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 1-5 of 5 results.

A319432 The first differences (A129761) of the tribonacci representation numbers (A003714 or A014417) consists of runs of 1's separated by the terms of the present sequence.

Original entry on oeis.org

2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 43, 2, 3, 6, 2, 11, 2, 3, 86, 2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 171, 2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 43, 2, 3, 6, 2, 11, 2, 3, 342, 2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 43, 2, 3, 6, 2, 11, 2, 3, 86, 2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 683, 2
Offset: 1

Views

Author

N. J. A. Sloane, Sep 30 2018

Keywords

Comments

The runs of 1's in A129761 have lengths that apparently are given by the Fibonacci word A005614 (with a slight change at the start). The present sequence shows the terms greater than 1. The structure of this sequence is (partially) analyzed in the comments in A129761.

Crossrefs

Programs

  • PARI
    p=0; for (n=1, 2052, if (bitand(n,2*n)==0, if (n-p>1, print1 (n-p ", ")); p=n)) \\ Rémy Sigrist, Oct 07 2018

A319952 Let M = A022342(n) be the n-th number whose Zeckendorf representation is even; then a(n) = A129761(M).

Original entry on oeis.org

1, 2, 3, 1, 6, 1, 2, 11, 1, 2, 3, 1, 22, 1, 2, 3, 1, 6, 1, 2, 43, 1, 2, 3, 1, 6, 1, 2, 11, 1, 2, 3, 1, 86, 1, 2, 3, 1, 6, 1, 2, 11, 1, 2, 3, 1, 22, 1, 2, 3, 1, 6, 1, 2, 171, 1, 2, 3, 1, 6, 1, 2, 11, 1, 2, 3, 1, 22, 1, 2, 3, 1, 6, 1
Offset: 2

Views

Author

Jeffrey Shallit and N. J. A. Sloane, Oct 03 2018

Keywords

Comments

The Zeckendorf representations of numbers are given in A014417. The even ones are specified by A022342.
The offset here is 2 (because A129761 should really have had offset 1 not 0).

Crossrefs

Programs

  • Maple
    with(combinat): F:=fibonacci:
    A130234 := proc(n)
        local i;
        for i from 0 do
            if F(i) >= n then
                return i;
            end if;
        end do:
    end proc:
    A014417 := proc(n)
        local nshi, Z, i ;
        if n <= 1 then
            return n;
        end if;
        nshi := n ;
        Z := [] ;
        for i from A130234(n) to 2 by -1 do
            if nshi >= F(i) and nshi > 0 then
                Z := [1, op(Z)] ;
                nshi := nshi-F(i) ;
            else
                Z := [0, op(Z)] ;
            end if;
        end do:
        add( op(i, Z)*10^(i-1), i=1..nops(Z)) ;
    end proc:
    A072649:= proc(n) local j; global F; for j from ilog[(1+sqrt(5))/2](n)
           while F(j+1)<=n do od; (j-1); end proc:
    A003714 := proc(n) global F; option remember; if(n < 3) then RETURN(n); else RETURN((2^(A072649(n)-1))+A003714(n-F(1+A072649(n)))); fi; end proc:
    A129761 := n -> A003714(n+1)-A003714(n):
    a:=[];
    for n from 1 to 120 do
       if (A014417(n) mod 2) = 0 then a:=[op(a), A129761(n-1)]; fi;
    od;
    a;

Formula

If the Zeckendorf representation of M ends with exactly k zeros, ...10^k, then a(n) = ceiling(2^k/3).

A276419 Non-Fibonacci numbers n such that A129761(n) = 1.

Original entry on oeis.org

6, 9, 11, 14, 16, 18, 19, 22, 24, 26, 27, 29, 30, 32, 35, 37, 39, 40, 42, 43, 45, 47, 48, 50, 52, 53, 56, 58, 60, 61, 63, 64, 66, 68, 69, 71, 73, 74, 76, 77, 79, 81, 82, 84, 85, 87, 90, 92, 94, 95, 97, 98, 100, 102, 103, 105, 107, 108, 110, 111, 113, 115, 116
Offset: 1

Views

Author

Altug Alkan, Sep 03 2016

Keywords

Comments

It appears that this is the intersection of A058065 with A001690 (non-Fibonacci). - Michel Marcus, Sep 05 2016

Crossrefs

Extensions

More terms from Michel Marcus, Sep 04 2016

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

A005578 a(2*n) = 2*a(2*n-1), a(2*n+1) = 2*a(2*n)-1.

Original entry on oeis.org

1, 1, 2, 3, 6, 11, 22, 43, 86, 171, 342, 683, 1366, 2731, 5462, 10923, 21846, 43691, 87382, 174763, 349526, 699051, 1398102, 2796203, 5592406, 11184811, 22369622, 44739243, 89478486, 178956971, 357913942, 715827883, 1431655766, 2863311531, 5726623062, 11453246123
Offset: 0

Views

Author

Keywords

Comments

Might be called the "Arima sequence" after Yoriyuki Arima who in 1769 constructed this sequence as the number of moves of the outer ring in the optimal solution for the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017
Let u(k), v(k), w(k) be the 3 sequences defined by u(1)=1, v(1)=0, w(1)=0 and u(k+1) = u(k) + v(k), v(k+1) = u(k) + w(k), w(k+1) = v(k) + w(k); let M(k) = Max(u(k), v(k), w(k)); then a(n) = M(n). - Benoit Cloitre, Mar 25 2002
Unimodal analog of Fibonacci numbers: a(n+1) = Sum_{k=0..n/2} A071922(n-k, n-2*k). Based on the observation that F_{n+1} = Sum_{k} binomial (n-k, k). - Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 30 2002
Numbers n at which the length of the symmetric signed digit expansion of n with q=2 (i.e., the length of the representation of n in the (-1,0,1)2 number system) increases. - _Ralf Stephan, Jun 30 2003
Row sums of Riordan array (1/(1-x), x/(1-2*x^2)). - Paul Barry, Apr 24 2005
For n > 0, record-values of A107910: a(n) = A107910(A023548(n)). - Reinhard Zumkeller, May 28 2005
2^(n+1) = 2*a(n) + 2*A001045(n) + A000975(n-1); e.g., 2^6 = 64 = 2*a(5) + 2*A001045(5) + 2*A000975(4) = 2*11 + 2*11 + 2*10. Let a(n), A001045(n) and A000975(n-1) = the legs of a triangle (a, b, c). Then a(n-1), A001045(n-1) and A000975(n-2) = (S-c), (S-b), (S-a), where S = the triangle semiperimeter. Example: a(5), A001045(5) and A000975(4) = triangle (a, b, c) = (11, 11, 10). Then a(4), A001045(4), A000975(3) = (S-c), (S-b), (S-a) = (6, 5, 5). - Gary W. Adamson, Dec 24 2007
a(n) is the number of length-n binary representations of a nonnegative integer that is divisible by 3. The initial digits are allowed to be 0's. a(4) = 6 because we have 0000, 0011, 0110, 1001, 1100, 1111. - Geoffrey Critzer, Jan 13 2014
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [1, 0, 1; 0, 1, 1; 1, 1, 0] or of the 3 X 3 matrix [1, 1, 0; 1, 0, 1; 0, 1, 1]. - R. J. Mathar, Feb 04 2014
With 0 prefixed, this sequence is an autosequence of the first kind because the sequence of first differences A001045 is. Its companion is A052950. - Paul Curtz, Dec 18 2018, edited by M. F. Hasler, Dec 21 2018
Apparently, the sequence gives the distinct values taken by A129761, the first differences of fibbinary numbers. - Rémy Sigrist, Oct 26 2019
The sequence with offset 1 can be generated in three steps starting with A158780. First, put in alternate signs (1, -1, 1, -2, 2, -4, ...) and take the inverse; getting (1, 1, 0, 1, 1, 2, 3, 5, 8, 13, 21, ...). Take the invert transform of the latter, resulting in the sequence. It follows from the inverti transform being 1, 1, 0, 1, 1, 2, 3, ... that (for example), a(9) = 171 = (1, 1, 0, 1, 1, 2, 3, 5, 8) dot (86, 43, 0, 11, 6, 6, 6, 5, 8) = (86 + 43 + 0 + 11 + 6 + 6 + 6 + 5 + 8). A similar procedure is shown in the Aug 08 2019 comment of A006356. - Gary W. Adamson, Feb 04 2022

References

  • R. K. Guy, Graphs and the strong law of small numbers. Graph theory, combinatorics and applications. Vol. 2 (Kalamazoo, MI, 1988), 597-614, Wiley-Intersci. Publ., Wiley, New York, 1991.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Bisections: A007583 and A047849.
Cf. also A000975, A001045 (first differences), A129761.
Cf. A006356.

Programs

  • GAP
    List([0..40],n->(2^(n+1)+3+(-1)^n)/6); # Muniru A Asiru, Dec 22 2018
    
  • Magma
    [(2^(n+1)+3+(-1)^n)/6: n in [0..40]]; // Vincenzo Librandi, Aug 14 2011
    
  • Maple
    A005578:=-(-1+z+z^2)/((z-1)*(2*z-1)*(z+1)); # Simon Plouffe in his 1992 dissertation
    with(combstruct):ZL0:=S=Prod(Sequence(Prod(a, Sequence(b))), a):ZL1:=Prod(begin_blockP, Z, end_blockP):ZL2:=Prod(begin_blockLR, Z, Sequence(Prod(mu_length, Z), card>=1), end_blockLR): ZL3:=Prod(begin_blockRL, Sequence(Prod(mu_length, Z), card>=1), Z, end_blockRL):Q:=subs([a=Union(ZL3), b=ZL3], ZL0), begin_blockP=Epsilon, end_blockP=Epsilon, begin_blockLR=Epsilon, end_blockLR=Epsilon, begin_blockRL=Epsilon, end_blockRL=Epsilon, mu_length=Epsilon:temp15:=draw([S, {Q}, unlabelled], size=15):seq(count([S, {Q}, unlabelled], size=n), n=2..34); # Zerinvary Lajos, Mar 08 2008
  • Mathematica
    a=0; Table[a=2^n-a;(a/2+1)/2,{n,5!}] (* Vladimir Joseph Stephan Orlovsky, Nov 22 2009 *)
    LinearRecurrence[{2,1,-2}, {1,1,2}, 40] (* G. C. Greubel, Aug 26 2019 *)
  • PARI
    a(n)=(2^(n+1)+3+(-1)^n)/6 \\ Charles R Greathouse IV, Mar 22 2016
    
  • Python
    print([1+2**n//3 for n in range(40)])  # Gennady Eremin, Feb 05 2022
  • Sage
    [(2^(n+1)+3+(-1)^n)/6 for n in (0..40)] # G. C. Greubel, Aug 26 2019
    

Formula

a(n) = ceiling(2^n/3).
a(n) = 1 + floor((2^n)/3) (proof by mathematical induction).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3).
From Paul Barry, Jul 20 2003: (Start)
a(n) = A001045(n) + A000035(n+1), where A000035 = (0, 1, 0, 1, ...).
G.f.: (1 - x - x^2)/((1-x^2)*(1-2*x)). [Guy, 1988];
E.g.f.: (exp(2*x) - exp(-x))/3 + cosh(x) = (2*exp(2*x) + 3*exp(x) + exp(-x))/6. (End)
The 30 listed terms are given by a(0)=1, a(1)=1 and, for n > 1, by a(n) = a(n-1) + a(n-2) + Sum_{i=0..n-4} Fibonacci(i)*a(n-4-i). - John W. Layman, Jan 07 2000
a(n) = (2^(n+1) + 3 + (-1)^n)/6. - Vladeta Jovovic, Jul 02 2002
Binomial transform of A001045(n-1)(-1)^n + 0^n/2. - Paul Barry, Apr 28 2004
a(n) = (1 + A001045(n+1))/2. - Paul Barry, Apr 28 2004
a(n) = Sum_{k=0..n} (-1)^k*Sum_{j=0..n-k} (if((j-k) mod 2)=0, binomial(n-k, j), 0). - Paul Barry, Jan 25 2005
Let M = the 6 X 6 adjacency matrix of a benzene ring, (reference): [0,1,0,0,0,1; 1,0,1,0,0,0; 0,1,0,1,0,0; 0,0,1,0,1,0; 0,0,0,1,0,1; 1,0,0,0,1,0]. Then a(n) = leftmost nonzero term of M^n * [1,0,0,0,0,0]. E.g.: a(6) = 22 since M^6 * [1,0,0,0,0,0] = [22,0,21,0,21,0]. - Gary W. Adamson, Jun 14 2006
Starting (1, 2, 3, 6, 11, 22, ...), = row sums of triangle A135229. - Gary W. Adamson, Nov 23 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0] = [A005578(n), A001045(n), A000975(n-1)]. - Gary W. Adamson, Dec 24 2007
a(n) = 1 + 2^(n-1) - a(n-1) = a(n-1) + 2*a(n-2) - 1 = a(n-2) + 2^(n-2). - Paul Curtz, Jan 31 2009
a(n) = A023105(n+1) - 1. - Carl Joshua Quines, Jul 17 2019

Extensions

Edited by N. J. A. Sloane, Jun 20 2015
Showing 1-5 of 5 results.