A003482 -id:A003482 - cp's OEIS frontend

A214216 List of minimal forbidden subwords of the Fibonacci word A003482.

22, 111, 21212, 11211211, 2121121211212, 112112121121121211211, 2121121211211212112121121121211212, 1121121211211212112121121121211211212112121121121211211, 21211212112112121121211211212112112121121211211212112121121121211211212112121121121211212

Offset: 1

N. J. A. Sloane, Jul 10 2012

nonn

If S is one of the terms of this sequence, then no word RS can appear as a subword of A003482.
Or, make a list of all words in {1,2}* that do not appear as factors of A003482 and discard any word which has a shorter word on the list as a right factor.
All the terms are palindromes.
Complementing the first and last digits of each term gives (essentially) A214217.

			A106750(3)=121 -> P=1 -> 111 = a(1).
A106750(4)=12112 -> P=121 -> 2121 = a(2).
A106750(5)=12112121 -> P=121121 -> 11211211 = a(3).

A. de Luca, Subwords of the Fibonacci word, MathOverflow, April 9, 2011.
F. Mignosi, A. Restivo, M. Sciortino, Words and forbidden factors, WORDS (Rouen, 1999). Theoret. Comput. Sci. 273 (2002), no. 1-2, 99--117. MR1872445 (2002m:68096).

Cf. A003842, A106750, A213975, A214217.

To get a(n), take A106750(n+2), delete last two digits, producing a palindrome P, say. Then a(n) = 1P1 if n is odd, or 2P2 if n is even.

A081016 a(n) = (Lucas(4n+3) + 1)/5, or Fibonacci(2n+1)Fibonacci(2n+2), or A081015(n)/5.

Original entry on oeis.org

1, 6, 40, 273, 1870, 12816, 87841, 602070, 4126648, 28284465, 193864606, 1328767776, 9107509825, 62423800998, 427859097160, 2932589879121, 20100270056686, 137769300517680, 944284833567073, 6472224534451830

Offset: 0

R. K. Guy, Mar 01 2003

a(n-1) is, together with b(n) := A089508(n), n >= 1, the solution to a binomial problem; see A089508.
Numbers k such that 1 - 2*k + 5*k^2 is a square. - Artur Jasinski, Oct 26 2008
Also solution y of Diophantine equation x^2 + 4*y^2 = h^2 for which x = y-1. - Carmine Suriano, Jun 23 2010

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 26.

G. C. Greubel, Table of n, a(n) for n = 0..1190
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081015.
Partial sums of A033889. Bisection of A001654. Equals A003482 + 1.
Cf. A145995, A178898.

List([0..30], n-> (Lucas(1,-1,4*n+3)[2] +1)/5 ); # G. C. Greubel, Jul 13 2019

[(Lucas(4*n+3) +1)/5: n in [0..30]]; // G. C. Greubel, Dec 18 2017

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+3)+1)/5) od: # James Sellers, Mar 03 2003

LinearRecurrence[{8,-8,1}, {1,6,40}, 30] (* Bruno Berselli, Aug 31 2017 *)

a(n)=([0,1,0; 0,0,1; 1,-8,8]^n*[1;6;40])[1,1] \\ Charles R Greathouse IV, Sep 28 2015

first(n) = Vec((1-2*x)/((1-x)*(1-7*x+x^2)) + O(x^n)) \\ Iain Fox, Dec 19 2017

[(lucas_number2(4*n+3,1,-1) +1)/5 for n in (0..30)] # G. C. Greubel, Jul 13 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: (1 - 2*x)/((1 - x)*(1 - 7*x + x^2)).
a(n) = F(1) + F(5) + F(9) +...+ F(4*n+1) = F(2*n)*F(2*n+3) + 1, where F(j) = Fibonacci(j).
a(n) = 7*a(n-1) - a(n-2) - 1, n >= 2. - R. J. Mathar, Nov 07 2015

A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

Original entry on oeis.org

0, 2, 15, 104, 714, 4895, 33552, 229970, 1576239, 10803704, 74049690, 507544127, 3478759200, 23843770274, 163427632719, 1120149658760, 7677619978602, 52623190191455, 360684711361584, 2472169789339634, 16944503814015855, 116139356908771352, 796030994547383610

Offset: 0

R. K. Guy, Mar 01 2003

Another interpretation of this sequence is: nonnegative integers k such that (k + 1)^2 + (2k)^2 is a perfect square. So apart from a(0) = 0, a(n) + 1 and 2a(n) form the legs of a Pythagorean triple. - Nick Hobson, Jan 13 2007
Also solution y of Diophantine equation x^2 + 4*y^2 = k^2 for which x=y+1. - Carmine Suriano, Jun 23 2010
Also the index of the first of two consecutive heptagonal numbers whose sum is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 20 2014
Nonnegative integers k such that G(x) = k for some rational number x where G(x) = x/(1-x-x^2) is the generating function of the Fibonacci numbers. - Tom Edgar, Aug 24 2015
The integer solutions of the equation a(b+1) = (a-b)(a-b-1) or, equivalently, binomial(a, b) = binomial(a-1, b+1) are given by (a, b) = (a(n+1), A003482(n)=Fibonacci(2*n) * Fibonacci(2*n+3)) (Lind and Singmaster). - Tomohiro Yamada, May 30 2018

			G.f. = 2*x + 15*x^2 + 104*x^3 + 714*x^4 + 4895*x^5 + 33552*x^6 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 28.
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Colin Barker, Table of n, a(n) for n = 0..1000
Pridon Davlianidze, Problem B-1279, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; An Unusual Generalization, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
Dae S. Hong, When is the Generating Function of the Fibonacci Numbers an Integer?, The College Mathematics Journal, Vol. 46, No. 2 (March 2015), pp. 110-112
D. A. Lind, The quadratic field Q(sqrt(5)) and a certain diophantine equation, Fibonacci Quart., Vol. 6, No. 3 (1968), pp. 86-93.
David Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quart., Vol. 13, No. 4 (1973), pp. 295-298.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081017.
Partial sums of A033891. Bisection of A001654 and A059840.
Equals A089508 + 1.
Cf. A001622, A081007, A081016, A178898.

List([0..30], n-> (Lucas(1,-1, 4*n+1)[2] -1)/5); # G. C. Greubel, Jul 14 2019

[(Lucas(4*n+1)-1)/5: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,(luc(4*n+1)-1)/5) od: # James Sellers, Mar 03 2003

(LucasL[4*Range[0,30]+1]-1)/5 (* or *) LinearRecurrence[{8,-8,1}, {0,2,15}, 30] (* G. C. Greubel, Aug 24 2015, modified Jul 14 2019 *)

concat(0, Vec(x*(2-x)/((1-x)*(1-7*x+x^2)) + O(x^30))) \\ Colin Barker, Dec 20 2014

[(lucas_number2(4*n+1,1,-1) -1)/5 for n in (0..30)] # G. C. Greubel, Jul 14 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(3) + Fibonacci(7) + Fibonacci(11) + ... + Fibonacci(4n+3).
G.f.: x*(2-x)/((1-x)*(1-7*x+x^2)). - Colin Barker, Mar 30 2012
E.g.f.: (1/5)^(3/2)*((1+phi^2)*exp(phi^4*x) - (1 + (1/phi^2))*exp(x/phi^4) - sqrt(5)*exp(x)), where 2*phi = 1 + sqrt(5). - G. C. Greubel, Aug 24 2015
From - Michael Somos, Aug 27 2015: (Start)
a(n) = -A081016(-1-n) for all n in Z.
0 = a(n) - 7*a(n+1) + a(n+2) - 1 for all n in Z.
0 = a(n)*a(n+2) - a(n+1)^2 + a(n+1) + 2 for all n in Z.
0 = a(n)*(a(n) -7*a(n+1) -1) + a(n+1)*(a(n+1) - 1) - 2 for all n in Z. (End)
a(n) = (k(n) + sqrt(k(n)*(4 + 5*k(n))))/2, where k(n) = A049684(n). - Stefano Spezia, Mar 11 2021
Product_{n>=1} (1 + 1/a(n)) = phi (A001622) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021

More terms from James Sellers, Mar 03 2003

A206351 a(n) = 7*a(n-1) - a(n-2) - 4 with a(1)=1, a(2)=3.

Original entry on oeis.org

1, 3, 16, 105, 715, 4896, 33553, 229971, 1576240, 10803705, 74049691, 507544128, 3478759201, 23843770275, 163427632720, 1120149658761, 7677619978603, 52623190191456, 360684711361585, 2472169789339635

Offset: 1

James R. Buddenhagen, Feb 06 2012

A Pell sequence related to Heronian triangles (rational triangles), see A206334. The connection is this: consider the problem of finding triangles with area a positive integer n, and with sides (a, b, n) where a, b are rational. Note that n is both the area and one side. For many values of n this is not possible, and the sequence of such numbers n is quite erratic (see A206334). Nonetheless, each term in this sequence is such a value of n. For example, for n = 105 you can take the other two sides, a and b, to be 10817/104, and 233/104 and the area will equal n, i.e., 105.

			G.f. = x + 3*x^2 + 16*x^3 + 105*x^4 + 715*x^5 + 4896*x^6 + 33553*x^7 + ... - _Michael Somos_, Jun 26 2018

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Pridon Davlianidze, Problem B-1279, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; An Unusual Generalization, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A003482, A089508, A094214.

Subsequence of A206334.

a206351 n = a206351_list !! (n-1)
a206351_list = 1 : 3 : map (subtract 4)
               (zipWith (-) (map (* 7) (tail a206351_list)) a206351_list)
-- Reinhard Zumkeller, Feb 08 2012

[Fibonacci(2*n)*Fibonacci(2*n-3): n in [1..30]]; // G. C. Greubel, Aug 12 2018

genZ := proc(n)
local start;
option remember;
    start := [1, 3];
    if n < 3 then start[n]
    else 7*genZ(n - 1) - genZ(n - 2) - 4
    end if
end proc:
seq(genZ(n),n=1..20);

LinearRecurrence[{8, -8, 1}, {1, 3, 16}, 50] (* Charles R Greathouse IV, Feb 07 2012 *)
RecurrenceTable[{a[1] == 1, a[2] == 3, a[n] == 7 a[n - 1] - a[n - 2] - 4}, a, {n, 20}] (* Bruno Berselli, Feb 07 2012 *)
a[ n_] := Fibonacci[2 n] Fibonacci[2 n - 3]; (* Michael Somos, Jun 26 2018 *)
nxt[{a_,b_}]:={b,7b-a-4}; NestList[nxt,{1,3},20][[;;,1]] (* Harvey P. Dale, Aug 29 2024 *)

Vec((1-5*x)/(1-8*x+8*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Feb 07 2012

{a(n) = fibonacci(2*n) * fibonacci(2*n - 3)}; /* Michael Somos, Jun 26 2018 */

From Bruno Berselli, Feb 07 2012: (Start)
G.f.: x*(1-5*x)/(1-8*x+8*x^2-x^3).
a(n) = A081018(n-1) + 1. (End)
a(n) = -A003482(-n) = Fibonacci(2*n)*Fibonacci(2*n-3). - Michael Somos, Jun 26 2018
a(n) = A089508(n-1) + 2 for n>1. - Bruno Berselli, Jun 20 2019 [Formula found by Umberto Cerruti]
Product_{n>=2} (1 - 1/a(n)) = 1/phi (A094214) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021
a(n) = (Fibonacci(2*n-2) + 1/Lucas(2*n-2))*(Fibonacci(2*n-1) + 1/Lucas(2*n-1)). - Peter Bala, Sep 03 2022

A337929 Numbers w such that (F(2n-1)^2, -F(2n)^2, w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

Original entry on oeis.org

1, 11, 79, 545, 3739, 25631, 175681, 1204139, 8253295, 56568929, 387729211, 2657535551, 18215019649, 124847601995, 855718194319, 5865179758241, 40200540113371, 275538601035359, 1888569667134145, 12944449068903659, 88722573815191471, 608113567637436641

Offset: 1

XU Pingya, Sep 30 2020

			2*(F(3)^2)^3 + 2*(-F(4)^2)^3 + 11^3 = 2*4^3 + 2*(-9)^3 + 11^3 = 1, 11 is a term.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045, A000578, A003482, A056573, A337928.

Table[(2*Fibonacci[2n]^6 - 2*Fibonacci[2n-1]^6 + 1)^(1/3), {n, 22}]
LinearRecurrence[{8,-8,1},{1,11,79},30] (* Harvey P. Dale, Aug 23 2021 *)

a(n) = (2*F(2*n)^6 - 2*F(2*n-1)^6 + 1)^(1/3).
From Colin Barker, Oct 01 2020: (Start)
G.f.: x*(1 + 3*x - x^2) / ((1 - x)*(1 - 7*x + x^2)).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3) for n>3.
(End)
a(n) = 2*A003482(n) + 1. - Hugo Pfoertner, Oct 01 2020
a(n) = A033888(n) - A064170(n+2). - Flávio V. Fernandes, Jan 10 2021

A288913 a(n) = Lucas(4*n + 3).

Original entry on oeis.org

4, 29, 199, 1364, 9349, 64079, 439204, 3010349, 20633239, 141422324, 969323029, 6643838879, 45537549124, 312119004989, 2139295485799, 14662949395604, 100501350283429, 688846502588399, 4721424167835364, 32361122672259149, 221806434537978679, 1520283919093591604

Offset: 0

Bruno Berselli, Jun 19 2017

a(n) mod 4 gives A101000.

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A101000.
Cf. A033891: fourth quadrisection of A000045.
Partial sums are in A081007 (after 0).
Positive terms of A098149, and subsequence of A001350, A002878, A016897, A093960, A068397.
Quadrisection of A000032: A056854 (first), A056914 (second), A246453 (third, without 11), this sequence (fourth).

[Lucas(4*n + 3): n in [0..30]]; // G. C. Greubel, Dec 22 2017

LucasL[4 Range[0, 21] + 3]
LinearRecurrence[{7,-1}, {4,29}, 30] (* G. C. Greubel, Dec 22 2017 *)

Vec((4 + x)/(1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017

from sympy import lucas
def a(n):  return lucas(4*n + 3)
print([a(n) for n in range(22)]) # Michael S. Branicky, Apr 29 2021

def L():
    x, y = -1, 4
    while True:
        yield y
        x, y = y, 7*y - x
r = L(); [next(r) for  in (0..21)] # _Peter Luschny, Jun 20 2017

G.f.: (4 + x)/(1 - 7*x + x^2).
a(n) = 7*a(n-1) - a(n-2) for n>1, with a(0)=4, a(1)=29.
a(n) = ((sqrt(5) + 1)^(4*n + 3) - (sqrt(5) - 1)^(4*n + 3))/(8*16^n).
a(n) = Fibonacci(4*n+4) + Fibonacci(4*n+2).
a(n) = 4*A004187(n+1) + A004187(n).
a(n) = 5*A003482(n) + 4 = 5*A081016(n) - 1.
a(n) = A002878(2*n+1) = A093960(2*n+3) = A001350(4*n+3) = A068397(4*n+3).
a(n+1)*a(n+k) - a(n)*a(n+k+1) = 15*Fibonacci(4*k). Example: for k=6, a(n+1)*a(n+6) - a(n)*a(n+7) = 15*Fibonacci(24) = 695520.

A260259 a(n) = F(n)*F(n+1) - (-1)^n, where F = A000045.

Original entry on oeis.org

-1, 2, 1, 7, 14, 41, 103, 274, 713, 1871, 4894, 12817, 33551, 87842, 229969, 602071, 1576238, 4126649, 10803703, 28284466, 74049689, 193864607, 507544126, 1328767777, 3478759199, 9107509826, 23843770273, 62423800999, 163427632718, 427859097161, 1120149658759

Offset: 0

Bruno Berselli, Oct 31 2015

Primes in sequence for n = 1, 3, 5, 6, 9, 24, 42, 48, 53, 71, 86, 102, 138, 182, 302, 438, 506, 926, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
A. Bremner, R. Høibakk, D. Lukkassen, Crossed ladders and Euler’s quartic, Annales Mathematicae et Informaticae, 36 (2009) pp. 29-41. See p. 33.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

First bisection of A111569.
Cf. A226205: numbers of the form F(n)*F(n+1)+(-1)^n.
Cf. A000045, A001654, A003482, A059929, A089508 (first bisection, without -1), A206351.

[Fibonacci(n)*Fibonacci(n+1)-(-1)^n: n in [0..30]];

with(combinat): A260259:=n->fibonacci(n)*fibonacci(n+1)-(-1)^n: seq(A260259(n), n=0..50); # Wesley Ivan Hurt, Feb 04 2017

Table[Fibonacci[n] Fibonacci[n + 1] - (-1)^n, {n, 0, 30}]

makelist(fib(n)*fib(n+1)-(-1)^n,n,0,30);

for(n=0, 30, print1(fibonacci(n)*fibonacci(n+1)-(-1)^n", "));

a(n) = round((2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 29 2016

Vec(-(1-4*x+x^2)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 29 2016

[fibonacci(n)*fibonacci(n+1)-(-1)^n for n in (0..30)]

G.f.: (-1 + 4*x - x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = -a(-n-1) = 2*a(n-1) + 2*a(n-2) - a(n-3) for all n in Z.
a(n) = F(n+2)^2 - 2*F(n+1)^2.
a(n) = A059929(n) - A059929(n-1) with A059929(-1)=1.
a(n) = -A001654(n+1) + 4*A001654(n) - A001654(n-1).
a(n) = A206351((n+2)/2)-2 for even n; a(n) = A003482((n-1)/2)+2 for odd n.
Sum_{i>=0} 1/a(i) = .754301907697893871765121109686...
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 29 2016

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

cp's OEIS Frontend

A214216 List of minimal forbidden subwords of the Fibonacci word A003482.

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A081016 a(n) = (Lucas(4*n+3) + 1)/5, or Fibonacci(2*n+1)*Fibonacci(2*n+2), or A081015(n)/5.

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A081018 a(n) = (Lucas(4n+1)-1)/5, or Fibonacci(2n)*Fibonacci(2n+1), or A081017(n)/5.

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A206351 a(n) = 7*a(n-1) - a(n-2) - 4 with a(1)=1, a(2)=3.

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A337929 Numbers w such that (F(2*n-1)^2, -F(2*n)^2, w) are primitive solutions of the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).

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A288913 a(n) = Lucas(4*n + 3).

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A081016 a(n) = (Lucas(4n+3) + 1)/5, or Fibonacci(2n+1)Fibonacci(2n+2), or A081015(n)/5.

A337929 Numbers w such that (F(2n-1)^2, -F(2n)^2, w) are primitive solutions of the Diophantine equation 2x^3 + 2y^3 + z^3 = 1, where F(n) is the n-th Fibonacci number (A000045).