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 1-10 of 33 results. Next

A182093 Partial sums of A005590.

Original entry on oeis.org

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

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Author

Reinhard Zumkeller, Apr 11 2012

Keywords

Comments

a(n+1) = a(n) + A005590(n+1).

Crossrefs

Cf. A005590.

Programs

  • Haskell
    a182093 n = a182093_list !! n
    a182093_list = scanl1 (+) a005590_list
  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := a[n] = If[OddQ@ n, a[(n - 1)/2 + 1] - a[(n - 1)/2], a[n/2]]; Accumulate@ Table[a@ n, {n, 0, 104}] (* after Jean-François Alcover at A005590, or *)
    Table[SeriesCoefficient[(x/(1 - x)) Product[(1 + x^(2^k) - x^(2^(k + 1))), {k, 0, n}], {x, 0, n}], {n, 0, 70}] (* Michael De Vlieger, Feb 27 2017 *)

Formula

a(2^n) = n + 1.
G.f.: (x/(1 - x))*Product_{k>=0} (1 + x^(2^k) - x^(2^(k+1))). - Ilya Gutkovskiy, Feb 27 2017

A000069 Odious numbers: numbers with an odd number of 1's in their binary expansion.

Original entry on oeis.org

1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 21, 22, 25, 26, 28, 31, 32, 35, 37, 38, 41, 42, 44, 47, 49, 50, 52, 55, 56, 59, 61, 62, 64, 67, 69, 70, 73, 74, 76, 79, 81, 82, 84, 87, 88, 91, 93, 94, 97, 98, 100, 103, 104, 107, 109, 110, 112, 115, 117, 118, 121, 122, 124, 127, 128
Offset: 1

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Author

Keywords

Comments

This sequence and A001969 give the unique solution to the problem of splitting the nonnegative integers into two classes in such a way that sums of pairs of distinct elements from either class occur with the same multiplicities [Lambek and Moser]. Cf. A000028, A000379.
In French: les nombres impies.
Has asymptotic density 1/2, since exactly 2 of the 4 numbers 4k, 4k+1, 4k+2, 4k+3 have an even sum of bits, while the other 2 have an odd sum. - Jeffrey Shallit, Jun 04 2002
Nim-values for game of mock turtles played with n coins.
A115384(n) = number of odious numbers <= n; A000120(a(n)) = A132680(n). - Reinhard Zumkeller, Aug 26 2007
Indices of 1's in the Thue-Morse sequence A010060. - Tanya Khovanova, Dec 29 2008
For any positive integer m, the partition of the set of the first 2^m positive integers into evil ones E and odious ones O is a fair division for any polynomial sequence p(k) of degree less than m, that is, Sum_{k in E} p(k) = Sum_{k in O} p(k) holds for any polynomial p with deg(p) < m. - Pietro Majer, Mar 15 2009
For n>1 let b(n) = a(n-1). Then b(b(n)) = 2b(n). - Benoit Cloitre, Oct 07 2010
Lexicographically earliest sequence of distinct nonnegative integers with no term being the binary exclusive OR of any terms. The equivalent sequence for addition or for subtraction is A005408 (the odd numbers) and for multiplication is A026424. - Peter Munn, Jan 14 2018
Numbers of the form m XOR (2*m+1) for some m >= 0. - Rémy Sigrist, Apr 14 2022

Examples

			For k=2, x=0 and x=0.2 we respectively have 1^2 + 2^2 + 4^2 + 7^2 = 0^2 + 3^2 + 5^2 + 6^2 = 70;
(1.2)^2 + (2.2)^2 + (4.2)^2 + (7.2)^2 = (0.2)^2 + (3.2)^2 + (5.2)^2 + (6.2)^2 = 75.76;
for k=3, x=1.8, we have (2.8)^3 + (3.8)^3 + (5.8)^3 + (8.8)^3 + (9.8)^3 + (12.8)^3 + (14.8)^3 + (15.8)^3 = (1.8)^3 + (4.8)^3 + (6.8)^3 + (7.8)^3 + (10.8)^3 + (11.8)^3 + (13.8)^3 + (16.8)^3 = 11177.856. - _Vladimir Shevelev_, Jan 16 2012
		

References

  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 433.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 22.
  • Vladimir S. Shevelev, On some identities connected with the partition of the positive integers with respect to the Morse sequence, Izv. Vuzov of the North-Caucasus region, Nature sciences 4 (1997), 21-23 (in Russian).
  • N. J. A. Sloane, A handbook of Integer Sequences, Academic Press, 1973 (including this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015.
Complement of A001969 (the evil numbers). Cf. A133009.
a(n) = 2*n + 1 - A010060(n) = A001969(n) + (-1)^A010060(n).
First differences give A007413.
Note that A000079, A083420, A002042, A002089, A132679 are subsequences.
See A027697 for primes, also A230095.
Cf. A005408 (odd numbers), A006068, A026424.

Programs

  • Haskell
    a000069 n = a000069_list !! (n-1)
    a000069_list = [x | x <- [0..], odd $ a000120 x]
    -- Reinhard Zumkeller, Feb 01 2012
    
  • Magma
    [ n: n in [1..130] | IsOdd(&+Intseq(n, 2)) ]; // Klaus Brockhaus, Oct 07 2010
    
  • Maple
    s := proc(n) local i,j,k,b,sum,ans; ans := [ ]; j := 0; for i while jA000069 := n->t1[n]; # s(k) gives first k terms.
    is_A000069 := n -> type(add(i,i=convert(n,base,2)),odd):
    seq(`if`(is_A000069(i),i,NULL),i=0..40); # Peter Luschny, Feb 03 2011
  • Mathematica
    Select[Range[300], OddQ[DigitCount[ #, 2][[1]]] &] (* Stefan Steinerberger, Mar 31 2006 *)
    a[ n_] := If[ n < 1, 0, 2 n - 1 - Mod[ Total @ IntegerDigits[ n - 1, 2], 2]]; (* Michael Somos, Jun 01 2013 *)
  • PARI
    {a(n) = if( n<1, 0, 2*n - 1 - subst( Pol(binary( n-1)), x, 1) % 2)}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    {a(n) = if( n<2, n==1, if( n%2, a((n+1)/2) + n-1, -a(n/2) + 3*(n-1)))}; /* Michael Somos, Jun 01 2013 */
    
  • PARI
    a(n)=2*n-1-hammingweight(n-1)%2 \\ Charles R Greathouse IV, Mar 22 2013
    
  • Python
    [n for n in range(1, 201) if bin(n)[2:].count("1") % 2] # Indranil Ghosh, May 03 2017
    
  • Python
    def A000069(n): return ((m:=n-1)<<1)+(m.bit_count()&1^1) # Chai Wah Wu, Mar 03 2023

Formula

G.f.: 1 + Sum_{k>=0} (t*(2+2t+5t^2-t^4)/(1-t^2)^2) * Product_{j=0..k-1} (1-x^(2^j)), t=x^2^k. - Ralf Stephan, Mar 25 2004
a(n+1) = (1/2) * (4*n + 1 + (-1)^A000120(n)). - Ralf Stephan, Sep 14 2003
Numbers n such that A010060(n) = 1. - Benoit Cloitre, Nov 15 2003
a(2*n+1) + a(2*n) = A017101(n) = 8*n+3. a(2*n+1) - a(2*n) gives the Thue-Morse sequence (1, 3 version): 1, 3, 3, 1, 3, 1, 1, 3, 3, 1, 1, 3, 1, ... A001969(n) + A000069(n) = A016813(n) = 4*n+1. - Philippe Deléham, Feb 04 2004
(-1)^a(n) = 2*A010060(n)-1. - Benoit Cloitre, Mar 08 2004
a(1) = 1; for n > 1: a(2*n) = 6*n-3 -a(n), a(2*n+1) = a(n+1) + 2*n. - Corrected by Vladimir Shevelev, Sep 25 2011
For k >= 1 and for every real (or complex) x, we have Sum_{i=1..2^k} (a(i)+x)^s = Sum_{i=1..2^k} (A001969(i)+x)^s, s=0..k.
For x=0, s <= k-1, this is known as Prouhet theorem (see J.-P. Allouche and Jeffrey Shallit, The Ubiquitous Prouhet-Thue-Morse Sequence). - Vladimir Shevelev, Jan 16 2012
a(n+1) mod 2 = 1 - A010060(n) = A010059(n). - Robert G. Wilson v, Jan 18 2012
A005590(a(n)) > 0. - Reinhard Zumkeller, Apr 11 2012
A106400(a(n)) = -1. - Reinhard Zumkeller, Apr 29 2012
a(n+1) = A006068(n) XOR (2*A006068(n) + 1). - Rémy Sigrist, Apr 14 2022

A001969 Evil numbers: nonnegative integers with an even number of 1's in their binary expansion.

Original entry on oeis.org

0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39, 40, 43, 45, 46, 48, 51, 53, 54, 57, 58, 60, 63, 65, 66, 68, 71, 72, 75, 77, 78, 80, 83, 85, 86, 89, 90, 92, 95, 96, 99, 101, 102, 105, 106, 108, 111, 113, 114, 116, 119, 120, 123, 125, 126, 129
Offset: 1

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Author

Keywords

Comments

This sequence and A000069 give the unique solution to the problem of splitting the nonnegative integers into two classes in such a way that sums of pairs of distinct elements from either class occur with the same multiplicities [Lambek and Moser]. Cf. A000028, A000379.
In French: les nombres païens.
Theorem: First differences give A036585. (Observed by Franklin T. Adams-Watters.)
Proof from Max Alekseyev, Aug 30 2006 (edited by N. J. A. Sloane, Jan 05 2021): (Start)
Observe that if the last bit of a(n) is deleted, we get the nonnegative numbers 0, 1, 2, 3, ... in order.
The last bit in a(n+1) is 1 iff the number of bits in n is odd, that is, iff A010060(n+1) is 1.
So, taking into account the different offsets here and in A010060, we have a(n) = 2*(n-1) + A010060(n-1).
Therefore the first differences of the present sequence equal 2 + first differences of A010060, which equals A036585. QED (End)
Integers k such that A010060(k-1)=0. - Benoit Cloitre, Nov 15 2003
Indices of zeros in the Thue-Morse sequence A010060 shifted by 1. - Tanya Khovanova, Feb 13 2009
Conjecture, checked up to 10^6: a(n) is also the sequence of numbers k representable as k = ror(x) XOR rol(x) (for some integer x) where ror(x)=A038572(x) is x rotated one binary place to the right, rol(x)=A006257(x) is x rotated one binary place to the left, and XOR is the binary exclusive-or operator. - Alex Ratushnyak, May 14 2016
From Charlie Neder, Oct 07 2018: (Start)
Conjecture is true: ror(x) and rol(x) have an even number of 1 bits in total (= 2 * A000120(x)), and XOR preserves the parity of this total, so the resulting number must have an even number of 1 bits. An x can be constructed corresponding to a(n) like so:
If the number of bits in a(n) is even, add a leading 0 so a(n) is 2k+1 bits long.
Do an inverse shuffle on a(n), then "divide" by 11, rotate the result k bits to the right, and shuffle to get x. (End)
Numbers of the form m XOR (2*m) for some m >= 0. - Rémy Sigrist, Feb 07 2021
The terms "evil numbers" and "odious numbers" were coined by Richard K. Guy, c. 1976 (Haque and Shallit, 2016) and appeared in the book by Berlekamp et al. (Vol. 1, 1st ed., 1982). - Amiram Eldar, Jun 08 2021

References

  • Elwyn R. Berlekamp, John H. Conway, Richard K. Guy, Winning Ways for Your Mathematical Plays, Volume 1, 2nd ed., A K Peters, 2001, chapter 14, p. 110.
  • Hugh L. Montgomery, Ten Lectures on the Interface Between Analytic Number Theory and Harmonic Analysis, Amer. Math. Soc., 1996, p. 208.
  • Donald J. Newman, A Problem Seminar, Springer; see Problem #89.
  • Vladimir S. Shevelev, On some identities connected with the partition of the positive integers with respect to the Morse sequence, Izv. Vuzov of the North-Caucasus region, Nature sciences 4 (1997), 21-23 (Russian).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Complement of A000069 (the odious numbers). Cf. A133009.
a(n)=2*n+A010060(n)=A000069(n)-(-1)^A010060(n). Cf. A018900.
The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015.
Cf. A036585 (differences), A010060, A006364.
For primes see A027699, also A130593.

Programs

  • Haskell
    a001969 n = a001969_list !! (n-1)
    a001969_list = [x | x <- [0..], even $ a000120 x]
    -- Reinhard Zumkeller, Feb 01 2012
    
  • Magma
    [ n : n in [0..129] | IsEven(&+Intseq(n,2)) ]; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 2006
    
  • Maple
    s := proc(n) local i,j,ans; ans := [ ]; j := 0; for i from 0 while jA001969 := n->t1[n]; # s(k) gives first k terms.
    # Alternative:
    seq(`if`(add(k, k=convert(n,base,2))::even, n, NULL), n=0..129); # Peter Luschny, Jan 15 2021
    # alternative for use outside this sequence
    isA001969 := proc(n)
        add(d,d=convert(n,base,2)) ;
        type(%,'even') ;
    end proc:
    A001969 := proc(n)
        option remember ;
        local a;
        if n = 0 then
            1;
        else
            for a from procname(n-1)+1 do
                if isA001969(a) then
                    return a;
                end if;
            end do:
        end if;
    end proc:
    seq(A001969(n),n=1..200) ; # R. J. Mathar, Aug 07 2022
  • Mathematica
    Select[Range[0,300], EvenQ[DigitCount[ #, 2][[1]]] &]
    a[ n_] := If[ n < 1, 0, With[{m = n - 1}, 2 m + Mod[-Total@IntegerDigits[m, 2], 2]]]; (* Michael Somos, Jun 09 2019 *)
  • PARI
    a(n)=n-=1; 2*n+subst(Pol(binary(n)),x,1)%2
    
  • PARI
    a(n)=if(n<1,0,if(n%2==0,a(n/2)+n,-a((n-1)/2)+3*n))
    
  • PARI
    a(n)=2*(n-1)+hammingweight(n-1)%2 \\ Charles R Greathouse IV, Mar 22 2013
    
  • Python
    def ok(n): return bin(n)[2:].count('1') % 2 == 0
    print(list(filter(ok, range(130)))) # Michael S. Branicky, Jun 02 2021
    
  • Python
    from itertools import chain, count, islice
    def A001969_gen(): # generator of terms
        return chain((0,),chain.from_iterable((sorted(n^ n<<1 for n in range(2**l,2**(l+1))) for l in count(0))))
    A001969_list = list(islice(A001969_gen(),30)) # Chai Wah Wu, Jun 29 2022
    
  • Python
    def A001969(n): return ((m:=n-1).bit_count()&1)+(m<<1) # Chai Wah Wu, Mar 03 2023

Formula

a(n+1) - A001285(n) = 2n-1 has been verified for n <= 400. - John W. Layman, May 16 2003 [This can be directly verified by comparing Ralf Stephan's formulas for this sequence (see below) and for A001285. - Jianing Song, Nov 04 2024]
Note that 2n+1 is in the sequence iff 2n is not and so this sequence has asymptotic density 1/2. - Franklin T. Adams-Watters, Aug 23 2006
a(n) = (1/2) * (4n - 3 - (-1)^A000120(n-1)). - Ralf Stephan, Sep 14 2003
G.f.: Sum_{k>=0} (t(3+2t+3t^2)/(1-t^2)^2) * Product_{l=0..k-1} (1-x^(2^l)), where t = x^2^k. - Ralf Stephan, Mar 25 2004
a(2*n+1) + a(2*n) = A017101(n-1) = 8*n-5.
a(2*n) - a(2*n-1) gives the Thue-Morse sequence (3, 1 version): 3, 1, 1, 3, 1, 3, 3, 1, 1, 3, .... A001969(n) + A000069(n) = A016813(n-1) = 4*n-3. - Philippe Deléham, Feb 04 2004
a(1) = 0; for n > 1: a(n) = 3*n-3 - a(n/2) if n even, a(n) = a((n+1)/2)+n-1 if n odd.
Let b(n) = 1 if sum of digits of n is even, -1 if it is odd; then Shallit (1985) showed that Product_{n>=0} ((2n+1)/(2n+2))^b(n) = 1/sqrt(2).
a(n) = 2n - 2 + A010060(n-1). - Franklin T. Adams-Watters, Aug 28 2006
A005590(a(n-1)) <= 0. - Reinhard Zumkeller, Apr 11 2012
A106400(a(n-1)) = 1. - Reinhard Zumkeller, Apr 29 2012
a(n) = (a(n-1) + 2) XOR A010060(a(n-1) + 2). - Falk Hüffner, Jan 21 2022
a(n+1) = A006068(n) XOR (2*A006068(n)). - Rémy Sigrist, Apr 14 2022

Extensions

More terms from Robin Trew (trew(AT)hcs.harvard.edu)

A003714 Fibbinary numbers: if n = F(i1) + F(i2) + ... + F(ik) is the Zeckendorf representation of n (i.e., write n in Fibonacci number system) then a(n) = 2^(i1 - 2) + 2^(i2 - 2) + ... + 2^(ik - 2). Also numbers whose binary representation contains no two adjacent 1's.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37, 40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85, 128, 129, 130, 132, 133, 136, 137, 138, 144, 145, 146, 148, 149, 160, 161, 162, 164, 165, 168, 169, 170, 256, 257, 258, 260, 261, 264
Offset: 0

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Author

Keywords

Comments

The name "Fibbinary" is due to Marc LeBrun.
"... integers whose binary representation contains no consecutive ones and noticed that the number of such numbers with n bits was fibonacci(n)". [posting to sci.math by Bob Jenkins (bob_jenkins(AT)burtleburtle.net), Jul 17 2002]
From Benoit Cloitre, Mar 08 2003: (Start)
A number m is in the sequence if and only if C(3m, m) (or equally, C(3m, 2m)) is odd.
a(n) == A003849(n) (mod 2). (End)
Numbers m such that m XOR 2*m = 3*m. - Reinhard Zumkeller, May 03 2005. [This implies that A003188(2*a(n)) = 3*a(n) holds for all n.]
Numbers whose base-2 representation contains no two adjacent ones. For example, m = 17 = 10001_2 belongs to the sequence, but m = 19 = 10011_2 does not. - Ctibor O. Zizka, May 13 2008
m is in the sequence if and only if the central Stirling number of the second kind S(2*m, m) = A007820(m) is odd. - O-Yeat Chan (math(AT)oyeat.com), Sep 03 2009
A000120(3*a(n)) = 2*A000120(a(n)); A002450 is a subsequence.
Every nonnegative integer can be expressed as the sum of two terms of this sequence. - Franklin T. Adams-Watters, Jun 11 2011
Subsequence of A213526. - Arkadiusz Wesolowski, Jun 20 2012
This is also the union of A215024 and A215025 - see the Comment in A014417. - N. J. A. Sloane, Aug 10 2012
The binary representation of each term m contains no two adjacent 1's, so we have (m XOR 2m XOR 3m) = 0, and thus a two-player Nim game with three heaps of (m, 2m, 3m) stones is a losing configuration for the first player. - V. Raman, Sep 17 2012
Positions of zeros in A014081. - John Keith, Mar 07 2022
These numbers are similar to Fibternary numbers A003726, Tribbinary numbers A060140 and Tribternary numbers. This sequence is a subsequence of Fibternary numbers A003726. The number of Fibbinary numbers less than any power of two is a Fibonacci number. We can generate this sequence recursively: start with 0 and 1; then, if x is in the sequence add 2x and 4x+1 to the sequence. The Fibbinary numbers have the property that the n-th Fibbinary number is even if the n-th term of the Fibonacci word is a. Respectively, the n-th Fibbinary number is odd (of the form 4x+1) if the n-th term of the Fibonacci word is b. Every number has a Fibbinary multiple. - Tanya Khovanova and PRIMES STEP Senior, Aug 30 2022
This is the ordered set S of numbers defined recursively by: 0 is in S; if x is in S, then 2*x and 4*x + 1 are in S. See Kimberling (2006) Example 3, in references below. - Harry Richman, Jan 31 2024

Examples

			From _Joerg Arndt_, Jun 11 2011: (Start)
In the following, dots are used for zeros in the binary representation:
  a(n)  binary(a(n))  n
    0:    .......     0
    1:    ......1     1
    2:    .....1.     2
    4:    ....1..     3
    5:    ....1.1     4
    8:    ...1...     5
    9:    ...1..1     6
   10:    ...1.1.     7
   16:    ..1....     8
   17:    ..1...1     9
   18:    ..1..1.    10
   20:    ..1.1..    11
   21:    ..1.1.1    12
   32:    .1.....    13
   33:    .1....1    14
   34:    .1...1.    15
   36:    .1..1..    16
   37:    .1..1.1    17
   40:    .1.1...    18
   41:    .1.1..1    19
   42:    .1.1.1.    20
   64:    1......    21
   65:    1.....1    22
(End)
		

References

  • Donald E. Knuth, The Art of Computer Programming: Fundamental Algorithms, Vol. 1, 2nd ed., Addison-Wesley, 1973, pp. 85, 493.

Crossrefs

A007088(a(n)) = A014417(n) (same sequence in binary). Complement: A004780. Char. function: A085357. Even terms: A022340, odd terms: A022341. First difference: A129761.
Other sequences based on similar restrictions on binary expansion: A003726 & A278038, A003754, A048715, A048718, A107907, A107909.
3*a(n) is in A001969.
Cf. A014081 (count 11 bits).

Programs

  • Haskell
    import Data.Set (Set, singleton, insert, deleteFindMin)
    a003714 n = a003714_list !! n
    a003714_list = 0 : f (singleton 1) where
       f :: Set Integer -> [Integer]
       f s = m : (f $ insert (4*m + 1) $ insert (2*m) s')
             where (m, s') = deleteFindMin s
    -- Reinhard Zumkeller, Jun 03 2012, Feb 07 2012
    
  • Maple
    A003714 := proc(n)
        option remember;
        if n < 3 then
            n ;
        else
            2^(A072649(n)-1) + procname(n-combinat[fibonacci](1+A072649(n))) ;
        end if;
    end proc:
    seq(A003714(n),n=0..10) ;
    # To produce a table giving n, a(n) (base 10), a(n) (base 2) - from N. J. A. Sloane, Sep 30 2018
    # binary: binary representation of n, in human order
    binary:=proc(n) local t1,L;
    if n<0 then ERROR("n must be nonnegative"); fi;
    if n=0 then return([0]); fi;
    t1:=convert(n,base,2); L:=nops(t1);
    [seq(t1[L+1-i],i=1..L)];
    end;
    for n from 0 to 100 do t1:=A003714(n); lprint(n, t1, binary(t1)); od:
  • Mathematica
    fibBin[n_Integer] := Block[{k = Ceiling[Log[GoldenRatio, n Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; FromDigits[fr, 2]]; Table[fibBin[n], {n, 0, 61}] (* Robert G. Wilson v, Sep 18 2004 *)
    Select[Range[0, 270], ! MemberQ[Partition[IntegerDigits[#, 2], 2, 1], {1, 1}] &] (* Harvey P. Dale, Jul 17 2011 *)
    Select[Range[256], BitAnd[#, 2 #] == 0 &] (* Alonso del Arte, Jun 18 2012 *)
    With[{r = Range[10^5]}, Pick[r, BitAnd[r, 2 r], 0]] (* Eric W. Weisstein, Aug 18 2017 *)
    Select[Range[0, 299], SequenceCount[IntegerDigits[#, 2], {1, 1}] == 0 &] (* Requires Mathematica version 10 or later. -- Harvey P. Dale, Dec 06 2018 *)
  • PARI
    msb(n)=my(k=1); while(k<=n, k<<=1); k>>1
    for(n=1,1e4,k=bitand(n,n<<1);if(k,n=bitor(n,msb(k)-1),print1(n", "))) \\ Charles R Greathouse IV, Jun 15 2011
    
  • PARI
    select( is_A003714(n)=!bitand(n,n>>1), [0..266])
    {(next_A003714(n,t)=while(t=bitand(n+=1,n<<1), n=bitor(n,1<A003714(t)) \\ M. F. Hasler, Nov 30 2021
    
  • Python
    for n in range(300):
        if 2*n & n == 0:
            print(n, end=",") # Alex Ratushnyak, Jun 21 2012
    
  • Python
    def A003714(n):
        tlist, s = [1,2], 0
        while tlist[-1]+tlist[-2] <= n:
            tlist.append(tlist[-1]+tlist[-2])
        for d in tlist[::-1]:
            s *= 2
            if d <= n:
                s += 1
                n -= d
        return s # Chai Wah Wu, Jun 14 2018
    
  • Python
    def fibbinary():
        x = 0
        while True:
            yield x
            y = ~(x >> 1)
            x = (x - y) & y # Falk Hüffner, Oct 23 2021
    (C++)
    /* start with x=0, then repeatedly call x=next_fibrep(x): */
    ulong next_fibrep(ulong x)
    {
        // 2 examples:         //  ex. 1             //  ex.2
        //                     // x == [*]0 010101   // x == [*]0 01010
        ulong y = x | (x>>1);  // y == [*]? 011111   // y == [*]? 01111
        ulong z = y + 1;       // z == [*]? 100000   // z == [*]? 10000
        z = z & -z;            // z == [0]0 100000   // z == [0]0 10000
        x ^= z;                // x == [*]0 110101   // x == [*]0 11010
        x &= ~(z-1);           // x == [*]0 100000   // x == [*]0 10000
        return x;
    }
    /* Joerg Arndt, Jun 22 2012 */
    
  • Scala
    (0 to 255).filter(n => (n & 2 * n) == 0) // Alonso del Arte, Apr 12 2020
    (C#)
    public static bool IsFibbinaryNum(this int n) => ((n & (n >> 1)) == 0) ? true : false; // Frank Hollstein, Jul 07 2021

Formula

No two adjacent 1's in binary expansion.
Let f(x) := Sum_{n >= 0} x^Fibbinary(n). (This is the generating function of the characteristic function of this sequence.) Then f satisfies the functional equation f(x) = x*f(x^4) + f(x^2).
a(0) = 0, a(1) = 1, a(2) = 2, a(n) = 2^(A072649(n) - 1) + a(n - A000045(1 + A072649(n))). - Antti Karttunen
It appears that this sequence gives m such that A082759(3*m) is odd; or, probably equivalently, m such that A037011(3*m) = 1. - Benoit Cloitre, Jun 20 2003
If m is in the sequence then so are 2*m and 4*m + 1. - Henry Bottomley, Jan 11 2005
A116361(a(n)) <= 1. - Reinhard Zumkeller, Feb 04 2006
A085357(a(n)) = 1; A179821(a(n)) = a(n). - Reinhard Zumkeller, Jul 31 2010
a(n)/n^k is bounded (but does not tend to a limit), where k = 1.44... = A104287. - Charles R Greathouse IV, Sep 19 2012
a(n) = a(A193564(n+1))*2^(A003849(n) + 1) + A003849(n) for n > 0. - Daniel Starodubtsev, Aug 05 2021
There are Fibonacci(n+1) terms with up to n bits in this sequence. - Charles R Greathouse IV, Oct 22 2021
Sum_{n>=1} 1/a(n) = 3.704711752910469457886531055976801955909489488376627037756627135425780134020... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022

Extensions

Edited by Antti Karttunen, Feb 21 2006
Cross reference to A007820 added (into O-Y.C. comment) by Jason Kimberley, Sep 14 2009
Typo corrected by Jeffrey Shallit, Sep 26 2014

A229817 Even bisection gives sequence a itself, n->a(2*(2*n+k)-1) gives k-th differences of a for k=1..2 with a(n)=n for n<2.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Sep 30 2013

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local m, q, r;
          m:= (irem(n, 4, 'q')+1)/2;
          `if`(n<2, n, `if`(irem(n, 2, 'r')=0, a(r),
          add(a(q+m-j)*(-1)^j*binomial(m, j), j=0..m)))
        end:
    seq(a(n), n=0..100);
  • Mathematica
    a[n_] := a[n] = Module[{m, q, r}, {q, m} = QuotientRemainder[n, 4]; m = (m + 1)/2; If[n<2, n, If[Mod[n, 2]==0, a[Quotient[n, 2]], Sum[a[q+m-j] * (-1)^j * Binomial[m, j], {j, 0, m}]]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Feb 22 2017, translated from Maple *)

Formula

a(2*n) = a(n),
a(4*n+1) = a(n+1) - a(n),
a(4*n+3) = a(n+2) - 2*a(n+1) + a(n).

A229818 Even bisection gives sequence a itself, n->a(2*(3*n+k)-1) gives k-th differences of a for k=1..3 with a(n)=n for n<2.

Original entry on oeis.org

0, 1, 1, -1, 1, -1, -1, 0, 1, -2, -1, 6, -1, -2, 0, 4, 1, -8, -2, 2, -1, -4, 6, 6, -1, -2, -2, 2, 0, -1, 4, 0, 1, 1, -8, -1, -2, 1, 2, 0, -1, -4, -4, 1, 6, -4, 6, 8, -1, -3, -2, 4, -2, 2, 2, 1, 0, 6, -1, -20, 4, 7, 0, -14, 1, 20, 1, -7, -8, 6, -1, -3, -2, -1
Offset: 0

Views

Author

Alois P. Heinz, Sep 30 2013

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local m, q, r;
          m:= (irem(n, 6, 'q')+1)/2;
          `if`(n<2, n, `if`(irem(n, 2, 'r')=0, a(r),
          add(a(q+m-j)*(-1)^j*binomial(m, j), j=0..m)))
        end:
    seq(a(n), n=0..100);
  • Mathematica
    a[n_] := a[n] = Module[{m, q, r, q2, r2}, {q, r} = QuotientRemainder[n, 6]; m = (r+1)/2; If[n<2, n, {q2, r2} = QuotientRemainder[n, 2]; If[r2 == 0, a[q2], Sum[a[q+m-j]*(-1)^j*Binomial[m, j], {j, 0, m}]]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Mar 08 2017, translated from Maple *)
  • PARI
    {M=Map(); a(n)= n&&n>>=valuation(n, 2); my(r); mapisdefined(M, n, &r) && return(r); r=if(n<2, n, my(m=n%6, k=n\6); if(1==m, a(k+1)-a(k), 3==m, a(k+2)-2*a(k+1)+a(k), a(k+3)-3*a(k+2)+3*a(k+1)-a(k))); mapput(~M, n, r); r;} \\ Ruud H.G. van Tol, Nov 19 2024

Formula

a(2*n) = a(n),
a(6*n+1) = a(n+1) - a(n),
a(6*n+3) = a(n+2) - 2*a(n+1) + a(n),
a(6*n+5) = a(n+3) - 3*a(n+2) + 3*a(n+1) - a(n).

A229819 Even bisection gives sequence a itself, n->a(2*(4*n+k)-1) gives k-th differences of a for k=1..4 with a(n)=n for n<2.

Original entry on oeis.org

0, 1, 1, -1, 1, -1, -1, 7, 1, 0, -1, -2, -1, 6, 7, -14, 1, -2, 0, 4, -1, -8, -2, 14, -1, 2, 6, -4, 7, 6, -14, 0, 1, -2, -2, 2, 0, 6, 4, -28, -1, 0, -8, 8, -2, -22, 14, 41, -1, 8, 2, -14, 6, 19, -4, -24, 7, -6, 6, 5, -14, -5, 0, 5, 1, -1, -2, 0, -2, 0, 2, 2, 0
Offset: 0

Views

Author

Alois P. Heinz, Sep 30 2013

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local m, q, r;
          m:= (irem(n, 8, 'q')+1)/2;
          `if`(n<2, n, `if`(irem(n, 2, 'r')=0, a(r),
          add(a(q+m-j)*(-1)^j*binomial(m, j), j=0..m)))
        end:
    seq(a(n), n=0..100);
  • Mathematica
    a[n_] := a[n] = Module[{m, q, r, q2, r2}, {q, r} = QuotientRemainder[n, 8]; m = (r+1)/2; If[n<2, n, {q2, r2} = QuotientRemainder[n, 2]; If[r2 == 0, a[q2], Sum[a[q+m-j]*(-1)^j*Binomial[m, j], {j, 0, m}]]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Mar 08 2017, translated from Maple *)

Formula

a(2*n) = a(n),
a(8*n+1) = a(n+1) - a(n),
a(8*n+3) = a(n+2) - 2*a(n+1) + a(n),
a(8*n+5) = a(n+3) - 3*a(n+2) + 3*a(n+1) - a(n).
a(8*n+7) = a(n+4) - 4*a(n+3) + 6*a(n+2) - 4*a(n+1) + a(n).

A229820 Even bisection gives sequence a itself, n->a(2*(5*n+k)-1) gives k-th differences of a for k=1..5 with a(n)=n for n<2.

Original entry on oeis.org

0, 1, 1, -1, 1, -1, -1, 7, 1, -21, -1, 0, -1, -2, 7, 6, 1, -14, -21, 28, -1, -2, 0, 4, -1, -8, -2, 14, 7, -14, 6, 2, 1, -4, -14, 6, -21, 0, 28, -28, -1, -2, -2, 2, 0, 6, 4, -28, -1, 48, -8, 0, -2, 8, 14, -22, 7, 20, -14, 40, 6, 8, 2, -14, 1, -2, -4, 60, -14
Offset: 0

Views

Author

Alois P. Heinz, Sep 30 2013

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local m, q, r;
          m:= (irem(n, 10, 'q')+1)/2;
          `if`(n<2, n, `if`(irem(n, 2, 'r')=0, a(r),
          add(a(q+m-j)*(-1)^j*binomial(m, j), j=0..m)))
        end:
    seq(a(n), n=0..100);
  • Mathematica
    a[n_] := a[n] = Module[{m, q, r, q2, r2}, {q, r} = QuotientRemainder[n, 10]; m = (r+1)/2; If[n<2, n, {q2, r2} = QuotientRemainder[n, 2]; If[r2 == 0, a[q2], Sum[a[q+m-j]*(-1)^j*Binomial[m, j], {j, 0, m}]]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Mar 08 2017, translated from Maple *)

Formula

a(2*n) = a(n),
a(2*(5*n+k)-1) = Sum_{j=0..k} (-1)^j * C(k,j) * a(n+k-j) for k=1..5.

A229821 Even bisection gives sequence a itself, n->a(2*(6*n+k)-1) gives k-th differences of a for k=1..6 with a(n)=n for n<2.

Original entry on oeis.org

0, 1, 1, -1, 1, -1, -1, 7, 1, -21, -1, 49, -1, 0, 7, -2, 1, 6, -21, -14, -1, 28, 49, -42, -1, -2, 0, 4, 7, -8, -2, 14, 1, -14, 6, -14, -21, 2, -14, -4, -1, 6, 28, 0, 49, -28, -42, 76, -1, -2, -2, 2, 0, 6, 4, -28, 7, 48, -8, -8, -2, 0, 14, 8, 1, -22, -14, 20, 6
Offset: 0

Views

Author

Alois P. Heinz, Sep 30 2013

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local m, q, r;
          m:= (irem(n, 12, 'q')+1)/2;
          `if`(n<2, n, `if`(irem(n, 2, 'r')=0, a(r),
          add(a(q+m-j)*(-1)^j*binomial(m, j), j=0..m)))
        end:
    seq(a(n), n=0..100);
  • Mathematica
    a[n_] := a[n] = Module[{m, q, r, q2, r2}, {q, r} = QuotientRemainder[n, 12]; m = (r+1)/2; If[n<2, n, {q2, r2} = QuotientRemainder[n, 2]; If[r2 == 0, a[q2], Sum[a[q+m-j]*(-1)^j*Binomial[m, j], {j, 0, m}]]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Mar 08 2017, translated from Maple *)

Formula

a(2*n) = a(n),
a(2*(6*n+k)-1) = Sum_{j=0..k} (-1)^j * C(k,j) * a(n+k-j) for k=1..6.

A229822 Even bisection gives sequence a itself, n->a(2*(7*n+k)-1) gives k-th differences of a for k=1..7 with a(n)=n for n<2.

Original entry on oeis.org

0, 1, 1, -1, 1, -1, -1, 7, 1, -21, -1, 49, -1, -91, 7, 0, 1, -2, -21, 6, -1, -14, 49, 28, -1, -42, -91, 28, 7, -2, 0, 4, 1, -8, -2, 14, -21, -14, 6, -14, -1, 90, -14, 2, 49, -4, 28, 6, -1, 0, -42, -28, -91, 76, 28, -84, 7, -2, -2, 2, 0, 6, 4, -28, 1, 48, -8
Offset: 0

Views

Author

Alois P. Heinz, Sep 30 2013

Keywords

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; local m, q, r;
          m:= (irem(n, 14, 'q')+1)/2;
          `if`(n<2, n, `if`(irem(n, 2, 'r')=0, a(r),
          add(a(q+m-j)*(-1)^j*binomial(m, j), j=0..m)))
        end:
    seq(a(n), n=0..100);
  • Mathematica
    a[n_] := a[n] = Module[{m, q, r, q2, r2}, {q, r} = QuotientRemainder[n, 14]; m = (r+1)/2; If[n<2, n, {q2, r2} = QuotientRemainder[n, 2]; If[r2 == 0, a[q2], Sum[a[q+m-j]*(-1)^j*Binomial[m, j], {j, 0, m}]]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Mar 08 2017, translated from Maple *)

Formula

a(2*n) = a(n),
a(2*(7*n+k)-1) = Sum_{j=0..k} (-1)^j * C(k,j) * a(n+k-j) for k=1..7.
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