Programs

  • Mathematica
    a[n_] := Mod[IntegerExponent[n, 2], 2] * 4 + 1; Array[a, 100] (* Amiram Eldar, Nov 28 2022 *)
  • PARI
    a(n)=numerator(2/n*(4^n-1)*bernfrac(2*n))%6
  • PARI
    a(n)=valuation(n,2)%2 * 4 + 1; \\ Andrew Howroyd, Aug 01 2018
  • Scheme
    (define (A035263 n) (let loop ((n n) (i 1)) (cond ((odd? n) (modulo i 2)) (else (loop (/ n 2) (+ 1 i))))))
(define (A089608 n) (- 5 (* 4 (A035263 n))))
;; Antti Karttunen, Sep 11 2017

Formula

a(n) = 5 - 4*A035263(n).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 7/3. - Amiram Eldar, Nov 28 2022
From Amiram Eldar, Jan 04 2023: (Start)
Multiplicative with a(2^e) = 5 if e is odd, and 1 if e is even, a(p^e) = 1 for p >= 3.
Dirichlet g.f.: zeta(s)*(2^s+5)/(2^s+1). (End)

Extensions

Keyword:mult added by Andrew Howroyd, Aug 01 2018

A103587 Sequence of run lengths in A103585.

Original entry on oeis.org

1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 1, 1
Offset: 1

Views

Author

N. J. A. Sloane, Mar 24 2005

Keywords

Comments

Sequence appears to have period 30. - Ralf Stephan, May 18 2007

Links

Crossrefs

Cf. A103585. Different from A056832!

A317956 Apply the morphism X -> XYX, where Y=1 if X contains an odd number of 1's, otherwise Y = 2 n times to 1, and concatenate the resulting string.

Original entry on oeis.org

1, 121, 1211121, 121112121211121, 1211121212111211121112121211121, 121112121211121112111212121112121211121212111211121112121211121
Offset: 0

Views

Author

N. J. A. Sloane, Aug 22 2018

Keywords

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)

Crossrefs

The limit of the terms is A056832.
A317957 is the same sequence over the alphabet {1,0}.

A078734 Start with 1,2, concatenate 2^k previous terms and change last term as follows: 1->2, 2->3, 3->1.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Benoit Cloitre, Dec 21 2002

Keywords

Examples

			Concatenating the first 2 terms 1,2 -> 1,2,1,2 and changing 2->3 gives the first 4 terms: 1,2,1,3.
Concatenating those first 4 terms ->1,2,1,3,1,2,1,3 and changing 3->1 gives the first 8 terms: 1,2,1,3,1,2,1,1.

Links

Crossrefs

Cf. A056832, A035263.

Programs

  • Magma
    [Valuation(n, 2) mod 3 + 1: n in [1..100]]; // Vincenzo Librandi, Aug 01 2018
  • Mathematica
    a[n_] := Mod[IntegerExponent[n, 2], 3] + 1; Array[a, 100] (* Amiram Eldar, Oct 28 2022 *)
  • PARI
    seq(n)={my(v=[1,2]); while(#v < n, v=concat(v,v); v[#v] = v[#v] % 3 + 1); vector(n, i, v[i])} \\ Andrew Howroyd, Jul 31 2018
  • PARI
    a(n) = valuation(n, 2) % 3 + 1; \\ Andrew Howroyd, Jul 31 2018

Formula

(Sum_{k=1..n} a(k))/n -> 1.57.... [This limit is the asymptotic mean of this sequence, 11/7. - Amiram Eldar, Oct 28 2022]
Multiplicative with a(2^e) = (e mod 3) + 1, a(p^e) = 1 for odd prime p. - Andrew Howroyd, Jul 31 2018
From Antti Karttunen, Dec 07 2021: (Start)
a(n) = 1 + A096271(n-1) = 1 + A010872(A007814(n)). [as above]
a(n) = A001511(A050985(n)).
(End)
Dirichlet g.f.: zeta(s)*(4^s+2^(s+1)+3)/(4^s+2^s+1). - Amiram Eldar, Dec 30 2022
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