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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A308197 Numbers m such that the tribonacci representation of m (A278038) ends in an even number of 0's.

Original entry on oeis.org

1, 3, 4, 5, 8, 10, 11, 12, 13, 14, 16, 17, 18, 21, 23, 25, 27, 28, 29, 32, 34, 35, 36, 37, 38, 40, 41, 42, 44, 45, 47, 48, 49, 52, 54, 55, 56, 57, 58, 60, 61, 62, 65, 67, 69, 71, 72, 73, 76, 78, 79, 80, 82, 84, 85, 86, 89, 91, 92, 93, 94, 95, 97, 98, 99, 102, 104, 106, 108, 109, 110, 113, 115, 116, 117, 118, 119, 121
Offset: 1

Views

Author

N. J. A. Sloane, Jun 22 2019

Keywords

Comments

The asymptotic density of this sequence is c/(c+1) = 0.647798..., where c = 1.839286... (A058265) is the tribonacci constant. - Amiram Eldar, Mar 04 2022

Crossrefs

Programs

  • Mathematica
    t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; q[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; EvenQ[Min[s] - 1]]; Select[Range[0, 121], q] (* Amiram Eldar, Mar 04 2022 *)

A308198 Numbers m such that the tribonacci representation of m (A278038) ends in an odd number of 0's.

Original entry on oeis.org

0, 2, 6, 7, 9, 15, 19, 20, 22, 24, 26, 30, 31, 33, 39, 43, 46, 50, 51, 53, 59, 63, 64, 66, 68, 70, 74, 75, 77, 81, 83, 87, 88, 90, 96, 100, 101, 103, 105, 107, 111, 112, 114, 120, 124, 127, 131, 132, 134, 140, 144, 145, 147, 151, 155, 156, 158, 164, 168, 169, 171, 173, 175, 179, 180, 182, 188, 192, 195, 199, 200, 202
Offset: 1

Views

Author

N. J. A. Sloane, Jun 22 2019

Keywords

Comments

The asymptotic density of this sequence is 1/(c+1) = 0.352201..., where c = 1.839286... (A058265) is the tribonacci constant. - Amiram Eldar, Mar 04 2022

Crossrefs

Programs

  • Mathematica
    t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; q[0] = True; q[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; OddQ[Min[s] - 1]]; Select[Range[0, 202], q] (* Amiram Eldar, Mar 04 2022 *)

A319430 First differences of the tribonacci representation numbers (A003726 or A278038).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 19, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5
Offset: 0

Views

Author

N. J. A. Sloane, Sep 30 2018

Keywords

Comments

This sequence appears to consist of runs of 1's of lengths given (essentially) by A275925, separated by single numbers > 1, which define the terms of A319431.
It would be nice to have a recurrence of some kind that produces A319431.

Crossrefs

Programs

  • Mathematica
    Differences@ Select[Range[0, 160], SequenceCount[IntegerDigits[#, 2], {1, 1, 1}] == 0 &] (* Michael De Vlieger, Dec 23 2019 *)

Formula

Conjecture: All terms are of the form ceiling(2^k/7) for some k (cf. A046630), and all numbers of the form ceiling(2^k/7) occur.
Conjecture (continued): Furthermore, new values of ceiling(2^k/7) (that is, new records) appear at n = 0, 6,12, 23, 43, 80, 148, 273, ..., which (apart from the start) are the tribonacci numbers minus 1, A000073 - 1, or A089068.
a(n) = ceiling(2^i/7) iff the Tribonacci representation of n+1 ends in i 0's. - Jeffrey Shallit, Oct 02 2018

A319431 The first differences (A319430) of the tribonacci representation numbers (A003726 or A278038) consists of runs of 1's separated by the terms of the present sequence.

Original entry on oeis.org

2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 37, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2, 3, 74, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 37, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 147, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2
Offset: 1

Views

Author

N. J. A. Sloane, Sep 30 2018

Keywords

Comments

The runs of 1's in A319430 have lengths that apparently are given by A275925 (with a slight change at the start). The present sequence shows the terms greater than 1.
Let b = "2,3,2", c = "2,3,2,5,2,3", d = "2,3,2,5,2". The sequence appears to consist of a word over the alphabet {b,c,d} interspersed with the sequence 10, 19, 10, 37, 10, 19, 74, 10, 19, 10, 37, 10, 147, 10, 19, 10, 37, 10, ...:
c, 10, d, 19, c, 10, b, 37, c, 10, d, 19, c, 74, c, 10, d, 19, c, 10, b, 37, c, 10, d, 147, c, 10, d, 19, c, 10, b, 37, c, 10, d, 19, c, 74, ...
The interspersed sequence 10, 19, 10, 37, 10, 19, 74, ... appears to have the same kind of structure.
It would be nice to have a recurrence of some kind that produces this sequence.

Examples

			0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 16, 17, 18, 19, 20, 21, 22, 24, 25, ... = A003726 (trib. repres. numbers)
1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 1, ... = A319430 (differences)
2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 5, 2, 19, 2, 3, 2, 5, 2, 3, 10, 2, 3, 2, 37, 2, ... (omit 1's, present sequence)
6, 1, 5, 1, 6, 1, 3, 1, 6, 1, 5, 1, 6, 1, 6, 1, 5, 1, 6, 1, 3, 1, 6, 1, 5, 1, ... = run lengths in differences
6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, ... = A275925 truncated (BISECTION of run lengths)
		

Crossrefs

Programs

  • Mathematica
    DeleteCases[Differences@ Select[Range[0, 1200], SequenceCount[IntegerDigits[#, 2], {1, 1, 1}] == 0 &] , 1] (* Michael De Vlieger, Dec 23 2019 *)

A308200 The tribonacci representation of a(n) is obtained by appending 0,0,0 to the tribonacci representation of n (cf. A278038).

Original entry on oeis.org

0, 7, 13, 20, 24, 31, 37, 44, 51, 57, 64, 68, 75, 81, 88, 94, 101, 105, 112, 118, 125, 132, 138, 145, 149, 156, 162, 169, 173, 180, 186, 193, 200, 206, 213, 217, 224, 230, 237, 243, 250, 254, 261, 267, 274, 281, 287, 294, 298, 305, 311, 318, 325, 331, 338, 342, 349, 355, 362, 368, 375, 379, 386
Offset: 0

Views

Author

N. J. A. Sloane, Jun 23 2019

Keywords

Crossrefs

Essentially partial sums of A276792.

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

A001590 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) with a(0)=0, a(1)=1, a(2)=0.

Original entry on oeis.org

0, 1, 0, 1, 2, 3, 6, 11, 20, 37, 68, 125, 230, 423, 778, 1431, 2632, 4841, 8904, 16377, 30122, 55403, 101902, 187427, 344732, 634061, 1166220, 2145013, 3945294, 7256527, 13346834, 24548655, 45152016, 83047505, 152748176, 280947697, 516743378, 950439251
Offset: 0

Views

Author

Keywords

Comments

The name "tribonacci number" is less well-defined than "Fibonacci number". The sequence A000073 (which begins 0, 0, 1) is probably the most important version, but the name has also been applied to A000213, A001590, and A081172. - N. J. A. Sloane, Jul 25 2024
Dimensions of the homogeneous components of the higher order peak algebra associated to cubic roots of unity (Hilbert series = 1 + 1*t + 2*t^2 + 3*t^3 + 6*t^4 + 11*t^5 ...). - Jean-Yves Thibon (jyt(AT)univ-mlv.fr), Oct 22 2006
Starting with offset 3: (1, 2, 3, 6, 11, 10, 37, ...) = row sums of triangle A145579. - Gary W. Adamson, Oct 13 2008
Starting (1, 2, 3, 6, 11, ...) = INVERT transform of the periodic sequence (1, 1, 0, 1, 1, 0, 1, 1, 0, ...). - Gary W. Adamson, May 04 2009
The comment of May 04 2009 is equivalent to: The numbers of ordered compositions of n using integers that are not multiples of 3 is equal to a(n+2) for n>=1, see [Hoggatt-Bicknell (1975) eq (2.7)]. - Gary W. Adamson, May 13 2013
Primes in the sequence are 2, 3, 11, 37, 634061, 7256527, ... in A231574. - R. J. Mathar, Aug 09 2012
Pisano period lengths: 1, 2, 13, 8, 31, 26, 48, 16, 39, 62,110,104,168, 48,403, 32, 96, 78, 360, 248, ... . - R. J. Mathar, Aug 10 2012
a(n+1) is the top left entry of the n-th power of any of 3 X 3 matrices [0, 1, 0; 1, 1, 1; 1, 0, 0], [0, 1, 1; 1, 1, 0; 0, 1, 0], [0, 1, 1; 0, 0, 1; 1, 0, 1] or [0, 0, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
a(n+3) equals the number of n-length binary words avoiding runs of zeros of lengths 3i+2, (i=0,1,2,...). - Milan Janjic, Feb 26 2015
Sums of pairs of successive terms of A000073. - N. J. A. Sloane, Oct 30 2016
The power Q^n, for n >= 0, of the tribonacci Q-matrix Q = matrix([1, 1, 1], [1, 0, 0], [0, 1, 0]) is, by the Cayley-Hamilton theorem, Q^n = matrix([a(n+2), a(n+1) + a(n), a(n+1)], [a(n+1), a(n) + a(n-1), a(n)], [a(n), a(n-1) + a(n-2), a(n-1)]), with a(-2) = -1 and a(-1) = 1. One can use a(n) = a(n-1) + a(n-2) + a(n-3) in order to obtain a(-1) and a(-2). - Wolfdieter Lang, Aug 13 2018
a(n+2) is the number of entries n, for n>=1, in the sequence {A278038(k)}A278038(0)%20=%201).%20-%20_Wolfdieter%20Lang">{k>=1} (without A278038(0) = 1). - _Wolfdieter Lang, Sep 11 2018
In terms of the tribonacci numbers T(n) = A000073(n) the nonnegative powers of the Q-matrix (from the Aug 13 2018 comment) are Q^n = T(n)*Q^2 + (T(n-1) + T(n-2))*Q + T(n-1)*1_3, for n >= 0, with T(-1) = 1, T(-2) = -1. This is equivalent to the powers t^n of the tribonacci constant t = A058255 (or also for powers of the complex solutions). - Wolfdieter Lang, Oct 24 2018

Examples

			a(12)=a(11)+a(10)+a(9): 230=125+68+37.
For n=5 the partitions of 5 are 1+1+1+1+1 (1 composition), 1+1+1+2 (4 compositions), 1+2+2 (3 compositions), 1+1+3 (not contrib because 3 is a part), 2+3 (no contrib because 3 is a part), 1+4 (2 compositions) and 5 (1 composition), total 1+4+3+2+1=11 =a(5+2) - _R. J. Mathar_, Jan 13 2023
		

References

  • Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
  • 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

Programs

  • GAP
    a:=[0,1,0];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Oct 24 2018
  • Magma
    I:=[0,1,0]; [n le 3 select I[n]  else Self(n-1)+Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Apr 19 2018
    
  • Maple
    seq(coeff(series(x*(1-x)/(1-x-x^2-x^3),x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Oct 24 2018
    # alternative
    A001590 := proc(n)
        option remember;
        if n <=2 then
            op(n+1,[0,1,0]) ;
        else
            procname(n-1)+procname(n-2)+procname(n-3) ;
        end if;
    end proc:
    seq(A001590(n),n=0..30) ;# R. J. Mathar, Nov 22 2024
  • Mathematica
    LinearRecurrence[{1,1,1}, {0,1,0}, 50] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)
    RecurrenceTable[{a[0]==0, a[1]==1, a[2]==0, a[n]==a[n-1]+a[n-2]+a[n-3]}, a, {n, 40}] (* Vincenzo Librandi, Apr 19 2018 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; 1,1,1]^n*[0;1;0])[1,1] \\ Charles R Greathouse IV, Jul 28 2015
    
  • Sage
    def A001590():
        W = [0, 1, 0]
        while True:
            yield W[0]
            W.append(sum(W))
            W.pop(0)
    a = A001590(); [next(a) for  in range(38)]  # _Peter Luschny, Sep 12 2016
    

Formula

G.f.: x*(1-x)/(1-x-x^2-x^3).
Limit a(n)/a(n-1) = t where t is the real solution of t^3 = 1 + t + t^2, t = A058265 = 1.839286755... . If T(n) = A000073(n) then t^n = T(n-1) + a(n)*t + T(n)*t^2, for n >= 0, with T(-1) = 1.
a(3*n) = Sum_{k+l+m=n} (n!/k!l!m!)*a(l+2*m). Example: a(12)=a(8)+4a(7)+10a(6)+16a(5)+19a(4)+16a(3)+10a(2)+4a(1)+a(0) The coefficients are the trinomial coefficients. T(n) and T(n-1) also satisfy this equation. (T(-1)=1)
From Reinhard Zumkeller, May 22 2006: (Start)
a(n) = A000073(n+1)-A000073(n);
a(n) = A000073(n-1)+A000073(n-2) for n>1;
A000213(n-2) = a(n+1)-a(n) for n>1. (End)
a(n) + a(n+1) = A000213(n). - Philippe Deléham, Sep 25 2006
If p[1]=0, p[i]=2, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n+1)=det A. - Milan Janjic, May 02 2010
For n>=4, a(n)=2*a(n-1)-a(n-4). - Bob Selcoe, Feb 18 2014
a(-1-n) = -A078046(n). - Michael Somos, Jun 01 2014
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(3*r+1). - Fabian Pereyra, Nov 22 2024

Extensions

Additional comments from Miklos Kristof, Jul 03 2002

A080843 Tribonacci word: limit S(infinity), where S(0) = 0, S(1) = 0,1, S(2) = 0,1,0,2 and for n >= 0, S(n+3) = S(n+2) S(n+1) S(n).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Mar 29 2003

Keywords

Comments

An Arnoux-Rauzy or episturmian word.
From N. J. A. Sloane, Jul 10 2018: (Start)
The initial terms in a form suitable for copying:
0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,0,0,1,0,2,0,1,
0,1,0,2,0,1,0,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,
0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,
0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,
2,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,0,0,1,0,2,0,
1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,
1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,
1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,
...
Let TTW(a,b,c) denote this sequence written over the alphabet {a,b,c}. It begins:
a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,b,a,a,b,a,c,a,b,
a,b,a,c,a,b,a,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,b,
a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,b,
a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,
c,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,b,a,a,b,a,c,a,
b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,
b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,
b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,
... (End)
From Wolfdieter Lang, Aug 14 2018: (Start)
The substitution sequence 0 -> 0, 1; 1-> 0, 2; 2 -> 0 read as an irregular triangle with rows l >= 1 and length T(l+2), with the tribonacci numbers T = A000073, leads to the tribonacci tree TriT with level TriT(l) for l >= 1 given by a(0), a(1), ..., a(T(l+2)-1).
E.g., l = 4: 0 1 0 2 0 1 0 with T(6) = 7 leaves (nodes). See the example below.
This tree can be used to find the tribonacci representation of nonnegative n given in A278038, call it ZTri(n) (Z for generalized Zeckendorf), by replacing every 2 by 1, and reading from bottom to top, omitting the final 0, except for n = 0 which is represented by 0. See the example below. (End)

Examples

			From _Joerg Arndt_, Mar 12 2013: (Start)
The first few steps of the substitution are
Start: 0
Rules:
  0 --> 01
  1 --> 02
  2 --> 0
-------------
0:   (#=1)
  0
1:   (#=2)
  01
2:   (#=4)
  0102
3:   (#=7)
  0102010
4:   (#=13)
  0102010010201
5:   (#=24)
  010201001020101020100102
6:   (#=44)
  01020100102010102010010201020100102010102010
7:   (#=81)
  010201001020101020100102010201001020101020100102010010201010201001020102010010201
(End)
From _Wolfdieter Lang_, Aug 14 2018: (Start)
The levels l of the tree TriT begin (the branches (edges) have been omitted):
Substitution rule: 0 -> 0 1; 1 -> 0 2; 2 -> 0.
l=1:                                 0
l=2:                  0                                 1
l=3:             0             1                  0             2
l=4:         0      1       0     2          0       1          0
l=5:      0    1  0   2   0   1   0        0   1   0   2      0    1
...
----------------------------------------------------------------------------------
n =       0    1  2   3   4   5   6        7   8   9  10     11   12
The tribonacci representation of n >= 0 (A278038; here at level 5 for n = 0.. 12) is obtained by reading from bottom to top (along the branches not shown) replacing 2 with 1, omitting the last 0 except for n = 0.
          0    1  0   1   0   1   0        0   1   0  1      0    1
                  1   1   0   0   1        0   0   1  1      0    0
                          1   1   1        0   0   0  0      1    1
                                           1   1   1  1      1    1
E.g., ZTri(9) = A278038(9) = 1010. (End)
		

References

  • The entry A092782 has a more complete list of references and links. - N. J. A. Sloane, Aug 17 2018
  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 246.

Crossrefs

Cf. A003849 (the Fibonacci word), A092782.
See A092782 for a version over the alphabet {1,2,3}.
See A278045 for another construction.
First differences: A317950. Partial sums: A319198.

Programs

  • Maple
    M:=17; S[1]:=`0`; S[2]:=`01`; S[3]:=`0102`;
    for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
    t0:=S[M]: l:=length(t0); for i from 1 to l do lprint(i,substring(t0,i..i)); od:
    # N. J. A. Sloane, Nov 01 2006
    # A version that uses the letters a,b,c:
    M:=10; B[1]:=`a`; B[2]:=`ab`; B[3]:=`abac`;
    for n from 4 to M do B[n]:=cat(B[n-1], B[n-2], B[n-3]); od:
    B[10]; # N. J. A. Sloane, Oct 30 2018
  • Mathematica
    Nest[Flatten[ # /. {0 -> {0, 1}, 1 -> {0, 2}, 2 -> {0}}] &, {0}, 8] (* updated by Robert G. Wilson v, Nov 07 2010 *)
    SubstitutionSystem[{0->{0,1},1->{0,2},2->{0}},{0},{8}]//Flatten (* Harvey P. Dale, Nov 21 2021 *)
  • PARI
    strsub(s, vv, off=0)=
    {
        my( nl=#vv, r=[], ct=1 );
        while ( ct <= #s,
            r = concat(r, vv[ s[ct] + (1-off) ] );
            ct += 1;
        );
        return( r );
    }
    t=[0];  for (k=1, 10, t=strsub( t, [[0,1], [0,2], [0]], 0 ) );  t
    \\ Joerg Arndt, Sep 14 2013

Formula

Fixed point of morphism 0 -> 0, 1; 1 -> 0, 2; 2 -> 0.
a(n) = A092782(n+1) - 1. - Joerg Arndt, Sep 14 2013

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 06 2003

A003144 Positions of letter a in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).

Original entry on oeis.org

1, 3, 5, 7, 8, 10, 12, 14, 16, 18, 20, 21, 23, 25, 27, 29, 31, 32, 34, 36, 38, 40, 42, 44, 45, 47, 49, 51, 52, 54, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 75, 76, 78, 80, 82, 84, 86, 88, 89, 91, 93, 95, 97, 99, 101, 102, 104, 106, 108, 110, 112, 113, 115, 117, 119, 121, 123, 125
Offset: 1

Views

Author

Keywords

Comments

From Philippe Deléham, Feb 27 2009: (Start)
A003144, A003145, A003146 may be defined as follows. Consider the morphism psi: a -> ab, b -> ac, c -> a. The image (or trajectory) of a under repeated application of this map is the infinite ternary tribonacci word a, b, a, c, a, b, a, a, b, a, c, a, b, a, b, a, c, ... (setting a = 1, b = 2, c = 3 gives A092782). The indices of a, b, c give respectively A003144, A003145, A003146. (End) [For the word with a -> 0, b -> 1, c -> 2 with offset 0 see A080843. - Wolfdieter Lang, Aug 10 2018]
The infinite word may also be defined as the limit S_oo where S_1 = a, S_n = psi(S_{n-1}). Or, by S_1 = a, S_2 = ab, S_3 = abac, and thereafter S_n = S_{n-1} S_{n-2} S_{n-3}. It is the unique word such that S_oo = psi(S_oo).
Also, indices of a in the sequence closed under a -> abac, b -> aba, c -> ab; starting with a(1) = a. - Philippe Deléham, Apr 16 2004
Theorem: A number m is in this sequence iff the tribonacci representation of m-1 ends with 0. [Duchene and Rigo, Remark 2.5] - N. J. A. Sloane, Nov 18 2016; corrected Mar 02 2019.

References

  • Eric Duchêne, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, Urban Larsson, Wythoff Visions, Games of No Chance, Vol. 5; MSRI Publications, Vol. 70 (2017), pages 101-153.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A003145, A003146, A080843, A092782, A058265, A275926, A276793, A276796, A278039 (subtract 1 from each term, and use offset 0).
First differences are A276788.
For tribonacci representations of numbers see A278038.

Programs

  • Maple
    M:=17; S[1]:=`a`; S[2]:=`ab`; S[3]:=`abac`;
    for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
    t0:=S[M]: l:=length(t0); t1:=[];
    for i from 1 to l do if substring(t0,i..i) = `a` then t1:=[op(t1),i]; fi; od: t1; # N. J. A. Sloane, Nov 01 2006
  • Mathematica
    A003144L = StringPosition[SubstitutionSystem[{"a" -> "ab", "b" -> "ac", "c" -> "a"}, "a", {#}][[1]], "a"][[All, 1]] &; A003144L[7] (* JungHwan Min, Dec 22 2016 *)

Formula

It appears that a(n) is always either floor(n*t) or floor(n*t)+1 for all n, where t is the tribonacci constant A058265. See A275926. - N. J. A. Sloane, Oct 28 2016. This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019

Extensions

More terms from Philippe Deléham, Apr 16 2004
Entry revised by N. J. A. Sloane, Oct 13 2016

A003145 Positions of letter b in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).

Original entry on oeis.org

2, 6, 9, 13, 15, 19, 22, 26, 30, 33, 37, 39, 43, 46, 50, 53, 57, 59, 63, 66, 70, 74, 77, 81, 83, 87, 90, 94, 96, 100, 103, 107, 111, 114, 118, 120, 124, 127, 131, 134, 138, 140, 144, 147, 151, 155, 158, 162, 164, 168, 171, 175, 179, 182, 186, 188, 192, 195, 199, 202, 206, 208
Offset: 1

Views

Author

Keywords

Comments

A003144, A003145, A003146 may be defined as follows. Consider the map psi: a -> ab, b -> ac, c -> a. The image (or trajectory) of a under repeated application of this map is the infinite word a, b, a, c, a, b, a, a, b, a, c, a, b, a, b, a, c, ... (setting a = 1, b = 2, c = 3 gives A092782). The indices of a, b, c give respectively A003144, A003145, A003146. - Philippe Deléham, Feb 27 2009
The infinite word may also be defined as the limit S_oo where S_1 = a, S_n = psi(S_{n-1}). Or, by S_1 = a, S_2 = ab, S_3 = abac, and thereafter S_n = S_{n-1} S_{n-2} S_{n-3}. It is the unique word such that S_oo = psi(S_oo).
Also indices of b in the sequence closed under a -> abac, b -> aba, c -> ab; starting with a(1) = a. - Philippe Deléham, Apr 16 2004
Theorem: A number m is in this sequence iff the tribonacci representation of m-1 ends with 01. [Duchene and Rigo, Remark 2.5] - N. J. A. Sloane, Mar 02 2019

References

  • Eric Duchêne, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, Urban Larsson, Wythoff Visions, Games of No Chance, Vol. 5; MSRI Publications, Vol. 70 (2017), pages 101-153.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

First differences give A276789. A278040 (subtract 1 from each term, and use offset 1).
For tribonacci representations of numbers see A278038.

Programs

  • Maple
    M:=17; S[1]:=`a`; S[2]:=`ab`; S[3]:=`abac`;
    for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
    t0:=S[M]: l:=length(t0); t1:=[];
    for i from 1 to l do if substring(t0,i..i) = `b` then t1:=[op(t1),i]; fi; od: # N. J. A. Sloane
  • Mathematica
    StringPosition[SubstitutionSystem[{"a" -> "ab", "b" -> "ac", "c" -> "a"}, "b", {#}][[1]], "b"][[All, 1]] &@9 (* Michael De Vlieger, Mar 30 2017, Version 10.2, after JungHwan Min at A003144 *)

Formula

It appears that a(n) = floor(n*t^2) + eps for all n, where t is the tribonacci constant A058265 and eps is 0, 1, or 2. See A276799. - N. J. A. Sloane, Oct 28 2016. This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019

Extensions

More terms from Philippe Deléham, Apr 16 2004
Corrected by T. D. Noe and N. J. A. Sloane, Nov 01 2006
Entry revised by N. J. A. Sloane, Oct 13 2016
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