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

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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

A060142 Ordered set S defined by these rules: 0 is in S and if x is in S then 2x+1 and 4x are in S.

Original entry on oeis.org

0, 1, 3, 4, 7, 9, 12, 15, 16, 19, 25, 28, 31, 33, 36, 39, 48, 51, 57, 60, 63, 64, 67, 73, 76, 79, 97, 100, 103, 112, 115, 121, 124, 127, 129, 132, 135, 144, 147, 153, 156, 159, 192, 195, 201, 204, 207, 225, 228, 231, 240, 243, 249, 252, 255, 256, 259, 265, 268, 271
Offset: 0

Views

Author

Clark Kimberling, Mar 05 2001

Keywords

Comments

After expelling 0 and 1, the numbers 4x occupy same positions in S that 1 occupies in the infinite Fibonacci word (A003849).
a(A026351(n)) = A219608(n); a(A004957(n)) = 4 * a(n). - Reinhard Zumkeller, Nov 26 2012
Apart from the initial term, this lists the indices of the 1's in A086747. - N. J. A. Sloane, Dec 05 2019
From Gus Wiseman, Jun 10 2020: (Start)
Numbers k such that the k-th composition in standard order has all odd parts, or numbers k such that A124758(k) is odd. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. For example, the sequence of all compositions into odd parts begins:
0: () 57: (1,1,3,1) 135: (5,1,1,1)
1: (1) 60: (1,1,1,3) 144: (3,5)
3: (1,1) 63: (1,1,1,1,1,1) 147: (3,3,1,1)
4: (3) 64: (7) 153: (3,1,3,1)
7: (1,1,1) 67: (5,1,1) 156: (3,1,1,3)
9: (3,1) 73: (3,3,1) 159: (3,1,1,1,1,1)
12: (1,3) 76: (3,1,3) 192: (1,7)
15: (1,1,1,1) 79: (3,1,1,1,1) 195: (1,5,1,1)
16: (5) 97: (1,5,1) 201: (1,3,3,1)
19: (3,1,1) 100: (1,3,3) 204: (1,3,1,3)
25: (1,3,1) 103: (1,3,1,1,1) 207: (1,3,1,1,1,1)
28: (1,1,3) 112: (1,1,5) 225: (1,1,5,1)
31: (1,1,1,1,1) 115: (1,1,3,1,1) 228: (1,1,3,3)
33: (5,1) 121: (1,1,1,3,1) 231: (1,1,3,1,1,1)
36: (3,3) 124: (1,1,1,1,3) 240: (1,1,1,5)
39: (3,1,1,1) 127: (1,1,1,1,1,1,1) 243: (1,1,1,3,1,1)
48: (1,5) 129: (7,1) 249: (1,1,1,1,3,1)
51: (1,3,1,1) 132: (5,3) 252: (1,1,1,1,1,3)
(End)
Numbers whose binary representation has the property that every run of consecutive 0's has even length. - Harry Richman, Jan 31 2024

Examples

			From _Harry Richman_, Jan 31 2024: (Start)
In the following, dots are used for zeros in the binary representation:
   n  binary(a(n))  a(n)
   0:    .......     0
   1:    ......1     1
   2:    .....11     3
   3:    ....1..     4
   4:    ....111     7
   5:    ...1..1     9
   6:    ...11..    12
   7:    ...1111    15
   8:    ..1....    16
   9:    ..1..11    19
  10:    ..11..1    25
  11:    ..111..    28
  12:    ..11111    31
  13:    .1....1    33
  14:    .1..1..    36
  15:    .1..111    39
  16:    .11....    48
  17:    .11..11    51
  18:    .111..1    57
  19:    .1111..    60
  20:    .111111    63
  21:    1......    64
  22:    1....11    67
(End)
		

Crossrefs

Cf. A003714 (no consecutive 1's in binary expansion).
Odd partitions are counted by A000009.
Numbers with an odd number of 1's in binary expansion are A000069.
Numbers whose binary expansion has odd length are A053738.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without odd parts are A062880.
- Sum is A070939.
- Product is A124758.
- Strict compositions are A233564.
- Heinz number is A333219.
- Number of distinct parts is A334028.

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a060142 n = a060142_list !! n
    a060142_list = 0 : f (singleton 1) where
       f s = x : f (insert (4 * x) $ insert (2 * x + 1) s') where
           (x, s') = deleteFindMin s
    -- Reinhard Zumkeller, Nov 26 2012
    
  • Mathematica
    Take[Nest[Union[Flatten[# /. {{i_Integer -> i}, {i_Integer -> 2 i + 1}, {i_Integer -> 4 i}}]] &, {1}, 5], 32]  (* Or *)
    Select[Range[124], FreeQ[Length /@ Select[Split[IntegerDigits[#, 2]], First[#] == 0 &], ?OddQ] &] (* _Birkas Gyorgy, May 29 2012 *)
  • PARI
    is(n)=if(n<3, n<2, if(n%2,is(n\2),n%4==0 && is(n/4))) \\ Charles R Greathouse IV, Oct 21 2013

Extensions

Corrected by T. D. Noe, Nov 01 2006
Definition simplified by Charles R Greathouse IV, Oct 21 2013

A060141 Ordered set S defined by these rules: 0 and 1 are in S and if x is a nonzero number in S, then 3x and 9x+2 are in S.

Original entry on oeis.org

0, 1, 3, 9, 11, 27, 29, 33, 81, 83, 87, 99, 101, 243, 245, 249, 261, 263, 297, 299, 303, 729, 731, 735, 747, 749, 783, 785, 789, 891, 893, 897, 909, 911, 2187, 2189, 2193, 2205, 2207, 2241, 2243, 2247, 2349, 2351, 2355, 2367, 2369, 2673, 2675, 2679, 2691
Offset: 0

Views

Author

Clark Kimberling, Mar 05 2001

Keywords

Comments

The numbers of the form 9x+1 occupy the same positions in S that 1 occupies in the infinite Fibonacci word (A003849).

Crossrefs

A207869 a(n) = Z(n,-1), where Z(n,x) is the n-th Zeckendorf polynomial.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Feb 21 2012

Keywords

Comments

The Zeckendorf polynomials Z(x,n) are defined and ordered at A207813.

Examples

			The first ten Zeckendorf polynomials are 1, x, x^2, x^2 + 1, x^3, x^3 + 1, x + x^3, x^4, 1 + x^4, x + x^4; their values at x=-1 are 1, -1, 1, 2, -1, 0, -2, 1, 2, 0, indicating initial terms for A207869 and A207870.
		

Crossrefs

Programs

  • Mathematica
    fb[n_] := 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--]; fr]; t = Table[fb[n],
          {n, 1, 500}];
    b[n_] := Reverse[Table[x^k, {k, 0, n}]]
    p[n_, x_] := t[[n]].b[-1 + Length[t[[n]]]]
    Table[p[n, x], {n, 1, 40}]
    Table[p[n, x] /. x -> 1, {n, 1, 120}]  (* A007895 *)
    Table[p[n, x] /. x -> 2, {n, 1, 120}]  (* A003714 *)
    Table[p[n, x] /. x -> 3, {n, 1, 120}]  (* A060140 *)
       t1 = Table[p[n, x] /. x -> -1,
       {n, 1, 420}]                        (* A207869 *)
    Flatten[Position[t1, 0]]               (* A207870 *)
    t2 = Table[p[n, x] /. x -> I, {n, 1, 420}];
    Flatten[Position[t2, 0]]               (* A207871 *)
    Denominator[Table[p[n, x] /. x -> 1/2, {n, 1, 120}]]  (* A207872 *)
    Numerator[Table[p[n, x] /. x -> 1/2, {n, 1, 120}]]    (* A207873 *)

A356823 Tribternary numbers.

Original entry on oeis.org

0, 1, 3, 4, 9, 10, 12, 27, 28, 30, 31, 36, 37, 81, 82, 84, 85, 90, 91, 93, 108, 109, 111, 112, 243, 244, 246, 247, 252, 253, 255, 270, 271, 273, 274, 279, 280, 324, 325, 327, 328, 333, 334, 336, 729, 730, 732, 733, 738, 739, 741, 756, 757, 759, 760, 765, 766, 810, 811, 813, 814, 819
Offset: 1

Views

Author

Tanya Khovanova and PRIMES STEP Senior group, Aug 29 2022

Keywords

Comments

These are numbers whose ternary representations consist only of zeros and ones and do not have three consecutive ones.
The sequence of Tribternary numbers can be constructed by writing out the Tribonacci representations of nonnegative integers and then evaluating the result in ternary.
These are Tribbinary numbers written in base 2 and evaluated in base 3.
These numbers are similar to Fibbinary numbers A003714, Fibternary numbers A003726, and Tribbinary numbers A060140.
Tribbinary numbers A060140 are a subsequence.
Subsequence of A005836.
The number of Tribternary numbers less than any power of three is a Tribonacci number.
We can generate this sequence recursively: start with 0 and 1; then, if x is in the sequence add 3x, 9x+1, and 27x+4 to the sequence.
The n-th Tribternary number is divisible by 3 if the n-th term of the Tribonacci word is a. Respectively, the n-th Tribbinary number is of the form 9x+1 if the n-th term of the Tribonacci word is b, and the n-th Tribbinary number is of the form 27x+4 if the n-th term of the Tribonacci word is c.
Every nonnegative integer can be written as a sum of three Tribternary numbers.
Every number has a Tribternary multiple.

Crossrefs

Programs

  • Mathematica
    Select[Range[0, 1000], SequenceCount[IntegerDigits[#, 3], {1, 1, 1}] == 0 && SequenceCount[IntegerDigits[#, 3], {2}] == 0 &]
  • Python
    import heapq
    from itertools import islice
    def agen(): # generator of terms, using recursion in Comments
        x, h = None, [0]
        while True:
            x, oldx = heapq.heappop(h), x
            if x != oldx:
                yield x
                for t in [3*x, 9*x+1, 27*x+4]: heapq.heappush(h, t)
    print(list(islice(agen(), 62))) # Michael S. Branicky, Aug 30 2022

A366347 a(n) has as many prime factors as the ternary expansion of n has runs of nonzero digits; if the k-th run corresponds to A032924(e) and appears after m-1 0's then the p-adic valuation of a(n) is e (where p corresponds to the m-th prime number).

Original entry on oeis.org

1, 2, 4, 3, 8, 16, 9, 32, 64, 5, 6, 12, 27, 128, 256, 81, 512, 1024, 25, 18, 36, 243, 2048, 4096, 729, 8192, 16384, 7, 10, 20, 15, 24, 48, 45, 96, 192, 125, 54, 108, 2187, 32768, 65536, 6561, 131072, 262144, 625, 162, 324, 19683, 524288, 1048576, 59049
Offset: 0

Views

Author

Rémy Sigrist, Oct 07 2023

Keywords

Comments

This sequence is a variant of the Doudna sequence (A005940); here we consider runs of nonzero digits in ternary expansions, there in binary expansions.
This sequence is a bijection from the nonnegative integers to the positive integers with inverse A366348.
We can devise a similar sequence for any fixed base b >= 2:
- the case b = 2 corresponds (up to the offset) to the Doudna sequence (A005940),
- the case b = 3 corresponds to the present sequence,
- the case b = 10 corresponds to A290389.

Examples

			For n = 46: the ternary expansion of 46 is "1201; we have two runs of nonzero digits: "12" (= 5 = A032924(4)) after 2-1 0's and "1" (= 1 = A032924(1)) after 1-1 0's; so a(46) = prime(2)^4 * prime(1)^1 = 3^4 * 2^1 = 162.
		

Crossrefs

Programs

  • PARI
    See Links section.

Formula

a(3*n) = A003961(a(n)).
a(3^k) = prime(1 + k) for any k >= 0.
a(2 * 3^k) = prime(1 + k)^2 for any k >= 0.
a(n) is squarefree iff n belongs to A060140.

A207870 Numbers k matched to Zeckendorf polynomials divisible by x+1.

Original entry on oeis.org

6, 10, 14, 16, 23, 26, 35, 37, 42, 51, 57, 60, 68, 74, 83, 90, 92, 97, 106, 110, 116, 120, 127, 132, 134, 146, 149, 157, 163, 172, 178, 184, 188, 192, 194, 206, 214, 217, 234, 236, 241, 250, 254, 260, 264, 271, 276, 278, 288, 294, 298, 302, 304, 311
Offset: 1

Views

Author

Clark Kimberling, Feb 21 2012

Keywords

Comments

The Zeckendorf polynomials Z(x,k) are defined and ordered at A207813.

Examples

			The first ten Zeckendorf polynomials are 1, x, x^2, x^2 + 1, x^3, x^3 + 1, x + x^3, x^4, 1 + x^4, x + x^4; their values at x=-1 are 1, -1, 1, 2, -1, 0, -2, 1, 2, 0, indicating initial terms for A207869 and this sequence.
		

Crossrefs

Programs

  • Mathematica
    fb[n_] := 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--]; fr]; t = Table[fb[n],
          {n, 1, 500}];
    b[n_] := Reverse[Table[x^k, {k, 0, n}]]
    p[n_, x_] := t[[n]].b[-1 + Length[t[[n]]]]
    Table[p[n, x], {n, 1, 40}]
    Table[p[n, x] /. x -> 1, {n, 1, 120}]  (* A007895 *)
    Table[p[n, x] /. x -> 2, {n, 1, 120}]  (* A003714 *)
    Table[p[n, x] /. x -> 3, {n, 1, 120}]  (* A060140 *)
       t1 = Table[p[n, x] /. x -> -1,
       {n, 1, 420}]                        (* A207869 *)
    Flatten[Position[t1, 0]]               (* this sequence *)
    t2 = Table[p[n, x] /. x -> I, {n, 1, 420}];
    Flatten[Position[t2, 0]]               (* A207871 *)
    Denominator[Table[p[n, x] /. x -> 1/2, {n, 1, 120}]]  (* A207872 *)
    Numerator[Table[p[n, x] /. x -> 1/2, {n, 1, 120}]]    (* A207873 *)
Showing 1-7 of 7 results.