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.

Showing 1-8 of 8 results.

A014577 The regular paper-folding sequence (or dragon curve sequence). Alphabet {1,0}.

Original entry on oeis.org

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

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Author

Keywords

Comments

a(n) is the complement of the bit to the left of the least significant "1" in the binary expansion of n+1. E.g., n = 3, n+1 = 4 = 100_2, so a(3) = (complement of bit to left of 1) = 1. - Robert L. Brown, Nov 28 2001 [Adjusted to match offset by N. J. A. Sloane, Apr 15 2021]
To construct the sequence: start from 1,(..),0,(..),1,(..),0,(..),1,(..),0,(..),1,(..),0,... and fill undefined places with the sequence itself. - Benoit Cloitre, Jul 08 2007
This is the characteristic function of A091072 - 1. - Gary W. Adamson, Apr 11 2010
Turns (by 90 degrees) of the Heighway dragon which can be rendered as follows: [Init] Set n=0 and direction=0. [Draw] Draw a unit line (in the current direction). Turn left/right if a(n) is zero/nonzero respectively. [Next] Set n=n+1 and goto (draw). See fxtbook link below. - Joerg Arndt, Apr 15 2010
Sequence can be obtained by the L-system with rules L->L1R, R->L0R, 1->1, 0->0, starting with L, and then dropping all L's and R's (see example). - Joerg Arndt, Aug 28 2011
From Gary W. Adamson, Jun 20 2012: (Start)
One half of the infinite Farey tree can be mapped one-to-one onto A014577 since both sequences can be derived directly from the binary. First few terms are
1, 1, 0, 1, 1, 0, 0, 1, 1, 1, ...
1/2 2/3 1/3 3/4 3/5 2/5 1/4 4/5 5/7 5/8, ...
Infinite Farey tree fractions can be derived from the binary by appending a repeat of rightmost binary term to the right, then recording the number of runs to obtain the continued fraction representation. Example: 9 = 1001 which becomes 10011 which becomes [1,2,2] = 5/7. (End)
The sequence can be considered as a binomial transform operator for a target sequence S(n). Replace the first 1 in A014577 with the first term in S(n), then for successive "1" term in A014577, map the next higher term in S(n). If "0" in A014577, map the next lower term in S(n). Using the sequence S(n) = (1, 3, 5, 7, ...), we obtain (1), (3, 1), (3, 5, 3, 1), (3, 5, 7, 5, 3, 5, 3, 1), .... Then parse the terms into subsequences of 2^k terms, adding the terms in each string. We obtain (1, 4, 12, 32, 80, ...), the binomial transform of (1, 3, 5, 7, ...). The 8-bit string has one 1, three 5's, three 7's and one 1) as expected, or (1, 3, 3, 1) dot (1, 3, 5, 7). - Gary W. Adamson, Jun 24 2012
From Gary W. Adamson, May 29 2013: (Start)
The sequence can be generated directly from the lengths of continued fraction representations of fractions in one half of the Stern-Brocot tree (fractions between 0 and 1):
1/2
1/3 2/3
1/4 2/5 3/5 3/4
1/5 2/7 3/8 3/7 4/7 5/8 5/7 4/5
...
and their corresponding continued fraction representations are:
[2]
[3] [1,2]
[4] [2,2] [1,1,2] [1,3]
[5] [3,2] [2,1,2] [2,3] [1,1,3] [1,1,1,2] [1,2,2] [1,4]
...
Record the lengths by rows then reverse rows, getting:
1,
2, 1,
2, 3, 2, 1,
2, 3, 4, 3, 2, 3, 2, 1,
...
start with "1" and if the next term is greater than the current term, record a 1, otherwise 0; getting the present sequence, the Harter-Heighway dragon curve. (End)
The paper-folding word "110110011100100111011000..." can be created by concatenating the terms of a fixed point of the morphism or string substitution rule: 00 -> 1000, 01 -> 1001, 10 -> 1100 & 11 -> 1101, beginning with "11". - Robert G. Wilson v, Jun 11 2015
From Gary W. Adamson, Jun 04 2021: (Start)
Since the Heighway dragon is composed of right angles, it can be mapped with unit trajectories (Right = 1, Left = (-1), Up = i and Down = -i) on the complex plane where i = sqrt(-1). The initial (0th) iterate is chosen in this case to be the unit line from (0,0) to (-1,0). Then follow the directions below as indicated, resulting in a reflected variant of the dragon curve shown at the Eric Weisstein link. The conjectured system of complex plane trajectories is:
0 -1
1 -1, i
2 -1, i, 1, i
3 -1, i, 1, i, 1, -i, 1, i
4 -1, i, 1, i, 1, -i, 1, i, 1, -i, -1, -i, 1, -i, 1, i
...
The conjecture succeeds through the 4th iterate. It appears that to generate the (n+1)-th row, bring down the n-th row as the left half of row (n+1). For the right half of row (n+1), bring down the n-th row but change the signs of the first half of row n. For example, to get the complex plane instructions for the third iterate of the dragon curve, bring down (-1, i, 1, i) as the left half, and the right half is (1, -i, 1, i). (End)
From Gary W. Adamson, Jun 09 2021: (Start)
Partial sums of the iterate trajectories produce a sequence of complex addresses for unit segments. Partial sums of row 4 are: -1, (-1+i), i, 2i, (1+2i), (1+i), (2+i), (2+2i), (3+2i), (3+i), (2+i), 2, 3, (3-i), (4-i), 4. (zeros are omitted with terms of the form a+0i). The reflected variant of the dragon curve has the 0th iterate from (0,0) to (1,0), and the respective addresses simply change the signs of the real terms. (End)

Examples

			1 + x + x^3 + x^4 + x^7 + x^8 + x^9 + x^12 + x^15 + x^16 + x^17 + x^19 + ...
From _Joerg Arndt_, Aug 28 2011: (Start)
Generation via string substitution:
Start: L
Rules:
  L --> L1R
  R --> L0R
  0 --> 0
  1 --> 1
-------------
0:   (#=1)
  L
1:   (#=3)
  L1R
2:   (#=7)
  L1R1L0R
3:   (#=15)
  L1R1L0R1L1R0L0R
4:   (#=31)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R
5:   (#=63)
  L1R1L0R1L1R0L0R1L1R1L0R0L1R0L0R1L1R1L0R1L1R0L0R0L1R1L0R0L1R0L0R
Drop all L and R to obtain 1101100111001001110110001100100
(End)
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, pp. 155, 182.
  • Chandler Davis and Donald E. Knuth, Number Representations and Dragon Curves -- I and II, Journal of Recreational Mathematics, volume 3, number 2, April 1970, pages 66-81, and number 3, July 1970, pages 133-149. Reprinted in Donald E. Knuth, Selected Papers on Fun and Games, CSLI Publications, 2010, pages 571-614.
  • Michel Dekking, Michel Mendes France, and Alf van der Poorten, "Folds", The Mathematical Intelligencer, 4.3 (1982): 130-138 & front cover, and 4:4 (1982): 173-181 (printed in two parts).
  • Martin Gardner, Mathematical Magic Show, New York: Vintage, pp. 207-209 and 215-220, 1978.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

Cf. A059125, A065339, A005811, A220466, A088748, A091072, A343173 (first differences), A343174 (partial sums).
The two bisections are A000035 and the sequence itself.
See A343181 for prefixes of length 2^k-1.

Programs

  • Magma
    [(1+KroneckerSymbol(-1,n))/2: n in [1..100]]; // Vincenzo Librandi, Aug 05 2015
    
  • Magma
    [Floor(1/2*(1+(-1)^(1/2*((n+1)/2^Valuation(n+1, 2)-1)))): n in [0..100]]; // Vincenzo Librandi, Aug 05 2015
    
  • Maple
    nmax:=98: 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) := (n+1) mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 28 2013
    a014577 := proc(n) local p,s1,s2,i;
    if n=0 then return(1); fi;
    s1:=convert(n,base,2); s2:=nops(s1);
    for i from 1 to s2 do if s1[i]=1 then p:=i; break; fi; od:
    if p <= s2-1 then 1-s1[p+1]; else 1; fi; end;
    [seq(a014577(i),i=1..120)]; # N. J. A. Sloane, Apr 08 2021
    # third Maple program:
    a:= n-> 1-irem(iquo((n+1)/2^padic[ordp](n+1, 2), 2), 2):
    seq(a(n), n=0..120);  # Alois P. Heinz, Apr 08 2021
  • Mathematica
    a[n_] := Boole[EvenQ[((n+1)/2^IntegerExponent[n+1, 2]-1)/2]]; Table[a[n], {n, 0, 98}] (* Jean-François Alcover, Feb 16 2012, after Gary W. Adamson, updated Nov 21 2014 *)
    Table[1-(((Mod[#1,2^(#2+2)]/2^#2)&[n,IntegerExponent[n,2]])-1)/2,{n,1,100,1}] (* WolframAlpha compatible code; Robert L. Brown, Jan 06 2015 *)
    MapThread[(a[x_/;IntegerQ[(x-#1)/4]]:= #2)&,{{1,3},{1,0}}];a[x_/;IntegerQ[x/2]]:=a[x/2];a/@ Range[100] (* Bradley Klee, Aug 04 2015 *)
    (1 + JacobiSymbol[-1, Range[100]])/2 (* Paolo Xausa, May 22 2024 *)
    Array[Boole[BitAnd[#, BitAnd[#, -#]*2] == 0] &, 100] (* Paolo Xausa, May 22 2024, after Joerg Arndt C++ code *)
  • PARI
    {a(n) = if( n%2, a(n\2), 1 - (n/2%2))} /* Michael Somos, Feb 05 2012 */
    
  • PARI
    a(n)=1/2*(1+(-1)^(1/2*((n+1)/2^valuation(n+1,2)-1))) \\ Ralf Stephan, Sep 02 2013
    
  • PARI
    a(n)=!bittest(n+1,valuation(n+1,2)+1); \\ Robert L. Brown, Jul 07 2025
    
  • Python
    def A014577(n):
        s = bin(n+1)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return 1-int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021

Formula

a(n) = (1+Jacobi(-1,n+1))/2 (cf. A034947). - N. J. A. Sloane, Jul 27 2012 [Adjusted to match offset by Peter Munn, Jul 01 2022]
Set a=1, b=0, S(0)=a, S(n+1) = S(n), a, F(S(n)), where F(x) reverses x and then interchanges a and b; sequence is limit S(infinity).
From Ralf Stephan, Jul 03 2003: (Start)
a(4*n) = 1, a(4*n+2) = 0, a(2*n+1) = a(n).
a(n) = 1 - A014707(n) = 2 - A014709(n) = A014710(n) - 1. (End)
Set a=1, b=0, S(0)=a, S(n+1)=S(n), a, M(S(n)), where M(S) is S but the bit in the middle position flipped. (Proof via isomorphism of both formulas to a modified string substitution.) - Benjamin Heiland, Dec 11 2011
a(n) = 1 if A005811(n+1) > A005811(n), otherwise a(n) = 0. - Gary W. Adamson, Jun 20 2012
a((2*n+1)*2^p-1) = (n+1) mod 2, p >= 0. - Johannes W. Meijer, Jan 28 2013
G.f. g(x) satisfies g(x) = x*g(x^2) + 1/(1-x^4). - Robert Israel, Jan 06 2015
a(n) = 1 - A065339(n+1) mod 2. - Peter Munn, Jun 29 2022
From A.H.M. Smeets, Mar 19 2023: (Start)
a(n) = 1 - A038189(n+1).
a(n) = A082410(n+2).
a(n) = 1 - A089013(n+1)
a(n) = (3 - A099545(n+1))/2.
a(n) = (A112347(n+1) + 1)/2.
a(n) = (A121238(n+1) + 1)/2.
a(n) = (A317335(n) + 2)/3.
a(n) = (A317336(n) + 10)/3. (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Sep 14 2024

Extensions

More terms from Ralf Stephan, Jul 03 2003

A034947 Jacobi (or Kronecker) symbol (-1/n).

Original entry on oeis.org

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, 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, 1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, -1, -1, 1, 1
Offset: 1

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Author

Keywords

Comments

Also the regular paper-folding sequence.
For a proof that a(n) equals the paper-folding sequence, see Allouche and Sondow, arXiv v4. - Jean-Paul Allouche and Jonathan Sondow, May 19 2015
It appears that, replacing +1 with 0 and -1 with 1, we obtain A038189. Alternatively, replacing -1 with 0 we obtain (allowing for offset) A014577. - Jeremy Gardiner, Nov 08 2004
Partial sums = A005811 starting (1, 2, 1, 2, 3, 2, 1, 2, 3, ...). - Gary W. Adamson, Jul 23 2008
The congruence in {-1,1} of the odd part of n modulo 4 (Cf. A099545). - Peter Munn, Jul 09 2022

Examples

			G.f. = x + x^2 - x^3 + x^4 + x^5 - x^6 - x^7 + x^8 + x^9 + x^10 - x^11 - x^12 + ...
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, pp. 155, 182.
  • H. Cohen, Course in Computational Number Theory, p. 28.

Crossrefs

Moebius transform of A035184.
Cf. A091072 (indices of 1), A091067 (indices of -1), A371594 (indices of run starts).
The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - N. J. A. Sloane, Jul 27 2012

Programs

  • Magma
    [KroneckerSymbol(-1,n): n in [1..100]]; // Vincenzo Librandi, Aug 16 2016
    
  • Maple
    with(numtheory): A034947 := n->jacobi(-1,n);
  • Mathematica
    Table[KroneckerSymbol[ -1, n], {n, 0, 100}] (* Corrected by Jean-François Alcover, Dec 04 2013 *)
  • PARI
    {a(n) = kronecker(-1, n)};
    
  • PARI
    for(n=1, 81, f=factor(n); print1((-1)^sum(s=1, omega(n), f[s, 2]*(Mod(f[s, 1], 4)==3)), ", ")); \\ Arkadiusz Wesolowski, Nov 05 2013
    
  • PARI
    a(n)=direuler(p=1,n,if(p==2,1/(1-kronecker(-4, p)*X)/(1-X),1/(1-kronecker(-4, p)*X))) /* Ralf Stephan, Mar 27 2015 */
    
  • PARI
    a(n) = if(n%2==0, a(n/2), (n+2)%4-2) \\ Peter Munn, Jul 09 2022
  • Python
    def A034947(n):
        s = bin(n)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return 1-2*int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021
    
  • Python
    def A034947(n): return -1 if n>>(-n&n).bit_length()&1 else 1 # Chai Wah Wu, Feb 26 2025
    

Formula

Multiplicative with a(2^e) = 1, a(p^e) = (-1)^(e*(p-1)/2) if p>2.
a(2*n) = a(n), a(4*n+1) = 1, a(4*n+3) = -1, a(-n) = -a(n). a(n) = 2*A014577(n-1)-1.
a(prime(n)) = A070750(n) for n > 1. - T. D. Noe, Nov 08 2004
This sequence can be constructed by starting with w = "empty string", and repeatedly applying the map w -> w 1 reverse(-w) [See Allouche and Shallit p. 182]. - N. J. A. Sloane, Jul 27 2012
a(n) = (-1)^A065339(n) = lambda(A097706(n)), where A065339(n) is number of primes of the form 4*m + 3 dividing n (counted with multiplicity) and lambda is Liouville's function, A008836. - Arkadiusz Wesolowski, Nov 05 2013 and Peter Munn, Jun 22 2022
Sum_{n>=1} a(n)/n = Pi/2, due to F. von Haeseler; more generally, Sum_{n>=1} a(n)/n^(2*d+1) = Pi^(2*d+1)*A000364(d)/(2^(2*d+2)-2)(2*d)! for d >= 0; see Allouche and Sondow, 2015. - Jean-Paul Allouche and Jonathan Sondow, Mar 20 2015
Dirichlet g.f.: beta(s)/(1-2^(-s)) = L(chi_2(4),s)/(1-2^(-s)). - Ralf Stephan, Mar 27 2015
a(n) = A209615(n) * (-1)^(v2(n)), where v2(n) = A007814(n) is the 2-adic valuation of n. - Jianing Song, Apr 24 2021
a(n) = 2 - A099545(n) == A000265(n) (mod 4). - Peter Munn, Jun 22 2022 and Jul 09 2022

A014707 a(4n) = 0, a(4n+2) = 1, a(2n+1) = a(n).

Original entry on oeis.org

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

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Author

Keywords

Comments

The regular paper-folding (or dragon curve) sequence.
It appears that the sequence of run lengths is A088431. - Dimitri Hendriks, May 06 2010
Runs of three consecutive ones appear around positions n = 22, 46, 54, 86, 94, 118, 150, 174, 182, ..., or for n of the form 2^(k+3)*(4*t+3)-2, k >= 0, t >= 0. - Vladimir Shevelev, Mar 19 2011

References

  • Guy Melançon, Factorizing infinite words using Maple, MapleTech journal, Vol. 4, No. 1, 1997, pp. 34-42, esp. p. 36.

Crossrefs

Equals 1 - A014577, which see for further references. Also a(n) = A038189(n+1).
The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410.

Programs

  • Haskell
    a014707 n = a014707_list !! n
    a014707_list = f 0 $ cycle [0,0,1,0] where
       f i (x:_:xs) = x : a014707 i : f (i+1) xs
    -- Reinhard Zumkeller, Sep 28 2011
    
  • Maple
    nmax:=92: 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) := n mod 2 od: od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Jan 28 2013
    # second Maple program:
    a:= proc(n) option remember;
         `if`(n::even, irem(n/2, 2), a((n-1)/2))
        end:
    seq(a(n), n=0..92);  # Alois P. Heinz, Jun 27 2022
  • Mathematica
    a[n_ /; Mod[n, 4] == 0] = 0; a[n_ /; Mod[n, 4] == 2] = 1; a[n_ /; Mod[n, 2] == 1] := a[n] = a[(n - 1)/2]; Table[a[n],{n,0,92}] (* Jean-François Alcover, May 17 2011 *)
    (1 - JacobiSymbol[-1, Range[100]])/2 (* Paolo Xausa, May 26 2024 *)
  • PARI
    a(n)=n+=1;my(h=bitand(n,-n));n=bitand(n,h<<1);n!=0; \\ Joerg Arndt, Apr 09 2021
  • Python
    def A014707(n):
        s = bin(n+1)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return int(s[m-i-2]) if m-i-2 >= 0 else 0 # Chai Wah Wu, Apr 08 2021
    
  • Python
    def A014707(n): n+=1; h=n&-n; n=n&(h<<1); return int(n!=0)
    print([A014707(n) for n in range(93)]) # Michael S. Branicky, Mar 29 2024 after Joerg Arndt
    

Formula

a(A091072(n)-1) = 0; a(A091067(n)-1) = 1. - Reinhard Zumkeller, Sep 28 2011 [Adjusted to match offset by Peter Munn, Jul 01 2022]
a(n) = (1-Jacobi(-1,n+1))/2 (cf. A034947). - N. J. A. Sloane, Jul 27 2012 [Adjusted to match offset by Peter Munn, Jul 01 2022]
Set a=0, b=1, S(0)=a, S(n+1) = S(n)aF(S(n)), where F(x) reverses x and then interchanges a and b; sequence is limit S(infinity).
a((2*n+1)*2^p-1) = n mod 2, p >= 0. - Johannes W. Meijer, Jan 28 2013
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Aug 31 2024

Extensions

More terms from Scott C. Lindhurst (ScottL(AT)alumni.princeton.edu)

A038189 Bit to left of least significant 1-bit in binary expansion of n.

Original entry on oeis.org

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

Views

Author

Fred Lunnon, Dec 11 1999

Keywords

Comments

Characteristic function of A091067.
Image, under the coding i -> floor(i/2), of the fixed point, starting with 0, of the morphism 0 -> 01, 1 -> 02, 2 -> 32, 3 -> 31. - Jeffrey Shallit, May 15 2016
Restricted to the positive integers, completely additive modulo 2. - Peter Munn, Jun 20 2022
If a(n) is defined as 1-a(-n) for all n<0, then a(n) = a(2*n), a(4*n+1) = 0, a(4*n+3) = 1 for all n in Z. - Michael Somos, Oct 05 2024

Examples

			a(6) = 1 since 6 = 110 and bit before rightmost 1 is a 1.
		

References

  • Jean-Paul Allouche and Jeffrey O. Shallit, Automatic sequences, Cambridge, 2003, sect. 5.1.6

Crossrefs

A014707(n)=a(n+1). A014577(n)=1-a(n+1).
The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - N. J. A. Sloane, Jul 27 2012
Related sequences A301848, A301849, A301850. - Fred Lunnon, Mar 27 2018

Programs

  • C
    int a(int n) { return (n & ((n&-n)<<1)) ? 1 : 0; } /* from Russ Cox */
    
  • Magma
    function a (n)
      if n eq 0 then return 0; // alternatively,  return 1;
      else while IsEven(n) do n := n div 2; end while; end if;
      return n div 2 mod 2; end function;
      nlo := 0; nhi := 32;
      [a(n) : n in [nlo..nhi] ]; //  Fred Lunnon, Mar 27 2018
    
  • Maple
    A038189 := proc(n)
        option remember;
        if n = 0 then
            0 ;
        elif type(n,'even') then
            procname(n/2) ;
        elif modp(n,4) = 1 then
            0 ;
        else
            1 ;
        end if;
    end proc:
    seq(A038189(n),n=0..100) ; # R. J. Mathar, Mar 30 2018
  • Mathematica
    f[n_] := Block[{id2 = Join[{0}, IntegerDigits[n, 2]]}, While[ id2[[-1]] == 0, id2 = Most@ id2]; id2[[-2]]]; f[0] = 0; Array[f, 105, 0] (* Robert G. Wilson v, Apr 14 2009 and fixed Feb 27 2014 *)
    f[n_] := f[n] = Switch[Mod[n, 4], 0, f[n/2], 1, 0, 2, f[n/2], 3, 1]; f[0] = 0; Array[f, 105, 0] (* Robert G. Wilson v, Apr 14 2009, fixed Feb 27 2014 *)
    a[ n_] := If[n==0, 0, Mod[(n/2^IntegerExponent[n, 2]-1)/2, 2]]; (* Michael Somos, Oct 05 2024 *)
  • PARI
    {a(n) = if(n==0, 0, ((n/2^valuation(n,2)-1)/2)%2)}; /* Michael Somos, Sep 22 2005 */
    
  • PARI
    a(n) = if(n<3, 0, prod(m=1,n, kronecker(-n,m)==kronecker(m,n))) /* Michael Somos, Sep 22 2005 */
    
  • PARI
    A038189(n)=bittest(n,valuation(n,2)+1) \\ M. F. Hasler, Aug 06 2015
    
  • PARI
    a(n)=my(h=bitand(n,-n));n=bitand(n,h<<1);n!=0; \\ Joerg Arndt, Apr 09 2021
    
  • Python
    def A038189(n):
        s = bin(n)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return int(s[m-i-2]) if m-i-2 >= 0 else 0 # Chai Wah Wu, Apr 08 2021
    
  • Python
    def a(n): return 1 if (n & ((n&-n)<<1)) else 0
    print([a(n) for n in range(108)]) # Michael S. Branicky, Feb 06 2025 after Russ Cox

Formula

a(0) = 0, a(2*n) = a(n) for n>0, a(4*n+1) = 0, a(4*n+3) = 1.
G.f.: Sum_{k>=0} t^3/(1-t^4), where t=x^2^k. Parity of A025480. For n >= 1, a(n) = 1/2 * (1 - (-1)^A025480(n-1)). - Ralf Stephan, Jan 04 2004 [index adjusted by Peter Munn, Jun 22 2022]
a(n) = 1 if Kronecker(-n,m)=Kronecker(m,n) for all m, otherwise a(n)=0. - Michael Somos, Sep 22 2005
a(n) = 1 iff A164677(n) < 0. - M. F. Hasler, Aug 06 2015
For n >= 1, a(n) = A065339(n) mod 2. - Peter Munn, Jun 20 2022
From A.H.M. Smeets, Mar 08 2023: (Start)
a(n+1) = 1 - A014577(n) for n >= 0.
a(n+1) = 2 - A014710(n) for n >= 0.
a(n) = (1 - A034947(n))/2 for n > 0. (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/2. - Amiram Eldar, Aug 30 2024

Extensions

More terms from David W. Wilson
Definition corrected by Russ Cox and Ralf Stephan, Nov 08 2004

A099545 Odd part of n, modulo 4.

Original entry on oeis.org

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

Views

Author

Ralf Stephan, Oct 23 2004

Keywords

Comments

The terms of this sequence are the even-indexed terms of A112658. - Alexandre Wajnberg, Jan 02 2006
Fractal sequence: odd terms are 1, 3, 1, 3,...; the even terms are the sequence itself: a(n)=a(2n)=a(4n)=a(8n)=a(16n)=... - Alexandre Wajnberg, Jan 02 2006
From Micah D. Tillman, Jan 29 2021: (Start)
Has the same structure as the regular paper-folding (dragon curve) sequence (A014577, A014709). We can interpret a(n) as the number of 90-degree rotations to make in a single direction at the n-th "turn" in the dragon curve. After all, making three 90-degree rotations to the left (turning a total of 270 degrees) is equivalent to making one 90-degree rotation to the right, and vice versa.
We can likewise produce the dragon curve by interpreting A000265(n), the whole odd part of n, as the number of 90-degree rotations to make in a single direction at the n-th "turn" in the curve. (End)

Examples

			a(100) = 1: the odd part of 100 is 100/4 = 25, and 25 mod 4 = 1.
		

Crossrefs

Programs

  • Mathematica
    Array[Mod[#/(2^IntegerExponent[#, 2]), 4] &, 105] (* Michael De Vlieger, Feb 24 2021 *)
  • PARI
    a(n)=bitand(n/(2^valuation(n,2)), 3); /* Joerg Arndt, Jul 18 2012 */
    
  • Python
    def A099545(n): return n>>(~n&n-1).bit_length()&3 # Chai Wah Wu, Feb 26 2025

Formula

a(n) = 2 * A038189(n) + 1.
(a(n)-1)/2 = A014707(n). - Alexandre Wajnberg, Jan 02 2006
a(n) = A010873(A000265(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2. - Amiram Eldar, Aug 29 2024

A014710 The regular paper-folding (or dragon curve) sequence. Alphabet {2,1}.

Original entry on oeis.org

2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2
Offset: 0

Views

Author

Keywords

Crossrefs

See A014577 for more references and more terms.
The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - N. J. A. Sloane, Jul 27 2012

Programs

  • Mathematica
    Nest[Append[#1, If[EvenQ[#2], 2 - BitAnd[1, #2/2], #1[[Ceiling[#2/2]]]]] & @@ {#, Length@ #} &, {2}, 105] (* Michael De Vlieger, Apr 08 2021 *)
  • PARI
    a(n)=if(n%2==0, 2-bitand(1,n\2), a(n\2) );
    for(n=0,122,print1(a(n),", "))
    
  • Python
    def A014710(n):
        s = bin(n+1)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return 2-int(s[m-i-2]) if m-i-2 >= 0 else 2 # Chai Wah Wu, Apr 08 2021

Formula

Set a=2, b=1, S(0)=a, S(n+1) = S(n)aF(S(n)), where F(x) reverses x and then interchanges a and b; sequence is limit S(infinity).
a(4*n) = 2, a(4*n+2) = 1, a(2*n+1) = a(n).

A082410 a(1)=0. Thereafter, the sequence is constructed using the rule: for any k >= 0, if a(1), a(2), ..., a(2^k+1) are known, the next 2^k terms are given as follows: a(2^k+1+i) = 1 - a(2^k+1-i) for 1 <= i <= 2^k.

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Apr 24 2003

Keywords

Comments

a(n) is A014577 shifted right twice (the definition here is similar to one of the constructions for A034947). - N. J. A. Sloane, Jul 27 2012
Complement of characteristic function of A060833.
From Tanya Khovanova, Apr 21 2020: (Start)
Suppose you have a deck of cards face down with 2^n cards such that the color pattern corresponds to this sequence: 0 for one color, 1 for the other. Then you proceed in the following manner: transfer to top card to the bottom of the deck, deal the next card, then repeat. The dealt cards will have alternating colors.
Even terms of this sequence alternate: 1, 0, 1, 0 and so on.
Removing even-indexed terms doesn't change the sequence. (End)

Examples

			First 3 terms are 0,1,1; therefore, a(4) = a(3+1) = 1 - a(3-1) = 1 - a(2) = 0, a(5) = a(3+2) = 1 - a(3-2) = 1 - a(1) = 1 and the sequence begins 0, 1, 1, 0, 1, ...
		

Crossrefs

The following are all essentially the same sequence: A014577, A014707, A014709, A014710, A034947, A038189, A082410. - N. J. A. Sloane, Jul 27 2012

Programs

  • Python
    def A082410(n):
        if n == 1:
            return 0
        s = bin(n-1)[2:]
        m = len(s)
        i = s[::-1].find('1')
        return 1-int(s[m-i-2]) if m-i-2 >= 0 else 1 # Chai Wah Wu, Apr 08 2021

Formula

For n >= 2, Sum_{k=1..n} a(k) = (n + A037834(n-1))/2.
a(1) = 0, a(4*n+2) = 1, a(4*n+4) = 0, a(2*n+1) = a(n+1) for n >= 0. - A.H.M. Smeets, Jul 27 2018

A317335 a(n) = A317332(n) - 8*n.

Original entry on oeis.org

1, 1, -2, 1, 1, -2, -2, 1, 1, 1, -2, -2, 1, -2, -2, 1, 1, 1, -2, 1, 1, -2, -2, -2, 1, 1, -2, -2, 1, -2, -2, 1, 1, 1, -2, 1, 1, -2, -2, 1, 1, 1, -2, -2, 1, -2, -2, -2, 1, 1, -2, 1, 1, -2, -2, -2, 1, 1, -2, -2, 1, -2, -2
Offset: 1

Views

Author

A.H.M. Smeets, Jul 26 2018

Keywords

Crossrefs

Programs

  • Python
    n, f, i, p, q = 1, 1, 0, 0, 1
    while i < 1000000:
        i, p, q = i+1, p*10, q*10
        if i == f:
            p, n = p+1, n+1
            f = f*n
    n, a, j = 0, 0, 0
    while p%q > 0:
        a, f, p, q = a+1, p//q, q, p%q
        if f == 8:
            n = n+1
            print(n, a-1-8*n)

Formula

a(4*n-3) = 1, a(4*n-1) = -2, a(2*n) = a(n) for n > 0.
abs(a(n)+1) = A014710(n-1).
abs(a(n)) = A014709(n-1).
Showing 1-8 of 8 results.