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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A274284 Numbers that are a product of distinct numbers in A006355.

Original entry on oeis.org

2, 4, 6, 8, 10, 12, 16, 20, 24, 26, 32, 40, 42, 48, 52, 60, 64, 68, 80, 84, 96, 104, 110, 120, 128, 136, 156, 160, 168, 178, 192, 208, 220, 240, 252, 260, 272, 288, 312, 320, 336, 356, 384, 408, 416, 420, 440, 466, 480, 504, 520, 544, 576, 624, 640, 660, 672
Offset: 1

Views

Author

Clark Kimberling, Jun 17 2016

Keywords

Comments

See the Comment on distinct-product sequences in A160009.

Examples

			8 = 2*4, 480 = 2*4*6*10.
		

Crossrefs

Cf. A160009.

Programs

  • Mathematica
    f[1] = 2; f[2] = 4; z = 32; f[n_] := f[n - 1] + f[n - 2]; f = Table[f[n], {n, 1, z}]; f
    s = {2}; Do[s = Union[s, Select[s*f[[i]], # <= f[[z]] &]], {i, z}]; s

A035513 Wythoff array read by falling antidiagonals.

Original entry on oeis.org

1, 2, 4, 3, 7, 6, 5, 11, 10, 9, 8, 18, 16, 15, 12, 13, 29, 26, 24, 20, 14, 21, 47, 42, 39, 32, 23, 17, 34, 76, 68, 63, 52, 37, 28, 19, 55, 123, 110, 102, 84, 60, 45, 31, 22, 89, 199, 178, 165, 136, 97, 73, 50, 36, 25, 144, 322, 288, 267, 220, 157, 118, 81, 58, 41, 27, 233, 521
Offset: 1

Views

Author

Keywords

Comments

T(0,0)=1, T(0,1)=2,...; y^2-x^2-xy
Inverse of sequence A064274 considered as a permutation of the nonnegative integers. - Howard A. Landman, Sep 25 2001
The Wythoff array W consists of all the Wythoff pairs (x(n),y(n)), where x=A000201 and y=A001950, so that W contains every positive integer exactly once. The differences T(i,2j+1)-T(i,2j) form the Wythoff difference array, A080164, which also contains every positive integer exactly once. - Clark Kimberling, Feb 08 2003
For n>2 the determinant of any n X n contiguous subarray of A035513 (as a square array) is 0. - Gerald McGarvey, Sep 18 2004
From Clark Kimberling, Nov 14 2007: (Start)
Except for initial terms in some cases:
(Row 1) = A000045
(Row 2) = A000032
(Row 3) = A006355
(Row 4) = A022086
(Row 5) = A022087
(Row 6) = A000285
(Row 7) = A022095
(Row 8) = A013655 (sum of Fibonacci and Lucas numbers)
(Row 9) = A022112
(Column 1) = A003622 = AA Wythoff sequence
(Column 2) = A035336 = BA Wythoff sequence
(Column 3) = A035337 = ABA Wythoff sequence
(Column 4) = A035338 = BBA Wythoff sequence
(Column 5) = A035339 = ABBA Wythoff sequence
(Column 6) = A035340 = BBBA Wythoff sequence
Main diagonal = A020941. (End)
The Wythoff array is the dispersion of the sequence given by floor(n*x+x-1), where x=(golden ratio). See A191426 for a discussion of dispersions. - Clark Kimberling, Jun 03 2011
If u and v are finite sets of numbers in a row of the Wythoff array such that (product of all the numbers in u) = (product of all the numbers in v), then u = v. See A160009 (row 1 products), A274286 (row 2), A274287 (row 3), A274288 (row 4). - Clark Kimberling, Jun 17 2016
All columns of the Wythoff array are compound Wythoff sequences. This follows from the main theorem in the 1972 paper by Carlitz, Scoville and Hoggatt. For an explicit expression see Theorem 10 in Kimberling's paper from 2008 in JIS. - Michel Dekking, Aug 31 2017
The Wythoff array can be viewed as an infinite graph over the set of nonnegative integers, built as follows: start with an empty graph; for all n = 0, 1, ..., create an edge between n and the sum of the degrees of all i < n. Finally, remove vertex 0. In the resulting graph, the connected components are chains and correspond to the rows of the Wythoff array. - Luc Rousseau, Sep 28 2017
Suppose that h < k are consecutive terms in a row of the Wythoff array. If k is in an even numbered column, then h = floor(k/tau); otherwise, h = -1 + floor(k/tau). - Clark Kimberling, Mar 05 2020
From Clark Kimberling, May 26 2020: (Start)
For k > = 0, column k shows the numbers m having F(k+1) as least term in the Zeckendorf representation of m. For n >= 1, let r(n,k) be the number of terms in column k that are <= n. Then n/r(n,k) = n/(F(k+1)*tau + F(k)*(n-1)), by Bottomley's formula, so that the limiting ratio is 1/(F(k+1)*tau + F(k)). Summing over all k gives Sum_{k>=0} 1/(F(k+1)*tau + F(k)) = 1. Thus, in the limiting sense:
38.19...% of the numbers m have least term 1;
23.60...% have least term 2;
14.58...% have least term 3;
9.01...% have least term 5, etc. (End)
Named after the Dutch mathematician Willem Abraham Wythoff (1865-1939). - Amiram Eldar, Jun 11 2021
From Clark Kimberling, Jun 04 2025: (Start)
Let u(n) = (T(n,1),T(n,2)) mod 2. The positive integers (A000027) are partitioned into 4 sets (sequences):
{n : u(n) = (0,0)} = (3, 5, 9, 15, 19, 25, 29,...) = 1 + 2*A190429
{n: u(n) = (0,1)} = (2, 6, 12, 16, 18, 22, 28,...) = A191331
{n : u(n) = (1,0)} = (1, 7, 11, 13, 17, 21, 23,...) = A086843
{n: u(n) = (1,1)} = (4, 8, 10, 14, 20, 24, 26,...) = A191330.
Let v(n) = (T(n,1),T(n,2)) mod 3. The positive integers are partitioned into 9 sets (sequences):
{n : v(n) = (0,0)} = (4, 13, 19, 28, 43, 52,...) = 1 + 3*A190434
{n: v(n) = (0,1)} = (3, 12, 27, 36, 42, 51,...) = 3*A140399
{n : v(n) = (0,2)} = (5, 11, 20, 35, 44, 50,...) = 2 + 3*A190439
{n: v(n) = (1,0)} = (9, 18, 24, 33, 48, 57,...) = 3*A140400
{n: v(n) = (1,1)} = (2, 8, 17, 26, 32, 41,...) = A384601
{n : v(n) = (1,2)} = (1, 10, 16, 25, 34, 40,...) = A384602
{n: v(n) = (2,0)} = (14, 23, 29, 38, 47, 53,...) = 2 + 3*A190438
{n: v(n) = (2,1)} = (7, 22, 31, 37, 46, 61,...) = 1 + 3*A190433
{n : v(n) = (2,2)} = (6, 15, 21, 30, 39, 45,...) = 3*A140398.
Conjecture: If m >= 2, then {(T(n,1), T(n,2)) mod m} has cardinality m^2. (End)

Examples

			The Wythoff array begins:
   1    2    3    5    8   13   21   34   55   89  144 ...
   4    7   11   18   29   47   76  123  199  322  521 ...
   6   10   16   26   42   68  110  178  288  466  754 ...
   9   15   24   39   63  102  165  267  432  699 1131 ...
  12   20   32   52   84  136  220  356  576  932 1508 ...
  14   23   37   60   97  157  254  411  665 1076 1741 ...
  17   28   45   73  118  191  309  500  809 1309 2118 ...
  19   31   50   81  131  212  343  555  898 1453 2351 ...
  22   36   58   94  152  246  398  644 1042 1686 2728 ...
  25   41   66  107  173  280  453  733 1186 1919 3105 ...
  27   44   71  115  186  301  487  788 1275 2063 3338 ...
  ...
The extended Wythoff array has two extra columns, giving the row number n and A000201(n), separated from the main array by a vertical bar:
0     1  |   1    2    3    5    8   13   21   34   55   89  144   ...
1     3  |   4    7   11   18   29   47   76  123  199  322  521   ...
2     4  |   6   10   16   26   42   68  110  178  288  466  754   ...
3     6  |   9   15   24   39   63  102  165  267  432  699 1131   ...
4     8  |  12   20   32   52   84  136  220  356  576  932 1508   ...
5     9  |  14   23   37   60   97  157  254  411  665 1076 1741   ...
6    11  |  17   28   45   73  118  191  309  500  809 1309 2118   ...
7    12  |  19   31   50   81  131  212  343  555  898 1453 2351   ...
8    14  |  22   36   58   94  152  246  398  644 1042 1686 2728   ...
9    16  |  25   41   66  107  173  280  453  733 1186 1919 3105   ...
10   17  |  27   44   71  115  186  301  487  788 1275 2063 3338   ...
11   19  |  30   49   79   ...
12   21  |  33   54   87   ...
13   22  |  35   57   92   ...
14   24  |  38   62   ...
15   25  |  40   65   ...
16   27  |  43   70   ...
17   29  |  46   75   ...
18   30  |  48   78   ...
19   32  |  51   83   ...
20   33  |  53   86   ...
21   35  |  56   91   ...
22   37  |  59   96   ...
23   38  |  61   99   ...
24   40  |  64   ...
25   42  |  67   ...
26   43  |  69   ...
27   45  |  72   ...
28   46  |  74   ...
29   48  |  77   ...
30   50  |  80   ...
31   51  |  82   ...
32   53  |  85   ...
33   55  |  88   ...
34   56  |  90   ...
35   58  |  93   ...
36   59  |  95   ...
37   61  |  98   ...
38   63  |     ...
   ...
Each row of the extended Wythoff array also satisfies the Fibonacci recurrence, and may be extended to the left using this recurrence backwards.
From _Peter Munn_, Jun 11 2021: (Start)
The Wythoff array appears to have the following relationship to the traditional Fibonacci rabbit breeding story, modified for simplicity to be a story of asexual reproduction.
Give each rabbit a number, 0 for the initial rabbit.
When a new round of rabbits is born, allocate consecutive numbers according to 2 rules (the opposite of many cultural rules for inheritance precedence): (1) newly born child of Rabbit 0 gets the next available number; (2) the descendants of a younger child of any given rabbit precede the descendants of an older child of the same rabbit.
Row n of the Wythoff array lists the children of Rabbit n (so Rabbit 0's children have the Fibonacci numbers: 1, 2, 3, 5, ...). The generation tree below shows rabbits 0 to 20. It is modified so that each round of births appears on a row.
                                                                 0
                                                                 :
                                       ,-------------------------:
                                       :                         :
                       ,---------------:                         1
                       :               :                         :
              ,--------:               2               ,---------:
              :        :               :               :         :
        ,-----:        3         ,-----:         ,-----:         4
        :     :        :         :     :         :     :         :
     ,--:     5     ,--:     ,---:     6     ,---:     7     ,---:
     :  :     :     :  :     :   :     :     :   :     :     :   :
  ,--:  8  ,--:  ,--:  9  ,--:  10  ,--:  ,--:  11  ,--:  ,--:  12
  :  :  :  :  :  :  :  :  :  :   :  :  :  :  :   :  :  :  :  :   :
  : 13  :  : 14  : 15  :  : 16   :  : 17  : 18   :  : 19  : 20   :
The extended array's nontrivial extra column (A000201) gives the number that would have been allocated to the first child of Rabbit n, if Rabbit n (and only Rabbit n) had started breeding one round early.
(End)
		

References

  • John H. Conway, Posting to Math Fun Mailing List, Nov 25 1996.
  • Clark Kimberling, "Stolarsky interspersions," Ars Combinatoria 39 (1995) 129-138.

Crossrefs

See comments above for more cross-references.
Cf. A003622, A064274 (inverse), A083412 (transpose), A000201, A001950, A080164, A003603, A265650, A019586 (row that contains n).
For two versions of the extended Wythoff array, see A287869, A287870.

Programs

  • Maple
    W:= proc(n,k) Digits:= 100; (Matrix([n, floor((1+sqrt(5))/2* (n+1))]). Matrix([[0,1], [1,1]])^(k+1))[1,2] end: seq(seq(W(n, d-n), n=0..d), d=0..10); # Alois P. Heinz, Aug 18 2008
    A035513 := proc(r, c)
        option remember;
        if c = 1 then
            A003622(r) ;
        else
            A022342(1+procname(r, c-1)) ;
        end if;
    end proc:
    seq(seq(A035513(r,d-r),r=1..d-1),d=2..15) ; # R. J. Mathar, Jan 25 2015
  • Mathematica
    W[n_, k_] := Fibonacci[k + 1] Floor[n*GoldenRatio] + (n - 1) Fibonacci[k]; Table[ W[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten
  • PARI
    T(n,k)=(n+sqrtint(5*n^2))\2*fibonacci(k+1) + (n-1)*fibonacci(k)
    for(k=0,9,for(n=1,k, print1(T(n,k+1-n)", "))) \\ Charles R Greathouse IV, Mar 09 2016
    
  • Python
    from sympy import fibonacci as F, sqrt
    import math
    tau = (sqrt(5) + 1)/2
    def T(n, k): return F(k + 1)*int(math.floor(n*tau)) + F(k)*(n - 1)
    for n in range(1, 11): print([T(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Apr 23 2017
    
  • Python
    from math import isqrt, comb
    from gmpy2 import fib2
    def A035513(n):
        a = (m:=isqrt(k:=n<<1))+(k>m*(m+1))
        x = n-comb(a,2)
        b, c = fib2(a-x+2)
        return b*(x+isqrt(5*x*x)>>1)+c*(x-1) # Chai Wah Wu, Jun 26 2025

Formula

T(n, k) = Fib(k+1)*floor[n*tau]+Fib(k)*(n-1) where tau = (sqrt(5)+1)/2 = A001622 and Fib(n) = A000045(n). - Henry Bottomley, Dec 10 2001
T(n,-1) = n-1. T(n,0) = floor(n*tau). T(n,k) = T(n,k-1) + T(n,k-2) for k>=1. - R. J. Mathar, Sep 03 2016

Extensions

Comments about the extended Wythoff array added by N. J. A. Sloane, Mar 07 2016

A001595 a(n) = a(n-1) + a(n-2) + 1, with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 3, 5, 9, 15, 25, 41, 67, 109, 177, 287, 465, 753, 1219, 1973, 3193, 5167, 8361, 13529, 21891, 35421, 57313, 92735, 150049, 242785, 392835, 635621, 1028457, 1664079, 2692537, 4356617, 7049155, 11405773, 18454929, 29860703, 48315633, 78176337
Offset: 0

Keywords

Comments

2-ranks of difference sets constructed from Segre hyperovals.
Sometimes called Leonardo numbers. - George Pollard, Jan 02 2008
a(n) is the number of nodes in the Fibonacci tree of order n. A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node (see the Knuth reference, p. 417). - Emeric Deutsch, Jun 14 2010
Also odd numbers whose index is a Fibonacci number: odd(Fib(k)). - Carmine Suriano, Oct 21 2010
This is the sequence A(1,1;1,1;1) of the family of sequences [a,b:c,d:k] considered by Gary Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 17 2010
In general, adding a constant to each successive term of a Horadam sequence with signature (c,d) will result in a third-order recurrence with signature (c+1, d-c,-d). - Gary Detlefs, Feb 01 2023

Examples

			a(7) = odd(F(7)) = odd(8) = 15. - _Carmine Suriano_, Oct 21 2010
		

References

  • E. W. Dijkstra, 'Fibonacci numbers and Leonardo numbers', circulated privately, July 1981.
  • E. W. Dijkstra, 'Smoothsort, an alternative for sorting in situ', Science of Computer Programming, 1(3): 223-233, 1982.
  • D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.
  • 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).
  • J. Ziegenbalg, Algorithmen, Spektrum Akademischer Verlag, 1996, p. 172.

Programs

  • GAP
    List([0..40], n-> 2*Fibonacci(n+1) -1); # G. C. Greubel, Jul 10 2019
    
  • Haskell
    a001595 n = a001595_list !! n
    a001595_list =
       1 : 1 : (map (+ 1) $ zipWith (+) a001595_list $ tail a001595_list)
    -- Reinhard Zumkeller, Aug 14 2011
    
  • Magma
    [2*Fibonacci(n+1)-1: n in [0..40]]; // G. C. Greubel, Jul 10 2019
    
  • Maple
    L := 1,3: for i from 3 to 40 do l := nops([ L ]): L := L,op(l,[ L ])+op(l-1,[ L ])+1: od: [ L ];
    A001595:=(1-z+z**2)/(z-1)/(z**2+z-1); # Simon Plouffe in his 1992 dissertation
    with(combinat): seq(fibonacci(n-1)+fibonacci(n+2)-1, n=0..40); # Zerinvary Lajos, Jan 31 2008
  • Mathematica
    Join[{1, 3}, Table[a[1]=1; a[2]=3; a[i]=a[i-1]+a[i-2]+1, {i, 3, 40} ] ]
    Table[2*Fibonacci[n+1]-1, {n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Oct 13 2009; modified by G. C. Greubel, Jul 10 2019 *)
    RecurrenceTable[{a[0]==a[1]==1,a[n]==a[n-1]+a[n-2]+1},a,{n,40}] (* or *) LinearRecurrence[{2,0,-1},{1,1,3},40] (* Harvey P. Dale, Aug 07 2012 *)
  • PARI
    a(n) = 2*fibonacci(n+1)-1 \\ Franklin T. Adams-Watters, Sep 30 2009
    
  • Python
    from sympy import fibonacci
    def A001595(n): return (fibonacci(n+1)<<1)-1 # Chai Wah Wu, Sep 10 2024
  • Sage
    [2*fibonacci(n+1)-1 for n in (0..40)] # G. C. Greubel, Jul 10 2019
    

Formula

a(n) = 2*Fibonacci(n+1) - 1 = A006355(n+2) - 1. - Richard L. Ollerton, Mar 22 2002
G.f.: (1-x+x^2)/(1-2x+x^3) = 2/(1-x-x^2) - 1/(1-x). [Conjectured by Simon Plouffe in his 1992 dissertation; this is readily verified.]
a(n) = (2/sqrt(5))*((1+sqrt(5))/2)^(n+1) - 2/sqrt(5)*((1-sqrt(5))/2)^(n+1) - 1.
a(n+1)/a(n) is asymptotic to Phi = (1+sqrt(5))/2. - Jonathan Vos Post, May 26 2005
For n >= 2, a(n+1) = ceiling(Phi*a(n)). - Franklin T. Adams-Watters, Sep 30 2009
a(n) = Sum_{k=0..n+1} A109754(n-k+1,k) - Sum_{k=0..n} A109754(n-k,k) = Sum_{k=0..n+1} A101220(n-k+1,0,k) - Sum_{k=0..n} A101220(n-k,0,k). - Ross La Haye, May 31 2006
a(n) = A000071(n+3) - A000045(n). - Vladimir Joseph Stephan Orlovsky, Oct 13 2009
a(n) = Fibonacci(n-1) + Fibonacci(n+2) - 1. - Zerinvary Lajos, Jan 31 2008, corrected by R. J. Mathar, Dec 17 2010
a(n) = 2*a(n-1) - a(n-3); a(0)=1, a(1)=1, a(2)=3. - Harvey P. Dale, Aug 07 2012
E.g.f.: 2*exp(x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5 - exp(x). - Stefano Spezia, Jan 23 2024

Extensions

Additional comments from Christian Krattenthaler (kratt(AT)ap.univie.ac.at)
Further edits from Franklin T. Adams-Watters, Sep 30 2009, and N. J. A. Sloane, Oct 03 2009

A054886 Layer counting sequence for hyperbolic tessellation by cuspidal triangles of angles (Pi/3,Pi/3,0) (this is the classical modular tessellation).

Original entry on oeis.org

1, 3, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634, 78176338, 126491972
Offset: 0

Author

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

Keywords

Comments

The layer sequence is the sequence of the cardinalities of the layers accumulating around a ( finite-sided ) polygon of the tessellation under successive side-reflections; see the illustration accompanying A054888.
Equivalently, coordination sequence for (3,3,infinity) tiling of hyperbolic plane. - N. J. A. Sloane, Dec 29 2015
Equivalently, spherical growth series for modular group.
Also, number of sequences of length n with terms 1, 2, and 3, with no adjacent terms equal, and no three consecutive terms (1, 2, 3) or (3, 2, 1). - Pontus von Brömssen, Jan 03 2022

References

  • P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 156.

Crossrefs

Essentially the same as A006355.

Programs

  • Mathematica
    Join[{1,3},2Fibonacci[Range[4,40]]] (* Harvey P. Dale, Jan 06 2012 *)
  • PARI
    my(x='x+O('x^50)); Vec((1+2*x+2*x^2+x^3)/(1-x-x^2)) \\ G. C. Greubel, Aug 06 2017

Formula

G.f.: (1+2*x+2*x^2+x^3)/(1-x-x^2) = (x^2+x+1)*(1+x)/(1-x-x^2).
a(n) = 2*F(n+2) for n >= 2, with F(n) the n-th Fibonacci number (cf. A000045).
E.g.f.: 2*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5 - 1 - x. - Stefano Spezia, Apr 18 2022

Extensions

Offset changed to 0 by N. J. A. Sloane, Jan 03 2022 at the suggestion of Pontus von Brömssen

A022086 Fibonacci sequence beginning 0, 3.

Original entry on oeis.org

0, 3, 3, 6, 9, 15, 24, 39, 63, 102, 165, 267, 432, 699, 1131, 1830, 2961, 4791, 7752, 12543, 20295, 32838, 53133, 85971, 139104, 225075, 364179, 589254, 953433, 1542687, 2496120, 4038807, 6534927, 10573734, 17108661, 27682395, 44791056, 72473451, 117264507
Offset: 0

Keywords

Comments

First differences of A111314. - Ross La Haye, May 31 2006
Pisano period lengths: 1, 3, 1, 6, 20, 3, 16, 12, 8, 60, 10, 6, 28, 48, 20, 24, 36, 24, 18, 60, ... . - R. J. Mathar, Aug 10 2012
For n>=6, a(n) is the number of edge covers of the union of two cycles C_r and C_s, r+s=n, with a single common vertex. - Feryal Alayont, Oct 17 2024

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 7,17.

Crossrefs

Essentially the same as A097135.
Sequences of the form Fibonacci(n+k) + Fibonacci(n-k) are listed in A280154.
Sequences of the form m*Fibonacci: A000045 (m=1), A006355 (m=2), this sequence (m=3), A022087 (m=4), A022088 (m=5), A022089 (m=6), A022090 (m=7), A022091 (m=8), A022092 (m=8), A022093 (m=10), A022345...A022366 (m=11...32).

Programs

  • Magma
    [3*Fibonacci(n): n in [0..40]]; // Vincenzo Librandi, Dec 31 2016
    
  • Maple
    BB := n->if n=0 then 3; > elif n=1 then 0; > else BB(n-2)+BB(n-1); > fi: > L:=[]: for k from 1 to 34 do L:=[op(L),BB(k)]: od: L; # Zerinvary Lajos, Mar 19 2007
    with (combinat):seq(sum((fibonacci(n,1)),m=1..3),n=0..32); # Zerinvary Lajos, Jun 19 2008
  • Mathematica
    LinearRecurrence[{1, 1}, {0, 3}, 40] (* Arkadiusz Wesolowski, Aug 17 2012 *)
    Table[Fibonacci[n + 4] + Fibonacci[n - 4] - 4 Fibonacci[n], {n, 0, 40}] (* Bruno Berselli, Dec 30 2016 *)
    Table[3 Fibonacci[n], {n, 0, 40}] (* Vincenzo Librandi, Dec 31 2016 *)
  • PARI
    a(n)=3*fibonacci(n) \\ Charles R Greathouse IV, Nov 06 2014
    
  • SageMath
    def A022086(n): return 3*fibonacci(n)
    print([A022086(n) for n in range(41)]) # G. C. Greubel, Apr 10 2025

Formula

a(n) = 3*Fibonacci(n).
a(n) = F(n-2) + F(n+2) for n>1, with F=A000045.
a(n) = round( ((6*phi-3)/5) * phi^n ) for n>2. - Thomas Baruchel, Sep 08 2004
a(n) = A119457(n+1,n-1) for n>1. - Reinhard Zumkeller, May 20 2006
G.f.: 3*x/(1-x-x^2). - Philippe Deléham, Nov 19 2008
a(n) = A187893(n) - 1. - Filip Zaludek, Oct 29 2016
E.g.f.: 6*sinh(sqrt(5)*x/2)*exp(x/2)/sqrt(5). - Ilya Gutkovskiy, Oct 29 2016
a(n) = F(n+4) + F(n-4) - 4*F(n), F = A000045. - Bruno Berselli, Dec 29 2016

A160009 Numbers that are the product of distinct Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 8, 10, 13, 15, 16, 21, 24, 26, 30, 34, 39, 40, 42, 48, 55, 63, 65, 68, 78, 80, 89, 102, 104, 105, 110, 120, 126, 130, 144, 165, 168, 170, 178, 195, 204, 208, 210, 233, 240, 267, 272, 273, 275, 288, 312, 315, 330, 336, 340, 377, 390, 432, 440, 442, 445
Offset: 1

Author

T. D. Noe, Apr 29 2009

Keywords

Comments

Starts the same as A049862, the product of two distinct Fibonacci numbers. This sequence has an infinite number of consecutive terms that are consecutive numbers (such as 15 and 16) because fib(k)*fib(k+3) and fib(k+1)*fib(k+2) differ by one for all k >= 0.
It follows from Carmichael's theorem that if u and v are finite sets of Fibonacci numbers such that (product of all the numbers in u) = (product of all the numbers in v), then u = v. The same holds for many other 2nd order linear recurrence sequences with constant coefficients. In the following guide to related "distinct product sequences", W = Wythoff array, A035513:
base sequence distinct-product sequence
A000045 (Fibonacci) A160009
A000032 (Lucas, without 2) A274280
A000032 (Lucas, with 2) A274281
A000285 (1,4,5,...) A274282
A022095 (1,5,6,...) A274283
A006355 (2,4,6,...) A274284
A013655 (2,5,7,...) A274285
A022086 (3,6,9,...) A274191
row 2 of W: (4,7,11,...) A274286
row 3 of W: (6,10,16,...) A274287
row 4 of W: (9,15,24,...) A274288
- Clark Kimberling, Jun 17 2016

Crossrefs

Programs

  • Mathematica
    s={1}; nn=30; f=Fibonacci[2+Range[nn]]; Do[s=Union[s,Select[s*f[[i]],#<=f[[nn]]&]], {i,nn}]; s=Prepend[s,0]

A128588 Expansion of g.f. x*(1+x+x^2)/(1-x-x^2).

Original entry on oeis.org

1, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634, 78176338
Offset: 1

Author

Gary W. Adamson, Mar 11 2007

Keywords

Comments

Previous name was: A007318 * A128587.
a(n)/a(n-1) tends to phi, 1.618... = A001622.
Regardless of initial two terms, any linearly recurring sequence with signature (1,1) will yield an a(n)/a(n+1) ratio tending to phi (the Golden Ratio). - Harvey P. Dale, Mar 29 2017
Apart from the initial term, double the Fibonacci numbers. O.g.f.: x*(1+x+x^2)/(1-x-x^2). a(n) gives the number of binary strings of length n-1 avoiding the substrings 000 and 111. a(n) also gives the number of binary strings of length n-1 avoiding the substrings 010 and 101. - Peter Bala, Jan 22 2008
Row lengths of triangle A232642. - Reinhard Zumkeller, May 14 2015
a(n) is the number of binary strings of length n-1 avoiding the substrings 000 and 111. - Allan C. Wechsler, Feb 13 2025

Crossrefs

Programs

  • GAP
    Concatenation([1], List([2..40], n-> 2*Fibonacci(n))); # G. C. Greubel, Jul 10 2019
  • Haskell
    a128588 n = a128588_list !! (n-1)
    a128588_list = 1 : cows where
                       cows = 2 : 4 : zipWith (+) cows (tail cows)
    -- Reinhard Zumkeller, May 14 2015
    
  • Magma
    [1] cat [2*Fibonacci(n): n in [2..40]]; // G. C. Greubel, Jul 10 2019
    
  • Maple
    a:= n-> `if`(n<2, n, 2*(<<0|1>, <1|1>>^n)[1,2]):
    seq(a(n), n=1..50);  # Alois P. Heinz, Apr 28 2018
  • Mathematica
    nn=40; a=(1-x^3)/(1-x); b=x*(1-x^2)/(1-x); CoefficientList[Series[a^2 /(1-b^2), {x,0,nn}], x]  (* Geoffrey Critzer, Sep 01 2012 *)
    LinearRecurrence[{1,1}, {1,2,4}, 40] (* Harvey P. Dale, Mar 29 2017 *)
    Join[{1}, 2*Fibonacci[Range[2,40]]] (* G. C. Greubel, Jul 10 2019 *)
  • PARI
    {a(n) = if( n<2, n==1, 2 * fibonacci(n))}; /* Michael Somos, Jul 18 2015 */
    
  • Sage
    [1]+[2*fibonacci(n) for n in (2..40)] # G. C. Greubel, Jul 10 2019
    

Formula

G.f.: x*(1+x+x^2)/(1-x-x^2).
Binomial transform of A128587; a(n+2) = a(n+1) + a(n), n>3.
a(n) = A068922(n-1), n>2. - R. J. Mathar, Jun 14 2008
For n > 1: a(n+1) = a(n) + if a(n) odd then max{a(n),a(n-1)} else min{a(n),a(n-1)}, see also A038754. - Reinhard Zumkeller, Oct 19 2015
E.g.f.: 4*exp(x/2)*sinh(sqrt(5)*x/2)/sqrt(5) - x. - Stefano Spezia, Feb 19 2023

Extensions

New name from Joerg Arndt, Feb 16 2024

A104161 G.f.: x*(1 - x + x^2)/((1-x)^2 * (1 - x - x^2)).

Original entry on oeis.org

0, 1, 2, 5, 10, 19, 34, 59, 100, 167, 276, 453, 740, 1205, 1958, 3177, 5150, 8343, 13510, 21871, 35400, 57291, 92712, 150025, 242760, 392809, 635594, 1028429, 1664050, 2692507, 4356586, 7049123
Offset: 0

Author

Creighton Dement, Mar 10 2005

Keywords

Comments

A floretion-generated sequence.
Floretion Algebra Multiplication Program, FAMP Code: 1vesrokseq[ (- .25'i - .25i' - .25'ii' + .25'jj' + .25'kk' + .25'jk' + .25'kj' - .25e)('i + i' + 'ji' + 'ki' + e) ] RokType: Y[sqa.Findk()] = Y[sqa.Findk()] + p.
Partial sums of Leonardo numbers A001595. - Jonathan Vos Post, Jan 01 2011

Crossrefs

Programs

  • GAP
    List([0..40], n-> 2*Fibonacci(n+2) -(n+2)); # G. C. Greubel, Jul 09 2019
  • Magma
    [2*Fibonacci(n+2) -(n+2): n in [0..40]]; // G. C. Greubel, Jul 09 2019
    
  • Mathematica
    a=0;b=1;Table[c=b+a+n; a=b; b=c, {n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2011 *)
    CoefficientList[Series[x*(1-x+x^2)/((1-x)^2*(1-x-x^2)),{x,0,40}],x] (* or *) LinearRecurrence[{3,-2,-1,1},{0,1,2,5},40] (* Harvey P. Dale, Sep 06 2012 *)
  • PARI
    my(x='x+O('x^40)); concat(0, Vec(x*(1-x+x^2)/((1-x)^2*(1-x-x^2)))) \\ G. C. Greubel, Sep 26 2017
    
  • SageMath
    [2*fibonacci(n+2) -(n+2) for n in (0..40)] # G. C. Greubel, Jul 09 2019
    

Formula

Superseeker results (incomplete): a(2) - 2a(n+1) + a(n) = A006355(n+1) (Number of binary vectors of length n containing no singletons); a(n+1) - a(n) = A001595(n) (2-ranks of difference sets constructed from Segre hyperovals); a(n) + n + 1 = A001595(n+1).
A107909(a(n)) = A000975(n). - Reinhard Zumkeller, May 28 2005
From Ross La Haye, Aug 03 2005: (Start)
a(n) = 2*(Fibonacci(n+2) - 1) - n.
a(n) = Sum_{k=0..n} A101220(n-k, 0, k). (End)
From Gary W. Adamson, Apr 02 2006: (Start)
a(n) = a(n-1) + a(n-2) + n-1.
a(n) = row sums of A117501, starting (1, 2, 5, 10, ...). (End)
a(n) = Sum_{k=0..n} A109754(n-k,k). - Ross La Haye, Apr 12 2006
a(n) = (Sum_{k=0..n} (n-k)*Fibonacci(k-1) + Fibonacci(k)) - n. - Ross La Haye, May 31 2006
From R. J. Mathar, Apr 18 2008: (Start)
a(n) = -2 - n + (-A094214)^n*(1-A010499/5) + (1+A010499/5)/A094214^n.
a(n) = A006355(n+3) - n - 2. (End)
a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4); a(0)=0, a(1)=1, a(2)=2, a(3)=5. - Harvey P. Dale, Sep 06 2012

A055389 a(0) = 1, then twice the Fibonacci sequence.

Original entry on oeis.org

1, 2, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634, 78176338
Offset: 0

Author

Robert G. Wilson v, Jul 05 2000

Keywords

Comments

a(n) is the number of sequences over the alphabet {0,1} such that all maximal blocks (of both 0's and 1's) have odd length. E.g., a(4) = 6 because we have 0001, 0101, 0111, 1000, 1010, 1110. - Geoffrey Critzer, Mar 06 2012

Crossrefs

Essentially the same as A006355.
Cf. A000045.

Programs

Formula

G.f.: (1 + x - x^2)/(1 - x - x^2).
E.g.f.: 1 + 4*exp(x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Stefano Spezia, Apr 18 2022

Extensions

More terms from James Sellers, Jul 07 2000

A072405 Triangle T(n, k) = C(n,k) - C(n-2,k-1) for n >= 3 and T(n, k) = 1 otherwise, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 4, 3, 1, 1, 4, 7, 7, 4, 1, 1, 5, 11, 14, 11, 5, 1, 1, 6, 16, 25, 25, 16, 6, 1, 1, 7, 22, 41, 50, 41, 22, 7, 1, 1, 8, 29, 63, 91, 91, 63, 29, 8, 1, 1, 9, 37, 92, 154, 182, 154, 92, 37, 9, 1, 1, 10, 46, 129, 246, 336, 336, 246, 129, 46, 10, 1, 1, 11, 56, 175, 375, 582, 672, 582, 375, 175, 56, 11, 1
Offset: 0

Author

Henry Bottomley, Jun 16 2002

Keywords

Comments

Starting 1,0,1,1,1,... this is the Riordan array ((1-x+x^2)/(1-x), x/(1-x)). Its diagonal sums are A006355. Its inverse is A106509. - Paul Barry, May 04 2005

Examples

			Rows start as:
  1;
  1, 1;
  1, 1,  1; (key row for starting the recurrence)
  1, 2,  2,  1;
  1, 3,  4,  3,  1;
  1, 4,  7,  7,  4, 1;
  1, 5, 11, 14, 11, 5, 1;
		

Crossrefs

Row sums give essentially A003945, A007283, or A042950.
Cf. A072406 for number of odd terms in each row.
Cf. A051597, A096646, A122218 (identical for n > 1).
Cf. A007318 (q=0), A072405 (q= -1), A173117 (q=1), A173118 (q=2), A173119 (q=3), A173120 (q=-4).

Programs

  • Magma
    T:= func< n,k | n lt 3 select 1 else Binomial(n,k) - Binomial(n-2,k-1) >;
    [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 28 2021
    
  • Mathematica
    t[2, 1] = 1; t[n_, n_] = t[, 0] = 1; t[n, k_] := t[n, k] = t[n-1, k-1] + t[n-1, k]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 28 2013, after Ralf Stephan *)
  • PARI
    A072405(n, k) = if(n>2, binomial(n, k)-binomial(n-2, k-1), 1) \\ M. F. Hasler, Jan 06 2024
  • Sage
    def T(n,k): return 1 if n<3 else binomial(n,k) - binomial(n-2,k-1)
    flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 28 2021
    

Formula

T(n, k) = C(n,k) - C(n-2,k-1) for n >= 3 and T(n, k) = 1 otherwise.
T(n, k) = T(n-1, k-1) + T(n-1, k) starting with T(2, 0) = T(2, 1) = T(2, 2) = 1 and T(n, 0) = T(n, n) = 1.
G.f.: (1-x^2*y) / (1 - x*(1+y)). - Ralf Stephan, Jan 31 2005
From G. C. Greubel, Apr 28 2021: (Start)
Sum_{k=0..n} T(n, k) = (n+1)*[n<3] + 3*2^(n-2)*[n>=3].
T(n, k, q) = q*[n=2] + Sum_{j=0..5} q^j*binomial(n-2*j, k-j)*[n>2*j] with T(n,0) = T(n,n) = 1 for q = -1. (End)
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