A022379 -id:A022379 - 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

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).

A210209 GCD of all sums of n consecutive Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 4, 1, 3, 2, 11, 1, 8, 1, 29, 2, 21, 1, 76, 1, 55, 2, 199, 1, 144, 1, 521, 2, 377, 1, 1364, 1, 987, 2, 3571, 1, 2584, 1, 9349, 2, 6765, 1, 24476, 1, 17711, 2, 64079, 1, 46368, 1, 167761, 2, 121393, 1, 439204, 1, 317811, 2, 1149851, 1, 832040

Offset: 0

Alonso del Arte, Mar 18 2012

Early on in the Posamentier & Lehmann (2007) book, the fact that the sum of any ten consecutive Fibonacci numbers is a multiple of 11 is presented as an interesting property of the Fibonacci numbers. Much later in the book a proof of this fact is given, using arithmetic modulo 11. An alternative proof could demonstrate that 11*F(n + 6) = Sum_{i=n..n+9} F(i).

			a(3) = 2 because all sums of three consecutive Fibonacci numbers are divisible by 2 (F(n) + F(n-1) + F(n-2) = 2F(n)), but since the GCD of 3 + 5 + 8 = 16 and 5 + 8 + 13 = 26 is 2, no number larger than 2 divides all sums of three consecutive Fibonacci numbers.
a(4) = 1 because the GCD of 1 + 1 + 2 + 3 = 7 and 1 + 2 + 3 + 5 = 11 is 1, so the sums of four consecutive Fibonacci numbers have no factors in common.

Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, New York (2007) p. 33.

Alois P. Heinz, Table of n, a(n) for n = 0..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.
I. D. Ruggles, Elementary problem B-1, Fibonacci Quarterly, Vol. 1, No. 1, February 1963, p. 73; Solution by Marjorie R. Bicknell, published in Vol. 1, No. 3, October 1963, pp. 76-77.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).

Cf. A000045, A000071, sum of the first n Fibonacci numbers, A001175 (Pisano periods). Cf. also A229339.
Bisections give: A005013 (even part), A131534 (odd part).
Sums of m consecutive Fibonacci numbers: A055389 (m = 3, ignoring the initial 1); A000032 (m = 4, these are the Lucas numbers); A013655 (m = 5); A022087 (m = 6); A022096 (m = 7); A022379 (m = 8).

a:= n-> (Matrix(7, (i, j)-> `if`(i=j-1, 1, `if`(i=7, [1, 0, -3, -1, 1, 3, 0][j], 0)))^iquo(n, 2, 'r'). `if`(r=0, <<0, 1, 1, 4, 3, 11, 8>>, <<1, 2, 1, 1, 2, 1, 1>>))[1, 1]: seq(a(n), n=0..80);  # Alois P. Heinz, Mar 18 2012

Table[GCD[Fibonacci[n + 1] - 1, Fibonacci[n]], {n, 1, 50}] (* Horst H. Manninger, Dec 19 2021 *)

a(n)=([0,1,0,0,0,0,0,0,0,0,0,0,0,0; 0,0,1,0,0,0,0,0,0,0,0,0,0,0; 0,0,0,1,0,0,0,0,0,0,0,0,0,0; 0,0,0,0,1,0,0,0,0,0,0,0,0,0; 0,0,0,0,0,1,0,0,0,0,0,0,0,0; 0,0,0,0,0,0,1,0,0,0,0,0,0,0; 0,0,0,0,0,0,0,1,0,0,0,0,0,0; 0,0,0,0,0,0,0,0,1,0,0,0,0,0; 0,0,0,0,0,0,0,0,0,1,0,0,0,0; 0,0,0,0,0,0,0,0,0,0,1,0,0,0; 0,0,0,0,0,0,0,0,0,0,0,1,0,0; 0,0,0,0,0,0,0,0,0,0,0,0,1,0; 0,0,0,0,0,0,0,0,0,0,0,0,0,1; 1,0,0,0,-3,0,-1,0,1,0,3,0,0,0]^n*[0;1;1;2;1;1;4;1;3;2;11;1;8;1])[1,1] \\ Charles R Greathouse IV, Jun 20 2017

G.f.: -x*(x^12-x^11+2*x^10-x^9-2*x^8-x^7-6*x^6+x^5-2*x^4+x^3+2*x^2+x+1) / (x^14-3*x^10-x^8+x^6+3*x^4-1) = -1/(x^4+x^2-1) + (x^2+1)/(x^4-x^2-1) + (x+2)/(6*(x^2+x+1)) + (x-2)/(6*(x^2-x+1)) - 2/(3*(x+1)) - 2/(3*(x-1)). - Alois P. Heinz, Mar 18 2012
a(n) = gcd(Fibonacci(n+1)-1, Fibonacci(n)). - Horst H. Manninger, Dec 19 2021
From Aba Mbirika, Jan 21 2022: (Start)
a(n) = gcd(F(n+1)-1, F(n+2)-1).
a(n) = Lcm_{A001175(m) divides n} m.
Proofs of these formulas are given in Theorems 15 and 25 of the Guyer-Mbirika paper. (End)

More terms from Alois P. Heinz, Mar 18 2012

A230448 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A226205(n+1), n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 3, 1, 2, 4, 5, 1, 3, 6, 9, 16, 1, 4, 9, 15, 25, 39, 1, 5, 13, 24, 40, 64, 105, 1, 6, 18, 37, 64, 104, 169, 272, 1, 7, 24, 55, 101, 168, 273, 441, 715, 1, 8, 31, 79, 156, 269, 441, 714, 1156, 1869, 1, 9, 39, 110, 235, 425, 710, 1155, 1870, 3025, 4896

Offset: 0

Johannes W. Meijer, Oct 19 2013

Triangle T(n, k) is related to the Kn1p sums of the ‘Races with Ties’ triangle A035317. See A230447 for the Kn1p sums and A180662 for the definitions of these sums.
The row sums equal ((-1)^n*3*A083581(n) + A022379(2*n+2))/15.
Note that the partial fraction expansion of the G.f. of the terms in the n-th row of the square array Tsq(n, k) = T(n+k, k) is related to A014334, the exponential convolution of the Fibonacci numbers with themselves, and to A000032, the Lucas numbers.

			The first few rows of triangle T(n, k), n >= 0 and 0 <= k <= n.
n/k 0   1   2    3    4     5     6     7
------------------------------------------------
0|  1
1|  1,  0
2|  1,  1,  3
3|  1,  2,  4,   5
4|  1,  3,  6,   9,  16
5|  1,  4,  9,  15,  25,   39
6|  1,  5, 13,  24,  40,   64,  105
7|  1,  6, 18,  37,  64,  104,  169,   272
The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.
n/k 0   1   2    3    4    5      6     7
------------------------------------------------
0|  1,  0,  3,   5,  16,  39,   105,  272
1|  1,  1,  4,   9,  25,  64,   169,  441
2|  1,  2,  6,  15,  40,  104,  273,  714
3|  1,  3,  9,  24,  64,  168,  441, 1155
4|  1,  4, 13,  37, 101,  269,  710, 1865
5|  1,  5, 18,  55, 156,  425, 1135, 3000
6|  1,  6, 24,  79, 235,  660, 1795, 4795
7|  1,  7, 31, 110, 345, 1005, 2800, 7595

Cf. (Triangle columns) A000012, A001477, A152950, A226205, A007598, A001654, A064831, A080145

T := proc(n, k) option remember: if k=0 then return(1) elif k=n then return(combinat[fibonacci](n+2)*combinat[fibonacci](n-1)) else procname(n-1, k-1) + procname(n-1, k) fi: end: seq(seq(T(n, k), k=0..n), n=0..10); # End first program.
T := proc(n, k): add(A035317(n+k-p-2, p), p=0..k) end: A035317 := proc(n, k): add((-1)^(i+k) * binomial(i+n-k+1, i), i=0..k) end: seq(seq(T(n, k), k=0..n), n=0..10); # End second program.

T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = F(n+2) * F(n-1) = A226205(n+1) with F(n) = A000045(n), the Fibonacci numbers, n >= 0 and 0 <= k <= n.
T(n, k) = sum(A035317(n+k-p-2, p), p=0..k), n >= 0 and 0 <= k <= n.
T(n+p+2, p-2) = A080239(n+2*p-1) - sum(A035317(n-k+p-1, k+p-1), k=0..floor(n/2)), n >= 0 and p >= 2.
The triangle as a square array Tsq(n, k) = T(n+k, k), n >= 0 and k >= 0.
Tsq(n, k) = sum(Tsq(n-1, i), i=0..k), n >= 1 and k >= 0, with Tsq(0, k) = A226205(k+1).
The two G.f.’s given below generate the terms in the n-th row of the square array Tsq(n, k). The remarkable second G.f. is the partial fraction expansion of the first G.f..
G.f.: 1/((1-x)^(n-2)*(1+x)*(x^2-3*x+1)), n >= 0.
G.f.: sum((-1)^(n+k-1)*A014334(k+2)/(2^(k+2)*(x-1)^(n-k-2)), k=0..n-3) + 1/(5*2^(n-2)*(1+x)) + (A000032(n+1) - A000032(n-1)*x)/(5*(x^2-3*x+1)), n >= 0.

A282465 a(n) = 11*Fibonacci(n+3) + Fibonacci(n-8) with n>=0.

Original entry on oeis.org

1, 46, 47, 93, 140, 233, 373, 606, 979, 1585, 2564, 4149, 6713, 10862, 17575, 28437, 46012, 74449, 120461, 194910, 315371, 510281, 825652, 1335933, 2161585, 3497518, 5659103, 9156621, 14815724, 23972345, 38788069, 62760414, 101548483, 164308897, 265857380, 430166277

Offset: 0

Bruno Berselli, Feb 20 2017

Similar sequences with the formula h*Fibonacci(n+k) + Fibonacci(n+k-h):
h= 1, k=-1: A000045;
h= 2, k= 1: A013655;
h= 3, k=-2: A118658 = 2*A212804;
h= 4, k= 2: A022379 = 3*A000204;
h= 5, k= 1: A022113;
h= 6, k= 2: A022125;
h= 7, k= 3: A097657;
h= 8, k= 2: A022355 = 21*A000045;
h= 9, k= 3: 10, 32, 42, 74, 116, 190, 306, 496, 802, ... =  2*A022140;
h=10, k= 3: 33, 22, 55, 77, 132, 209, 341, 550, 891, ... = 11*A013655;
h=11, k= 3: this sequence.

Indranil Ghosh, Table of n, a(n) for n = 0..4769
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000045, A013655, A022113, A022125, A022355, A022379, A022379, A097657, A118658.

Cf. sequences with g.f. (1 + r*x)/(1 - x - x^2) for r = 2..31, respectively: A000204, A000285, A022095 - A022110, A022391 - A022402.

[11*Fibonacci(n+3)+Fibonacci(n-8): n in [0..40]];

Table[11 Fibonacci[n + 3] + Fibonacci[n - 8], {n, 0, 40}]
LinearRecurrence[{1,1},{1,46},36] (* or *) CoefficientList[Series[(1 + 45*x)/(1 - x - x^2) , {x,0,35}],x] (* Indranil Ghosh, Feb 22 2017 *)

a(n) = 11*fibonacci(n+3) + fibonacci(n-8) \\ Indranil Ghosh, Feb 23 2017

G.f.: (1 + 45*x)/(1 - x - x^2).
a(n) = a(n-1) + a(n-2).
a(n) = a(i)*Fibonacci(n-i+1) + a(i-1)*Fibonacci(n-i). Examples:
for i= 3,  a(3)=93,  a(2)= 47: a(n) = 93*Fibonacci(n-2) + 47*Fibonacci(n-3);
for i=-1, a(-1)=45, a(-2)=-44: a(n) = 45*Fibonacci(n+2) - 44*Fibonacci(n+1).
Other formulae:
a(n) = 44*Fibonacci(n) + Fibonacci(n+2),
a(n) = 45*Fibonacci(n) + Fibonacci(n+1),
a(n) = 46*Fibonacci(n) + Fibonacci(n-1),
a(n) = 47*Fibonacci(n) - Fibonacci(n-2).
a(n) = ((91 + sqrt(5))*((1 + sqrt(5))/2)^n - (91 - sqrt(5))*((1 - sqrt(5))/2)^n)/sqrt(20).

A347351 Triangle read by rows: T(n,k) is the number of links of length k in a set of all necklaces A000358 of length n, 1 <= k <= n.

Original entry on oeis.org

1, 2, 1, 3, 0, 1, 4, 2, 0, 1, 5, 1, 1, 0, 1, 6, 4, 2, 1, 0, 1, 7, 3, 2, 1, 1, 0, 1, 8, 8, 3, 3, 1, 1, 0, 1, 9, 8, 7, 3, 2, 1, 1, 0, 1, 10, 18, 9, 5, 4, 2, 1, 1, 0, 1, 11, 21, 13, 8, 5, 3, 2, 1, 1, 0, 1, 12, 40, 24, 16, 8, 6, 3, 2, 1, 1, 0, 1, 13, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1, 0, 1

Offset: 0

Maxim Karimov and Vladislav Sulima, Aug 28 2021

Definitions:
1. A link is any 0 in any necklace from A000358 and all 1s following this 0 in this necklace to right until another 0 is encountered.
2. Length of the link is the number of elements in the link.
Sum of all elements n-row is Fibonacci(n-1)+n iff n=1 or n=p (follows from the identity for the sum of the Fibonacci numbers and the formula for the triangle T(n,k)).

			For k > 0:
   n\k |  1   2   3   4   5   6   7   8   9  10  ...
  -----+---------------------------------------
   1   |  1
   2   |  2   1
   3   |  3   0   1
   4   |  4   2   0   1
   5   |  5   1   1   0   1
   6   |  6   4   2   1   0   1
   7   |  7   3   2   1   1   0   1
   8   |  8   8   3   3   1   1   0   1
   9   |  9   8   7   3   2   1   1   0   1
  10   | 10  18   9   5   4   2   1   1   0   1
  ...
If we continue the calculation for nonpositive k, we get a table in which each row is a Fibonacci sequence, in which term(0) = A113166, term(1) = A034748.
For k <= 0:
   n\k |  0   -1   -2   -3   -4   -5   -6   -7   -8   -9 ...
  -----+------------------------------------------------
   1   |  0    1    1    2    3    5    8   13   21   34 ... A000045
   2   |  1    2    3    5    8   13   21   34   55   89 ... A000045
   3   |  1    4    5    9   14   23   37   60   97  157 ... A000285
   4   |  3    6    9   15   24   39   63  102  165  267 ... A022086
   5   |  3    9   12   21   33   54   87  141  228  369 ... A022379
   6   |  8   14   22   36   58   94  152  246  398  644 ... A022112
   7   |  8   19   27   46   73  119  192  311  503  814 ... A206420
   8   | 17   30   47   77  124  201  325  526  851 1377 ... A022132
   9   | 23   44   67  111  178  289  467  756 1223 1979 ... A294116
  10   | 41   68  109  177  286  463  749 1212 1961 3173 ... A022103
  ...

Cf. A000010, A000027, A000045, A000358, A113166, A034748.

function [res] = calcLinks(n,k)
if k==1
    res=n;
else
    d=divisors(n);
    res=0;
    for i=1:length(d)
        if d (i) >= k
            res=res+eulerPhi(n/d(i))*fiboExt(d(i)-k-1);
        end
    end
end
function [s] = fiboExt(m) % extended fibonacci function (including negative arguments)
m=sym(m); % for large fibonacci numbers
if m>=0 || mod(m,2)==1
    s=fibonacci(abs(m));
else
    s=fibonacci(abs(m))*(-1);
end

T(n, k) = if (k==1, n, sumdiv(n, d, if (d>=k, eulerphi(n/d)*fibonacci(d-k-1)))); \\ Michel Marcus, Aug 29 2021

If k=1, T(n,k)=n, otherwise T(n,k) = Sum_{d>=k, d|n} Phi(n/d)*Fibonacci(d-k-1), where Phi=A000010.

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cp's OEIS Frontend

A035513 Wythoff array read by falling antidiagonals.

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

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A210209 GCD of all sums of n consecutive Fibonacci numbers.

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A230448 T(n, k) = T(n-1, k-1) + T(n-1, k) with T(n, 0) = 1 and T(n, n) = A226205(n+1), n >= 0 and 0 <= k <= n.

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A282465 a(n) = 11*Fibonacci(n+3) + Fibonacci(n-8) with n>=0.

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A347351 Triangle read by rows: T(n,k) is the number of links of length k in a set of all necklaces A000358 of length n, 1 <= k <= n.

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