A022112 -id:A022112 - cp's OEIS frontend

A035513 Wythoff array read by falling antidiagonals.

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

N. J. A. Sloane

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

(Row 10-19) = A022113, A022120, A022121, A022379, A022130, A022382, A022088, A022136, A022137, A022089

(Row 20-28) = A022388, A022096, A022090, A022389, A022097, A022091, A022390, A022098, A022092

(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)

			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)

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

Alois P. Heinz, Table of n, a(n) for n = 1..5151
Peter G. Anderson, More Properties of the Zeckendorf Array, Fib. Quart. 52-5 (2014), 15-21.
L. Carlitz, Richard Scoville, and V. E. Hoggatt, Jr., Fibonacci representations, Fib. Quart., Vol. 10, No. 1 (1972), pp. 1-28.
Code Golf Stack Exchange, New Order #5: where Fibonacci and Beatty meet at Wythoff.
J. H. Conway, Allan Wechsler, Marc LeBrun, Dan Hoey, and N. J. A. Sloane, On Kimberling Sums and Para-Fibonacci Sequences, Correspondence and Postings to Math-Fun Mailing List, Nov 1996 to Jan 1997.
John Conway and Alex Ryba, The extra Fibonacci series and the Empire State Building, Math. Intelligencer, Vol. 38, No. 1 (2016), pp. 41-48.
Larry Ericksen and Peter G. Anderson, Patterns in differences between rows in k-Zeckendorf arrays, The Fibonacci Quarterly, Vol. 50, No. 1 (February 2012), pp. 11-18. - _N. J. A. Sloane_, Jun 10 2012
Clark Kimberling, Interspersions.
Clark Kimberling, The Zeckendorf array equals the Wythoff array, Fibonacci Quarterly, Vol. 33, No. 1 (1995), pp. 3-8.
Clark Kimberling, Complementary equations and Wythoff Sequences, JIS, Vol. 11 (2008), Article 08.3.3.
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
Clark Kimberling and Kenneth B. Stolarsky, Slow Beatty sequences, devious convergence, and partitional divergence, Amer. Math. Monthly, 123, No. 2 (2016), pp. 267-273.
Stéphane Legendre, Labeled Fibonacci Trees, Fibonacci Quart. 53 (2015), no. 2, 152-167.
A. J. Macfarlane, On the fibbinary numbers and the Wythoff array, arXiv:2405.18128 [math.CO], 2024. See pages 1-2.
Casey Mongoven, Sonification of multiple Fibonacci-related sequences, Annales Mathematicae et Informaticae, Vol. 41 (2013), pp. 175-192.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane, Classic Sequences
Sam Vandervelde, On the divisibility of Fibonacci sequences by primes of index two, The Fibonacci Quarterly, Vol. 50, No. 3 (2012), pp. 207-216. See Figure 1.
Eric Weisstein's World of Mathematics, Wythoff Array.
Wikipedia, Wythoff array.
Pedro Zanetti, C++ code snippet, for generating this sequence.
Index entries for sequences that are permutations of the natural numbers

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.

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

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

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

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

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

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

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

A028313 Elements in the 5-Pascal triangle (by row).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 6, 6, 1, 1, 7, 12, 7, 1, 1, 8, 19, 19, 8, 1, 1, 9, 27, 38, 27, 9, 1, 1, 10, 36, 65, 65, 36, 10, 1, 1, 11, 46, 101, 130, 101, 46, 11, 1, 1, 12, 57, 147, 231, 231, 147, 57, 12, 1, 1, 13, 69, 204, 378, 462, 378, 204, 69, 13, 1, 1, 14, 82, 273, 582, 840, 840, 582, 273, 82, 14, 1

Offset: 0

Mohammad K. Azarian

			Triangle begins as:
  1;
  1,  1;
  1,  5,  1;
  1,  6,  6,   1;
  1,  7, 12,   7,   1;
  1,  8, 19,  19,   8,   1;
  1,  9, 27,  38,  27,   9,   1;
  1, 10, 36,  65,  65,  36,  10,  1;
  1, 11, 46, 101, 130, 101,  46, 11,  1;
  1, 12, 57, 147, 231, 231, 147, 57, 12,  1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000007, A005009, A010892, A022112, A028275, A028314, A028315.
Cf. A028316, A028317, A028318, A028319, A028320, A028321, A028322.
Cf. A028323, A028324, A028325, A029653, A051472, A051936, A051937.
Cf. A178915.

[n le 1 select 1 else Binomial(n,k) +3*Binomial(n-2,k-1): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 05 2024

Table[If[n<2, 1, Binomial[n,k] +3*Binomial[n-2,k-1]], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 05 2024 *)

def A028313(n,k): return 1 if n<2 else binomial(n,k) + 3*binomial(n-2,k-1)
flatten([[A028313(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 05 2024

From Ralf Stephan, Jan 31 2005: (Start)
T(n, k) = C(n, k) + 3*C(n-2, k-1), with T(0, k) = T(1, k) = 1.
G.f.: (1 + 3*x^2*y)/(1 - x*(1+y)). (End)
From G. C. Greubel, Jan 05 2024: (Start)
T(n, n-k) = T(n, k).
T(n, n-1) = n + 3*(1 - [n=1]) = A178915(n+3), n >= 1.
T(n, n-2) = A051936(n+2), n >= 2.
T(n, n-3) = A051937(n+1), n >= 3.
T(2*n, n) = A028322(n).
Sum_{k=0..n} T(n, k) = A005009(n-2) - (3/4)*[n=0] - (3/2)*[n=1].
Sum_{k=0..n} (-1)^k * T(n, k) = A000007(n) - 3*[n=2].
Sum_{k=0..floor(n/2)} T(n-k, k) = A022112(n-2) + 3*([n=0] - [n=1]).
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k, k) = 4*A010892(n) - 3*([n=0] + [n=1]). (End)

More terms from Sam Alexander (pink2001x(AT)hotmail.com)

A022319 a(n) = a(n-1) + a(n-2) + 1 for n > 1, a(0)=1, a(1)=5.

Original entry on oeis.org

1, 5, 7, 13, 21, 35, 57, 93, 151, 245, 397, 643, 1041, 1685, 2727, 4413, 7141, 11555, 18697, 30253, 48951, 79205, 128157, 207363, 335521, 542885, 878407, 1421293, 2299701, 3720995, 6020697, 9741693

Offset: 0

N. J. A. Sloane

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Diana Savin and Elif Tan, On Companion sequences associated with Leonardo quaternions: Applications over finite fields, arXiv:2403.01592 [math.CO], 2024. See p. 2.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A192762 (partial sums).

a022319 n = a022319_list !! (n-1)
a022319_list = 1 : 5 : zipWith (+)
   (map (+ 1) a022319_list) (tail a022319_list)
-- Reinhard Zumkeller, May 26 2013

with(combinat): seq(fibonacci(n-2)+fibonacci(n+4)-1, n=0..29); # Zerinvary Lajos, Feb 01 2008

LinearRecurrence[{2, 0, -1}, {1, 5, 7}, 40] (* Harvey P. Dale, Nov 19 2014 *)

x='x+O('x^50); Vec((1 +3*x -3*x^2)/((1-x)*(1 -x -x^2))) \\ G. C. Greubel, Aug 25 2017

a(n) = Fibonacci(n-2) + Fibonacci(n+4) - 1. - Zerinvary Lajos, Feb 01 2008
From R. J. Mathar, Apr 07 2011: (Start)
G.f.: (1 + 3*x - 3*x^2) / ((1-x)*(1 - x - x^2)).
a(n) = A022112(n) - 1. (End)

A028314 Elements in the 5-Pascal triangle A028313 that are not 1.

Original entry on oeis.org

5, 6, 6, 7, 12, 7, 8, 19, 19, 8, 9, 27, 38, 27, 9, 10, 36, 65, 65, 36, 10, 11, 46, 101, 130, 101, 46, 11, 12, 57, 147, 231, 231, 147, 57, 12, 13, 69, 204, 378, 462, 378, 204, 69, 13, 14, 82, 273, 582, 840, 840, 582, 273, 82, 14, 15, 96, 355, 855, 1422, 1680, 1422, 855, 355, 96, 15

Offset: 0

Mohammad K. Azarian

			Triangle begins as:
   5;
   6,  6;
   7, 12,   7;
   8, 19,  19,   8;
   9, 27,  38,  27,   9;
  10, 36,  65,  65,  36,  10;
  11, 46, 101, 130, 101,  46,  11;
  12, 57, 147, 231, 231, 147,  57,  12;
  13, 69, 204, 378, 462, 378, 204,  69,  13;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000027, A010892, A022112, A028313, A028322,

Cf. A051936, A051937, A121262, A176448.

A028314:= func< n,k | Binomial(n+2,k+1) + 3*Binomial(n,k) >;
[A028314(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 06 2024

A028314[n_, k_]:= Binomial[n+2,k+1] + 3*Binomial[n,k];
Table[A028314[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 06 2024 *)

def A028314(n,k): return binomial(n+2,k+1) + 3*binomial(n,k)
flatten([[A028314(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 06 2024

From G. C. Greubel, Jan 06 2024: (Start)
T(n, k) = binomial(n+2, k+1) + 3*binomial(n, k).
T(n, n-k) = T(n, k).
T(n, 0) = T(n, n) = A000027(n+5).
T(n, 1) = T(n, n-1) = A051936(n+4).
T(n, 2) = T(n, n-2) = A051937(n+3).T(2*n, n) = A028322(n+1).
Sum_{k=0..n} T(n, k) = A176448(n).
Sum_{k=0..n} (-1)^k * T(n, k) = 1 + (-1)^n + 3*[n=0].
Sum_{k=0..n} T(n-k, k) = A022112(n+1) - (3-(-1)^n)/2.
Sum_{k=0..n} (-1)^k * T(n-k, k) = 4*A010892(n) - 2*A121262(n+1) - (3 - (-1)^n)/2. (End)
G.f.: (5 - 4*x - 4*x*y + 3*x^2*y)/((1 - x)*(1 - x*y)*(1 - x - x*y)). - Stefano Spezia, Dec 06 2024

More terms from James Sellers

A258160 a(n) = 8*Lucas(n).

Original entry on oeis.org

16, 8, 24, 32, 56, 88, 144, 232, 376, 608, 984, 1592, 2576, 4168, 6744, 10912, 17656, 28568, 46224, 74792, 121016, 195808, 316824, 512632, 829456, 1342088, 2171544, 3513632, 5685176, 9198808, 14883984, 24082792, 38966776, 63049568, 102016344, 165065912

Offset: 0

Bruno Berselli, May 22 2015

Bruno Berselli, Table of n, a(n) for n = 0..300
Tanya Khovanova, Recursive Sequences: a(n) = a(n-1)+a(n-2).
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A022112, A039834, A156279.
Cf. A022091: 8*Fibonacci(n).
Cf. A022352: Fibonacci(n+6) + Fibonacci(n-6).
Cf. sequences with the formula Fibonacci(n+k)-Fibonacci(n-k): A000045 (k=1); A000032 (k=2); A022087 (k=3); A022379 (k=4, without initial 6); A022345 (k=5); this sequence (k=6); A022363 (k=7).

Magma
```
[8*Lucas(n): n in [0..40]];
```

Table[8 LucasL[n], {n, 0, 40}]
CoefficientList[Series[8*(2 - x)/(1 - x - x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 21 2017 *)

a(n)=([0,1; 1,1]^n*[16;8])[1,1] \\ Charles R Greathouse IV, Oct 07 2015

[8*lucas_number2(n, 1, -1) for n in (0..40)]

G.f.: 8*(2 - x)/(1 - x - x^2).
a(n)   = Fibonacci(n+6) - Fibonacci(n-6), where Fibonacci(-6..-1) = -8, 5, -3, 2, -1, 1 (see similar sequences listed in Crossrefs).
a(n)   = Lucas(n+4) + Lucas(n) + Lucas(n-4), where Lucas(-4..-1) = 7, -4, 3, -1.
a(n)   = a(n-1) + a(n-2) for n>1, a(0)=16, a(1)=8.
a(n)   = 2*A156279(n).
a(n+1) = 4*A022112(n).

A280154 a(n) = 5*Lucas(n).

Original entry on oeis.org

10, 5, 15, 20, 35, 55, 90, 145, 235, 380, 615, 995, 1610, 2605, 4215, 6820, 11035, 17855, 28890, 46745, 75635, 122380, 198015, 320395, 518410, 838805, 1357215, 2196020, 3553235, 5749255, 9302490, 15051745, 24354235, 39405980, 63760215, 103166195, 166926410, 270092605, 437019015

Offset: 0

Bruno Berselli, Dec 27 2016

Fibonacci sequence beginning 10, 5.
After 5, the sequence provides the 3rd column of the rectangular array in A213590.
After 5, all terms belong to A191921 because a(n) = Lucas(n+4) - 3*Lucas(n-1).
From G. C. Greubel, Dec 27 2016: (Start)
{a(n) mod 3} yields (1,2,0,2,2,1,0,1), repeated, and is given as A082115.
{a(n) mod 6} yields (4,5,3,2,5,1,0,1,1,2,3,5,2,1,3,4,1,5,0,5,5,4,3,1) and is given as A082117. (End)

Bruno Berselli, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

Subsequence of A084176.
Cf. A022088: 5*Fibonacci(n).
Cf. A022359: Lucas(n+5) + Lucas(n-5).
Cf. A000032, A000045, A191921, A213590.
Cf. sequences with formula Fibonacci(n+k) + Fibonacci(n-k): A006355 (k=0, without the initial 1), A000032 (k=1), A022086 (k=2), A022112 (k=3, with an initial 4), A022090 (k=4), this sequence (k=5), A022352 (k=6).

Magma
```
[5*Lucas(n): n in [0..40]];
```

F := n -> combinat:-fibonacci(n):
seq(F(n+5) + F(n-5), n=0..38); # Peter Luschny, Dec 29 2016

Mathematica
```
Table[5 LucasL[n], {n, 0, 40}]
```

vector(40, n, n--; fibonacci(n+5)+fibonacci(n-5))

def A280154():
    x, y = 10, 5
    while True:
        yield x
        x, y = y, x + y
a = A280154(); print([next(a) for  in range(39)]) # _Peter Luschny, Dec 29 2016

G.f.: 5*(2 - x)/(1 - x - x^2).
a(n) = a(n-1) + a(n-2) for n>1.
a(n) = Fibonacci(n+5) + Fibonacci(n-5), with Fibonacci(-k) = -(-1)^k*Fibonacci(k) for the negative indices.

A229339 GCD of all sums of n consecutive Lucas numbers.

Original entry on oeis.org

1, 1, 2, 5, 1, 4, 1, 15, 2, 11, 1, 40, 1, 29, 2, 105, 1, 76, 1, 275, 2, 199, 1, 720, 1, 521, 2, 1885, 1, 1364, 1, 4935, 2, 3571, 1, 12920, 1, 9349, 2, 33825, 1, 24476, 1, 88555, 2, 64079, 1, 231840, 1, 167761, 2, 606965, 1, 439204, 1, 1589055, 2, 1149851, 1, 4160200, 1, 3010349, 2

Offset: 1

Alonso del Arte, Sep 23 2013

The sum of two consecutive Lucas number is the sum of four consecutive Fibonacci numbers, which is verified easily enough with the identity L(n) = F(n - 1) + F(n + 1). Therefore a(1) = a(2) = A210209(4).

			a(3) = 2 because any sum of three consecutive Lucas numbers is an even number.
a(4) = 5 because all sums of four consecutive Lucas numbers are divisible by 5.
a(5) = 1 because some sums of five consecutive Lucas numbers are coprime.

Colin Barker, Table of n, a(n) for n = 1..1000
Dan Guyer and aBa Mbirika, GCD of sums of k consecutive Fibonacci, Lucas, and generalized Fibonacci numbers, Journal of Integer Sequences, 24 No.9, Article 21.9.8 (2021), 25pp; arXiv preprint, arXiv:2104.12262 [math.NT], 2021.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).

Cf. A210209, A022112, A022088, A022098, A106291 (Pisano periods of the Lucas sequence).

a[n_] := a[n] = If[n <= 14, {1, 1, 2, 5, 1, 4, 1, 15, 2, 11, 1, 40, 1, 29}[[n]], 3*a[n - 4] + a[n - 6] - a[n - 8] - 3*a[n - 10] + a[n - 14]]; Array[a, 64] (* Giovanni Resta, Oct 04 2013 *)
CoefficientList[Series[(x^12 - x^11 + 2 x^10 - 5 x^9 - 2 x^8 - x^7 - 6 x^6 + x^5 - 2 x^4 + 5 x^3 + 2 x^2 + x + 1) / (-x^14 + 3 x^10 + x^8 - x^6 - 3 x^4 + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 09 2014 *)
LinearRecurrence[{0,0,0,3,0,1,0,-1,0,-3,0,0,0,1},{1,1,2,5,1,4,1,15,2,11,1,40,1,29},70] (* Harvey P. Dale, Jul 21 2021 *)
Table[GCD[LucasL[n + 1] - 2, LucasL[n] + 1], {n, 0, 50}] (* Horst H. Manninger, Dec 25 2021 *)

Vec(x*(x^12 -x^11 +2*x^10 -5*x^9 -2*x^8 -x^7 -6*x^6 +x^5 -2*x^4 +5*x^3 +2*x^2 +x +1) / (-x^14 +3*x^10 +x^8 -x^6 -3*x^4 +1) + O(x^100)) \\ Colin Barker, Nov 09 2014

a(n) = 3*a(n-4) + a(n-6) - a(n-8) - 3*a(n-10) + a(n-14) for n > 14. - Giovanni Resta, Oct 04 2013
G.f.: x*(x^12 -x^11 +2*x^10 -5*x^9 -2*x^8 -x^7 -6*x^6 +x^5 -2*x^4 +5*x^3 +2*x^2 +x +1) / (-x^14 +3*x^10 +x^8 -x^6 -3*x^4 +1). - Colin Barker, Nov 09 2014
From Aba Mbirika, Jan 04 2022: (Start)
a(n) = gcd(L(n+1)-1, L(n+2)-3).
a(n) = Lcm_{A106291(m) divides n} m.
Proofs of these formulas are given in Theorems 15 and 25 of the Guyer-Mbirika paper. (End)

A254884 a(n) = Fibonacci(2n) + ((-1)^n-1)Fibonacci(n).

Original entry on oeis.org

0, -1, 3, 4, 21, 45, 144, 351, 987, 2516, 6765, 17533, 46368, 120927, 317811, 830820, 2178309, 5699693, 14930352, 39079807, 102334155, 267892404, 701408733, 1836254589, 4807526976, 12586118975, 32951280099, 86267178436, 225851433717, 591285701421, 1548008755920

Offset: 0

Peter Luschny, Mar 09 2015

Index entries for linear recurrences with constant coefficients, signature (3, 2, -9, 2, 3, -1).

Cf. A000032, A000045, A005522, A006172, A022112, A033888.

gf := x -> x/(x^2-3*x+1) + x/(x^2-x-1) + x/(x^2+x-1):
seq(coeff(series(gf(x),x,n+1),x,n), n=0..30);

LinearRecurrence[{4,-1,-11,11,1,-4,1}, {0,-1,3,4,21,45,144}, 31]
LinearRecurrence[{3, 2, -9, 2, 3, -1},{0, -1, 3, 4, 21, 45},31] (* Ray Chandler, Aug 03 2015 *)

A254884 = lambda n: fibonacci(2*n) + ((-1)^n-1)*fibonacci(n)
[A254884(n) for n in range(31)]

Let phi = (1+sqrt(5))/2, p(n) = phi^n - (-phi)^(-n) and FL(n) = 1 + (p(n-1) + p(n+1) + p(2*n-1)) / sqrt(5).
a(n) = FL(-n) - FL(n). By this definition a(n) is a doubly infinite sequence.
a(n) = -a(-n) for all n in Z.
a(n) = A006172(n) - A005522(n).
a(2*n) = A033888(n).
G.f.: x/(x^2-3*x+1) + x/(x^2-x-1) + x/(x^2+x-1).
a(n) = 4*a(n-1) - a(n-2) - 11*a(n-3) + 11*a(n-4) + a(n-5) - 4*a(n-6) + a(n-7).

A294116 Fibonacci sequence beginning 2, 21.

Original entry on oeis.org

2, 21, 23, 44, 67, 111, 178, 289, 467, 756, 1223, 1979, 3202, 5181, 8383, 13564, 21947, 35511, 57458, 92969, 150427, 243396, 393823, 637219, 1031042, 1668261, 2699303, 4367564, 7066867, 11434431, 18501298, 29935729, 48437027, 78372756, 126809783, 205182539, 331992322, 537174861

Offset: 0

Bruno Berselli, Oct 23 2017

Steven Vajda, Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications, Dover Publications (2008), page 24 (formula 8).

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000032, A000045.
Subsequence of A047201, A047592, A113763.
Sequences of the type g(2,k;n): A118658 (k=0), A000032 (k=1), 2*A000045 (k=2,4), A020695 (k=3), A001060 (k=5), A022112 (k=6), A022113 (k=7), A294157 (k=8), A022114 (k=9), A022367 (k=10), A022115 (k=11), A022368 (k=12), A022116 (k=13), A022369 (k=14), A022117 (k=15), A022370 (k=16), A022118 (k=17), A022371 (k=18), A022119 (k=19), A022372 (k=20), this sequence (k=21), A022373 (k=22); A022374 (k=24); A022375 (k=26); A022376 (k=28), A190994 (k=29), A022377 (k=30); A022378 (k=32).

a0:=2; a1:=21; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]];

Mathematica
```
LinearRecurrence[{1, 1}, {2, 21}, 40]
```

Vec((2 + 19*x)/(1 - x - x^2) + O(x^40)) \\ Colin Barker, Oct 25 2017

a = BinaryRecurrenceSequence(1, 1, 2, 21)
print([a(n) for n in range(38)]) # Peter Luschny, Oct 25 2017

G.f.: (2 + 19*x)/(1 - x - x^2).
a(n) = a(n-1) + a(n-2).
Let g(r,s;n) be the n-th generalized Fibonacci number with initial values r, s. We have:
a(n) =     Lucas(n) + g(0,20;n), see A022354;
a(n) = Fibonacci(n) + g(2,20;n), see A022372;
a(n) =  2*g(1,21;n) - g(0,21;n);
a(n) =     g(1,k;n) + g(1,21-k;n) for all k in Z.
a(h+k) = a(h)*Fibonacci(k-1) + a(h+1)*Fibonacci(k) for all h, k in Z (see S. Vajda in References section). For h=0 and k=n:
a(n) = 2*Fibonacci(n-1) + 21*Fibonacci(n).
Sum_{j=0..n} a(j) = a(n+2) - 21.
a(n) = (2^(-n)*((1-sqrt(5))^n*(-20+sqrt(5)) + (1+sqrt(5))^n*(20+sqrt(5)))) / sqrt(5). - Colin Barker, Oct 25 2017

A153263 a(n) = A014217(n+3) - A014217(n).

Original entry on oeis.org

3, 5, 9, 13, 23, 35, 59, 93, 153, 245, 399, 643, 1043, 1685, 2729, 4413, 7143, 11555, 18699, 30253, 48953, 79205, 128159, 207363, 335523, 542885, 878409, 1421293, 2299703, 3720995, 6020699, 9741693, 15762393, 25504085, 41266479, 66770563

Offset: 0

Paul Curtz, Dec 22 2008

The least significant digits are a sequence of period length 4: 3,5,9,3.

One could extend A014217 using its recurrence to define A014217(-1)=-1. This would add a(-1)=3 here by definition, and the least significant digits would still follow the (same, wrapped) period of length 4: 3,3,5,9.

Index entries for linear recurrences with constant coefficients, signature (0,2,1).

Cf. A022112.

LinearRecurrence[{0,2,1},{3,5,9},40] (* Harvey P. Dale, Jun 23 2022 *)

a(2n+2) = a(2n+1) + a(2n) + 1. a(2n+3) = a(2n+2) + a(2n+1) - 1.
From R. J. Mathar, Feb 07 2009, Apr 18 2009: (Start)
a(n) = 2*a(n-2) + a(n-3) = (-1)^n + 2*A000032(n+1).
G.f.: (3+5x+3x^2)/ ((1+x)(1-x-x^2)). (End)
a(n) + a(n+1) = A022112(n+2). - R. J. Mathar, Feb 25 2013
a(n) = ((-2)^n + (1 - sqrt(5))^(1+n) + (1 + sqrt(5))^(1+n))/2^n. - Stefano Spezia, Dec 25 2021

More terms from R. J. Mathar, Feb 07 2009

Edited by R. J. Mathar, Apr 18 2009

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A035513 Wythoff array read by falling antidiagonals.

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A028313 Elements in the 5-Pascal triangle (by row).

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A022319 a(n) = a(n-1) + a(n-2) + 1 for n > 1, a(0)=1, a(1)=5.

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A028314 Elements in the 5-Pascal triangle A028313 that are not 1.

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A258160 a(n) = 8*Lucas(n).

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A280154 a(n) = 5*Lucas(n).

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A254884 a(n) = Fibonacci(2n) + ((-1)^n-1)Fibonacci(n).