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 1-7 of 7 results.

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

Views

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

A003269 a(n) = a(n-1) + a(n-4) with a(0) = 0, a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 19, 26, 36, 50, 69, 95, 131, 181, 250, 345, 476, 657, 907, 1252, 1728, 2385, 3292, 4544, 6272, 8657, 11949, 16493, 22765, 31422, 43371, 59864, 82629, 114051, 157422, 217286, 299915, 413966, 571388, 788674, 1088589, 1502555, 2073943
Offset: 0

Views

Author

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 0..m-1. The generating function is 1/(1-x-x^m). Also a(n) = Sum_{i=0..n/m} binomial(n-(m-1)*i, i). This family of binomial summations or recurrences gives the number of ways to cover (without overlapping) a linear lattice of n sites with molecules that are m sites wide. Special case: m=1: A000079; m=4: A003269; m=5: A003520; m=6: A005708; m=7: A005709; m=8: A005710.
For this family of sequences, a(n+1) is the number of compositions of n+1 into parts 1 and m. For n>=m, a(n-m+1)is the number of compositions of n in which each part is greater than m or equivalently, in which parts 1 through m are excluded. - Gregory L. Simay, Jul 14 2016
For this family of sequences, let a(m,n) = a(n-1) + a(n-m). Then the number of compositions of n having m as a least summand is a(m, n-m) - a(m+1, n-m-1). - Gregory L. Simay, Jul 14 2016
For n>=3, a(n-3) = number of compositions of n in which each part is >=4. - Milan Janjic, Jun 28 2010
For n>=1, number of compositions of n into parts == 1 (mod 4). Example: a(8)=5 because there are 5 compositions of 8 into parts 1 or 5: (1,1,1,1,1,1,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1), (5,1,1,1). - Adi Dani, Jun 16 2011
a(n+1) is the number of compositions of n into parts 1 and 4. - Joerg Arndt, Jun 25 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=4, 2*a(n-3) equals the number of 2-colored compositions of n with all parts >= 4, such that no adjacent parts have the same color. - Milan Janjic, Nov 27 2011
Number of permutations satisfying -k<=p(i)-i<=r and p(i)-i not in I, i=1..n, with k=1, r=3, I={1,2}. - Vladimir Baltic, Mar 07 2012
a(n+4) equals the number of binary words of length n having at least 3 zeros between every two successive ones. - Milan Janjic, Feb 07 2015
From Clark Kimberling, Jun 13 2016: (Start)
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*.
Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2,x}, g(3) = {3, 2*x, x+1, x^2}, etc.
Let T(r) be the tree obtained by substituting r for x.
If N is a positive integer such that r = N^(1/4) is not an integer, then the number of (not necessarily distinct) integers in g(n) is A003269(n), for n > = 1. See A274142. (End)

Examples

			G.f.: x + x^2 + x^3 + x^4 + 2*x^5 + 3*x^6 + 4*x^7 + 5*x^8 + 7*x^9 + 10*x^10 + ...
The number of compositions of 12 having 4 as a least summand is a(4, 12 -4 + 1) - a(5, 12 - 5 + 1) = A003269(9) - A003520(8) = 7-4 = 3. The compositions are (84), (48) and (444). - _Gregory L. Simay_, Jul 14 2016
		

References

  • A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 120.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

See A017898 for an essentially identical sequence.
Row sums of A180184.

Programs

  • Haskell
    a003269 n = a003269_list !! n
    a003269_list = 0 : 1 : 1 : 1 : zipWith (+) a003269_list
                                              (drop 3 a003269_list)
    -- Reinhard Zumkeller, Feb 27 2011
    
  • Magma
    I:=[0,1,1,1]; [n le 4 select I[n] else Self(n-1) + Self(n-4) :n in [1..50]]; // Marius A. Burtea, Sep 13 2019
    
  • Maple
    with(combstruct): SeqSetU := [S, {S=Sequence(U), U=Set(Z, card > 3)}, unlabeled]: seq(count(SeqSetU, size=j), j=4..51);
    seq(add(binomial(n-3*k,k),k=0..floor(n/3)),n=0..47); # Zerinvary Lajos, Apr 03 2007
    A003269:=z/(1-z-z**4); # Simon Plouffe in his 1992 dissertation
    ZL:=[S, {a = Atom, b = Atom, S = Prod(X,Sequence(Prod(X,b))), X = Sequence(b,card >= 3)}, unlabelled]: seq(combstruct[count](ZL, size=n), n=3..50); # Zerinvary Lajos, Mar 26 2008
    M:= Matrix(4, (i,j)-> if j=1 then [1,0,0,1][i] elif (i=j-1) then 1 else 0 fi); a:= n-> (M^(n))[1,2]; seq(a(n), n=0..48); # Alois P. Heinz, Jul 27 2008
  • Mathematica
    a[0]= 0; a[1]= a[2]= a[3]= 1; a[n_]:= a[n]= a[n-1] + a[n-4]; Table[a[n], {n,0,50}]
    CoefficientList[Series[x/(1-x-x^4), {x,0,50}], x] (* Zerinvary Lajos, Mar 29 2007 *)
    Table[Sum[Binomial[n-3*i-1,i], {i,0,(n-1)/3}], {n,0,50}]
    LinearRecurrence[{1,0,0,1}, {0,1,1,1}, 50] (* Robert G. Wilson v, Jul 12 2014 *)
    nxt[{a_,b_,c_,d_}]:={b,c,d,a+d}; NestList[nxt,{0,1,1,1},50][[;;,1]] (* Harvey P. Dale, May 27 2024 *)
  • PARI
    {a(n) = polcoeff( if( n<0, (1 + x^3) / (1 + x^3 - x^4), 1 / (1 - x - x^4)) + x * O(x^abs(n)), abs(n))} /* Michael Somos, Jul 12 2003 */
    
  • SageMath
    @CachedFunction
    def a(n): return ((n+2)//3) if (n<4) else a(n-1) + a(n-4) # a = A003269
    [a(n) for n in (0..50)] # G. C. Greubel, Jul 25 2022

Formula

G.f.: x/(1-x-x^4).
G.f.: -1 + 1/(1-Sum_{k>=0} x^(4*k+1)).
a(n) = a(n-3) + a(n-4) + a(n-5) + a(n-6) for n>4.
a(n) = floor(d*c^n + 1/2) where c is the positive real root of -x^4+x^3+1 and d is the positive real root of 283*x^4-18*x^2-8*x-1 (c=1.38027756909761411... and d=0.3966506381592033124...). - Benoit Cloitre, Nov 30 2002
Equivalently, a(n) = floor(c^(n+3)/(c^4+3) + 1/2) with c as defined above (see A086106). - Greg Dresden and Shuer Jiang, Aug 31 2019
a(n) = term (1,2) in the 4 X 4 matrix [1,1,0,0; 0,0,1,0; 0,0,0,1; 1,0,0,0]^n. - Alois P. Heinz, Jul 27 2008
From Paul Barry, Oct 20 2009: (Start)
a(n+1) = Sum_{k=0..n} C((n+3*k)/4,k)*((1+(-1)^(n-k))/2 + cos(Pi*n/2))/2;
a(n+1) = Sum_{k=0..n} C(k,floor((n-k)/3))(2*cos(2*Pi*(n-k)/3)+1)/3. (End)
a(n) = Sum_{j=0..(n-1)/3} binomial(n-1-3*j,j) (cf. A180184). - Vladimir Kruchinin, May 23 2011
A017817(n) = a(-4 - n) * (-1)^n. - Michael Somos, Jul 12 2003
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(2*k+1 + x^3)/( x*(2*k+2 + x^3) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 29 2013
Appears a(n) = hypergeometric([1/4-n/4,1/2-n/4,3/4-n/4,1-n/4], [1/3-n/3,2/3-n/3,1-n/3], -4^4/3^3) for n>=10. - Peter Luschny, Sep 18 2014

Extensions

Additional comments from Yong Kong (ykong(AT)curagen.com), Dec 16 2000
Initial 0 prepended by N. J. A. Sloane, Apr 09 2008

A048715 Binary expansion matches (100(0)*)*(0|1|10)?; or, Zeckendorf-like expansion of n using recurrence f(n) = f(n-1) + f(n-3).

Original entry on oeis.org

0, 1, 2, 4, 8, 9, 16, 17, 18, 32, 33, 34, 36, 64, 65, 66, 68, 72, 73, 128, 129, 130, 132, 136, 137, 144, 145, 146, 256, 257, 258, 260, 264, 265, 272, 273, 274, 288, 289, 290, 292, 512, 513, 514, 516, 520, 521, 528, 529, 530, 544, 545, 546, 548, 576, 577, 578, 580
Offset: 0

Views

Author

Antti Karttunen, Mar 30 1999

Keywords

Comments

No more than one 1-bit in each bit triple.
All terms satisfy A048727(n) = 7*n.
Constructed from A000930 in the same way as A003714 is constructed from A000045.
It appears that n is in the sequence if and only if C(7n,n) is odd (cf. A003714). - Benoit Cloitre, Mar 09 2003
The conjecture by Benoit is correct. This is easily proved using the well-known result that the multiplicity with which a prime p divides C(n+m,n) is the number of carries when adding n+m in base p. - Franklin T. Adams-Watters, Oct 06 2009
Appears to be the set of numbers x such that (x AND 5*x) = x and (x OR 3*x)/x = 3. - Gary Detlefs, Jun 08 2024

Crossrefs

Programs

  • Mathematica
    Reap[Do[If[OddQ[Binomial[7n, n]], Sow[n]], {n, 0, 400}]][[2, 1]]
    (* Second program: *)
    filterQ[n_] := With[{bb = IntegerDigits[n, 2]}, !MatchQ[bb, {_, 1, 0, 1, _}|{_, 1, 1, _}]];
    Select[Range[0, 580], filterQ] (* Jean-François Alcover, Dec 31 2020 *)
  • PARI
    is(n)=!bitand(n, 6*n) \\ Charles R Greathouse IV, Oct 03 2016
    
  • Perl
    for my $k (0..580) { print "$k, " if sprintf("%b", $k) =~ m{^(100(0)*)*(0|1|10)?$}; } # Georg Fischer, Jun 26 2021
    
  • Python
    import re
    def ok(n): return re.fullmatch('(100(0)*)*(0|1|10)?', bin(n)[2:]) != None
    print(list(filter(ok, range(581)))) # Michael S. Branicky, Jun 26 2021

Formula

a(0) = 0, a(n) = (2^(invfoo(n)-1))+a(n-foo(invfoo(n))), where foo(n) is foo(n-1) + foo(n-3) (A000930) and invfoo is its "integral" (floored down) inverse.
a(n) XOR 6*a(n) = 7*a(n); 3*a(n) XOR 4*a(n) = 7*a(n); 3*a(n) XOR 5*a(n) = 6*a(n); (conjectures). - Paul D. Hanna, Jan 22 2006
The conjectures can be verified using the Walnut theorem-prover (see links). - Sebastian Karlsson, Dec 31 2022

Extensions

Definition corrected by Georg Fischer, Jun 26 2021

A048719 Binary expansion matches ((0)*0011)*(0*).

Original entry on oeis.org

0, 3, 6, 12, 24, 48, 51, 96, 99, 102, 192, 195, 198, 204, 384, 387, 390, 396, 408, 768, 771, 774, 780, 792, 816, 819, 1536, 1539, 1542, 1548, 1560, 1584, 1587, 1632, 1635, 1638, 3072, 3075, 3078, 3084, 3096
Offset: 0

Views

Author

Antti Karttunen, Mar 30 1999

Keywords

Comments

1-bits occur only in pairs, separated from other such pairs by at least two 0-bits.
All terms satisfy both A048727(n) = 3*n and A048725(n) = 5*n.

Crossrefs

Intersection of A048716 and A048717.

Programs

  • Mathematica
    filterQ[n_] := With[{bb = IntegerDigits[n, 2]}, !MatchQ[bb, {1}|{1, 0, _}|{_, 0, 1}|{_, 0, 1, 0, _}|{_, 1, 1, 1, _}|{_, 1, 0, 1, _}]];
    Select[Range[0, 3096], filterQ] (* Jean-François Alcover, Dec 31 2020 *)
  • PARI
    is(n)=n%3==0 && !bitand(n/3, 14*n/3) \\ Charles R Greathouse IV, Oct 03 2016

Formula

a(n) = 3*A048718(n).

A115424 Integers n > 0 such that n XOR 62*n = 63*n.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 65, 128, 129, 130, 256, 257, 258, 260, 512, 513, 514, 516, 520, 1024, 1025, 1026, 1028, 1032, 1040, 2048, 2049, 2050, 2052, 2056, 2064, 2080, 4096, 4097, 4098, 4100, 4104, 4112, 4128, 4160, 4161, 8192, 8193, 8194, 8196, 8200, 8208
Offset: 1

Views

Author

Paul D. Hanna, Jan 22 2006

Keywords

Crossrefs

Cf. A003714 (Fibbinary numbers), A048715, A048718, A115422, A115423.

Programs

Formula

This sequence also seems to satisfy:
3*a(n) XOR 41*a(n) = 42*a(n);
5*a(n) XOR 35*a(n) = 38*a(n);
6*a(n) XOR 35*a(n) = 37*a(n);
7*a(n) XOR 35*a(n) = 36*a(n); etc.

A115422 Integers n > 0 such that n XOR 20*n = 21*n.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 32, 33, 36, 48, 64, 65, 66, 67, 72, 73, 96, 97, 128, 129, 130, 131, 132, 134, 144, 146, 192, 193, 194, 195, 256, 257, 258, 259, 260, 262, 264, 265, 268, 288, 289, 292, 384, 385, 386, 387, 388, 390, 512, 513, 514, 515, 516, 518
Offset: 1

Views

Author

Paul D. Hanna, Jan 22 2006

Keywords

Comments

n is in the sequence iff 2*n is in the sequence. - Robert Israel, Nov 11 2016

Crossrefs

Cf. A003714 (Fibbinary numbers), A048715, A048718, A115423, A115424.

Programs

Formula

This sequence also seems to satisfy:
5*a(n) XOR 16*a(n) = 21*a(n);
5*a(n) XOR 17*a(n) = 20*a(n); etc.
a(A224809(n+4)) = 2^n. - Gheorghe Coserea, Nov 11 2016

A115423 Integers n > 0 such that n XOR 30*n = 31*n.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 33, 64, 65, 66, 128, 129, 130, 132, 256, 257, 258, 260, 264, 512, 513, 514, 516, 520, 528, 1024, 1025, 1026, 1028, 1032, 1040, 1056, 1057, 2048, 2049, 2050, 2052, 2056, 2064, 2080, 2081, 2112, 2113, 2114, 4096, 4097, 4098, 4100, 4104, 4112
Offset: 1

Views

Author

Paul D. Hanna, Jan 22 2006

Keywords

Crossrefs

Cf. A003714 (Fibbinary numbers), A048715, A048718, A115422, A115424.

Programs

Formula

This sequence also seems to satisfy:
3*a(n) XOR 21*a(n) = 22*a(n);
5*a(n) XOR 19*a(n) = 22*a(n);
6*a(n) XOR 19*a(n) = 21*a(n); etc.
a(A003520(n+4)) = 2^n. - Gheorghe Coserea, Nov 11 2016
Showing 1-7 of 7 results.