A005248 -id:A005248 - cp's OEIS frontend

A061171 One half of second column of Lucas bisection triangle (odd part).

3, 19, 79, 283, 940, 2982, 9171, 27581, 81557, 237995, 687158, 1966764, 5588259, 15780103, 44323195, 123920827, 345062176, 957403026, 2647935987, 7302634865, 20087869313, 55128445259, 150971982314

Offset: 0

Wolfdieter Lang, Apr 20 2001

Numerator of g.f. is on half of row polynomial Sum_{m=0..2} A061187(1,m) * x^m.

G. C. Greubel, Table of n, a(n) for n = 0..1000
É. Czabarka, R. Flórez, L. Junes, A Discrete Convolution on the Generalized Hosoya Triangle, Journal of Integer Sequences, 18 (2015), #15.1.6.
Index entries for linear recurrences with constant coefficients, signature (6,-11,6,-1).

Cf. A000032, A000045, A001906, A002878, A005248.

I:=[3,19,79,283]; [n le 4 select I[n] else 6*Self(n-1) - 11*Self(n-2) + 6*Self(n-3) - Self(n-4): n in [1..30]]; // G. C. Greubel, Dec 21 2017

CoefficientList[Series[(1+x)(3-2x)/(1-3x+x^2)^2,{x,0,30}],x] (* or *) LinearRecurrence[{6,-11,6,-1},{3,19,79,283},30] (* Harvey P. Dale, Oct 11 2012 *)

my(x='x+O('x^30)); Vec((1+x)*(3-2*x)/(1-3*x+x^2)^2) \\ G. C. Greubel, Dec 21 2017

2*a(n) = A060924(n+1, 1).
G.f.: (1+x)*(3-2*x)/(1-3*x+x^2)^2.
a(n) = 6*a(n-1) - 11*a(n-2) + 6*a(n-3) - a(n-4), with a(0)=3, a(1)=19, a(2)=79, a(3)=283. - Harvey P. Dale, Oct 11 2012
a(n) = Fibonacci(2*n+4) + n*Lucas(2*n+3). - Lechoslaw Ratajczak, May 06 2020

A065034 a(n) = Lucas(2*n) + 1.

Original entry on oeis.org

3, 4, 8, 19, 48, 124, 323, 844, 2208, 5779, 15128, 39604, 103683, 271444, 710648, 1860499, 4870848, 12752044, 33385283, 87403804, 228826128, 599074579, 1568397608, 4106118244, 10749957123, 28143753124, 73681302248, 192900153619, 505019158608, 1322157322204

Offset: 0

N. J. A. Sloane, Nov 04 2001

Harry J. Smith, Table of n, a(n) for n = 0..200
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 8.
Markus Grassl and Andrew J. Scott, Fibonacci-Lucas SIC-POVMs, arXiv:1707.02944 [quant-ph], 2017.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000032, A000045, A005248, A005592.

Cf. A002878 (first differences). - R. J. Mathar, Jul 18 2009

[ Lucas(2*n) + 1: n in [0..210]]; // Vincenzo Librandi, Apr 15 2011

a:= n-> (<<0|1>, <1|1>>^(2*n). <<2,1>>)[1, 1]+1:
seq(a(n), n=0..30);  # Alois P. Heinz, Nov 01 2016

LucasL[2 Range[30]]+1 (* Harvey P. Dale, Oct 21 2011 *)
LinearRecurrence[{4, -4, 1}, {3, 4, 8}, 30] (* Jean-François Alcover, Jan 08 2019 *)

a(n) = { fibonacci(2*n + 1) + fibonacci(2*n - 1) + 1 } \\ Harry J. Smith, Oct 03 2009

Vec((3-2*x)*(1-2*x)/((1-x)*(1-3*x+x^2)) + O(x^40)) \\ Colin Barker, Nov 01 2016

a(n) = F(2*n+1) + F(2*n-1) + 1 = A005248(n) + 1.
From R. J. Mathar, Jul 18 2009: (Start)
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
G.f.: 1/(1-x) + (2-3*x)/(1-3*x+x^2). (End)
a(n) = 1 + ((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n. - Colin Barker, Nov 01 2016

a(0)=3 prepended by Joerg Arndt, Nov 01 2016

A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

Original entry on oeis.org

2, 10, 98, 970, 9602, 95050, 940898, 9313930, 92198402, 912670090, 9034502498, 89432354890, 885289046402, 8763458109130, 86749292044898, 858729462339850, 8500545331353602, 84146723851196170, 832966693180608098, 8245520207954884810

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 11 2003

a(n+1)/a(n) converges to (5+sqrt(24)) = 9.8989794... a(0)/a(1)=2/10; a(1)/a(2)=10/98; a(2)/a(3)=98/970; a(3)/a(4)=970/9602; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.10102051... = 1/(5+sqrt(24)) = (5-sqrt(24)).
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 96 = 0. - Colin Barker, Feb 25 2014
A triangle whose sides are a(n) - 1, a(n) and a(n) + 1 is nearly Fleenor-Heronian since its area is the product of an integer and the square root of 2. See A003500. - Charlie Marion, Dec 18 2020

			a(4) = 9602 = 10*a(3) - a(2) = 10*970 - 98 = (5+sqrt(24))^4 + (5-sqrt(24))^4.

T. D. Noe, Table of n, a(n) for n = 0..200
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A005248, A006242, A036336, A086927, A135927, A174503.

I:=[2,10]; [n le 2 select I[n] else 10*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Nov 07 2018

a[0] = 2; a[1] = 10; a[n_] := 10a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{10,-1}, {2,10}, 30] (* G. C. Greubel, Nov 07 2018 *)

polsym(x^2 - 10*x + 1,20) \\ Charles R Greathouse IV, Jun 11 2011

{a(n) = 2 * real( (5 + 2 * quadgen(24))^n )}; /* Michael Somos, Feb 25 2014 */

[lucas_number2(n,10,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.
G.f.: (2-10*x)/(1-10*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 5 - sqrt(24). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.09989 80642 72052 68138 ... = 2 + 1/(10 + 1/(98 + 1/(970 + ...))).
Also F(-alpha) = 0.89989 78538 78393 34715 ... has the continued fraction representation 1 - 1/(10 - 1/(98 - 1/(970 - ...))) and the simple continued fraction expansion 1/(1 + 1/((10-2) + 1/(1 + 1/((98-2) + 1/(1 + 1/((970-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((10^2-4) + 1/(1 + 1/((98^2-4) + 1/(1 + 1/((970^2-4) + 1/(1 + ...))))))). Cf. A174503 and A005248. (End)
a(-n) = a(n). - Michael Somos, Feb 25 2014
From Peter Bala, Oct 16 2019: (Start)
8*Sum_{n >= 1} 1/(a(n) - 12/a(n)) = 1.
12*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 8/a(n)) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/8 - 12*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 12))  and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/12 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 8)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(5-sqrt(24)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(sqrt(24)-5))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499. (End)
E.g.f.: 2*exp(5*x)*cosh(2*sqrt(6)*x). - Stefano Spezia, Oct 18 2019
From Peter Bala, Mar 29 2022: (Start)
a(n) = 2*T(n,5), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(2^n) = A135927(n+1) and a(3^n) = A006242(n+1), both for n >= 0. (End)

More terms from Colin Barker, Feb 25 2014

A173027 Numbers of rows R of the Wythoff array such that R is the n-th multiple of a tail of the Fibonacci sequence.

Original entry on oeis.org

1, 3, 4, 5, 16, 19, 22, 25, 28, 31, 34, 97, 105, 113, 121, 129, 137, 145, 153, 161, 169, 177, 185, 193, 201, 209, 217, 225, 233, 631, 652, 673, 694, 715, 736, 757, 778, 799, 820, 841, 862, 883, 904, 925, 946, 967, 988, 1009, 1030, 1051, 1072, 1093, 1114, 1135

Offset: 1

Clark Kimberling, Feb 07 2010

nonn

Row 1 of the array A173028.
Contribution from K. G. Stier, Dec 08 2012: (Start)
It appears that the numbers of this sequence form groups of m members respectively with same distance d of two consecutive values a(n) such that d is equal to even-indexed Fibonacci numbers (A001906) while m is equal to even-indexed Lucas numbers (A005248). Example: from n=1365 to 3571 d=987 and m=2207;
Also of interest are the gaps between two consecutive groups which appear to be sums of Fibonacci numbers F(2*n) plus F(4*n-2). Example: gap 5 after a(76) is 2639 = F(10) + F(18) = 55 + 2584.
Likewise, the tail (as mentioned in this sequence's name) of the Fibonacci sequence is chopped off by two initial terms at each of the gap positions. (End)

			Referring to rows of the Wythoff array (A035513),
Row 1: (1,2,3,5,...) = 1*(1,2,3,...)
Row 3: (6,10,16,...) = 2*(3,5,8,...)
Row 4: (9,15,24,...) = 3*(3,5,8,...)
Row 5: (12,20,32,...) = 4*(3,5,8,...)
Row 16: (40,65,105...) = 8*(5,13,21,...).

K. G. Stier, Table of n, a(n) for n = 1..10000

Cf. A000045, A035513, A173028, A220249.

A198636 One half of total number of round trips, each of length 2n, on the graph P_6 (o-o-o-o-o-o).

Original entry on oeis.org

3, 5, 13, 38, 117, 370, 1186, 3827, 12389, 40169, 130338, 423065, 1373466, 4459278, 14478659, 47011093, 152642789, 495626046, 1609284589, 5225309458, 16966465802, 55089756851, 178875298901, 580804419201, 1885860059450, 6123349080945

Offset: 0

Wolfdieter Lang, Nov 03 2011

See the array and triangle A198632 for the general graph P_N case (there N is n and the length is l=2*k).

			With the graph P_6 as 1-2-3-4-5-6:
n=0: a(0)=3 because w(6,0)=6, the number of vertices.
n=2: a(2)=5 because the 10 round trips of length 2 are 121, 212, 232, 323, 343, 434, 454, 545, 565 and 656.

M. F. Hasler, Table of n, a(n) for n = 0..499
S. Barbero, Dickson Polynomials, Chebyshev Polynomials, and Some Conjectures of Jeffery, Journal of Integer Sequences, 17 (2014), #14.3.8.
S. Barbero, U. Cerruti, N. Murru, Identities Involving Zeros of Ramanujan and Shanks Cubic Polynomials, J. Integer Seq., Vol. 16 (2013), Article 13.8.1, pp. 10-12.
Index entries for linear recurrences with constant coefficients, signature (5,-6,1).

Cf. A198632, A198633, A005248, A198635.

Table[7 (Binomial[2 n - 1, n - 1] + Sum[Binomial[2 n, n - 7 k], {k, Floor[n/7]}]) - 2^(2 n - 1) - (7/2) Boole[n == 0], {n, 0, 25}] (* Michael De Vlieger, Jul 17 2017 *)

vec_A198636(Nmax)=Vec((3-10*x+6*x^2)/(1-5*x+6*x^2-x^3)+O(x^Nmax)) \\ Indices will start at 1 in this vector. - M. F. Hasler, Nov 03 2013

{a(n) = if( n<0, n=-n; polcoeff( (3 - 12*x + 5*x^2) / (1 - 6*x + 5*x^2 - x^3) + x * O(x^n), n), polcoeff( (3 - 10*x + 6*x^2) / (1 - 5*x + 6*x^2 -x^3) + x * O(x^n), n))}; /* Michael Somos, Jul 17 2017 */

a(n) = w(6,2*n)/2, n>=0, with w(6,l) the total number of closed walks on the graph P_6 (the simple path with 6 points (vertices) and 5 lines (or edges)).
O.g.f. for w(6,l) (with zeros for odd l): y*(d/dy)S(6,y)/S(6,y) with y=1/x and Chebyshev S-polynomials (coefficients A049310). See also A198632 for a rewritten form.
O.g.f.: (3-10*x+6*x^2)/(1-5*x+6*x^2-x^3). - Colin Barker, Jan 02 2012
Conjecture: a(n) = 2^(2*n)*(sum_{k=1,2,3} (cos(k*Pi/7))^(2*n)). - L. Edson Jeffery, Jan 21 2012 (in fact this conjecture was recently proved in [Barbero, et al.])
a(n) = 7*(binomial(2n-1,n-1) + sum_{k = 1..floor(n/7)} binomial(2n,n-7k)) - 2^(2n-1). - M. Lawrence Glasser, Feb 20 2013
Let r,s,t be the roots of x^3 + x^2 - 2x - 1; then apparently a(n) = r^(2n) + s^(2n) + t^(2n). - James R. Buddenhagen, Nov 03 2013 [This is equivalent to the conjecture by L. Edson Jeffery.]
a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3). - M. F. Hasler, Nov 05 2013
G.f.: F(x) = (sum_{r=0..2} ((3-r)*(-1)^r*binomial(6-r,r))*x^r)/(sum_{s=0..3} ((-1)^s*binomial(6-s,s))*x^s). - L. Edson Jeffery, Nov 23 2013

A237132 Values of x in the solutions to x^2 - 3xy + y^2 + 11 = 0, where 0 < x < y.

Original entry on oeis.org

3, 4, 5, 9, 12, 23, 31, 60, 81, 157, 212, 411, 555, 1076, 1453, 2817, 3804, 7375, 9959, 19308, 26073, 50549, 68260, 132339, 178707, 346468, 467861, 907065, 1224876, 2374727, 3206767, 6217116, 8395425, 16276621, 21979508, 42612747, 57543099, 111561620

Offset: 1

Colin Barker, Feb 04 2014

The corresponding values of y are given by a(n+2).

Positive values of x (or y) satisfying x^2 - 18xy + y^2 + 704 = 0.

			9 is in the sequence because (x, y) = (9, 23) is a solution to x^2 - 3xy + y^2 + 11 = 0.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-1).

Cf. A001519, A005248, A055819, A237133, A218735.

Vec(-x*(x-1)*(3*x^2+7*x+3)/((x^2-x-1)*(x^2+x-1)) + O(x^100))

a(n) = 3*a(n-2)-a(n-4).
G.f.: -x*(x-1)*(3*x^2+7*x+3) / ((x^2-x-1)*(x^2+x-1)).
a(n) = F(n+2) + (-1)^n*F(n-3), n>1, with F the Fibonacci numbers (A000045). - Ralf Stephan, Feb 05 2014
Let h(n) = hypergeom([(1 - n)/2, n mod 2 - n/2], [1 - n], -4) then a(n) = h(n-1) + h(n) for n > 3. - Peter Luschny, Sep 03 2019

A047946 a(n) = 5F(n)^2 + 3(-1)^n where F(n) are the Fibonacci numbers A000045.

Original entry on oeis.org

3, 2, 8, 17, 48, 122, 323, 842, 2208, 5777, 15128, 39602, 103683, 271442, 710648, 1860497, 4870848, 12752042, 33385283, 87403802, 228826128, 599074577, 1568397608, 4106118242, 10749957123, 28143753122, 73681302248, 192900153617, 505019158608, 1322157322202

Offset: 0

John W. Layman, May 21 1999

Form the matrix A=[1,1,1;2,1,0;1,0,0]. a(n)=trace(A^n). - Paul Barry, Sep 22 2004
The set of prime divisors of elements of this sequence with the exception of 3 is the set of primes that do not divide odd Fibonacci numbers. - Tanya Khovanova, May 19 2008
If a(n) is prime then n is a power of 3 (Boase, 1998). The only values of k not exceeding 12 for which a(3^k) is prime are 0 and 1. - Amiram Eldar, Jun 19 2022

Mansur Boase, Problem 1558, Mathematics Magazine, Vol. 71, No. 4 (1998), p. 316; Primes in a Recursively Defined Sequence, Solution to Problem 1558 by TAMUK Problem Solvers, ibid., Vol. 72, No. 4 (1999), pp. 330-331.
Tanya Khovanova, Divisibility of Odd Fibonaccis, 2008.
Claudio de Jesús Pita Ruiz Velasco, On s-Fibonomials, Journal of Integer Sequences, Vol. 14 (2011), Article 11.3.7.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A000032, A005248.
Second row of array A028412.
Cf. A133247 (prime numbers p such that no odd Fibonacci number is divisible by p).

Table[LucasL[n]^2 - (-1)^n, {n, 0, 30}] (* Amiram Eldar, Feb 02 2022 *)

PARI
```
a(n)=5*fibonacci(n)^2+3*(-1)^n
```

from sympy import fibonacci
def A047946(n): return 5*fibonacci(n)**2+(-3 if n&1 else 3) # Chai Wah Wu, Jul 29 2022

a(n) = F(3n)/F(n), n>0.
a(n) = 2*a(n-1)+2*a(n-2)-a(n-3).
a(n) = 3a(n-1)-a(n-2)+5(-1)^n.
a(n) = A005248(n) + (-1)^n.
G.f.: ( 3-4*x-2*x^2 ) / ( (1+x)*(x^2-3*x+1) ).
for n>0 a(n) = A000045(3n)/A000045(n) - Benoit Cloitre, Aug 30 2003
For n>0, the linear recurrence for the sequence F(n*k)^2 has signature (a(n),a(n),-1) for n odd, and (a(n),-a(n), 1) for n even. For example, the linear recurrence for the sequence F(3*k)^2 has signature (17, 17, -1) (conjectured). - Greg Dresden, Aug 30 2021
a(n) = Lucas(n)^2 - (-1)^n. - Amiram Eldar, Feb 02 2022

Entry improved by comments from Michael Somos.

A104626 Numbers having three 1's in their base-phi representation.

Original entry on oeis.org

4, 5, 6, 8, 19, 48, 124, 323, 844, 2208, 5779, 15128, 39604, 103683, 271444, 710648, 1860499, 4870848, 12752044, 33385283, 87403804, 228826128, 599074579, 1568397608, 4106118244, 10749957123, 28143753124, 73681302248

Offset: 1

Eric W. Weisstein, Mar 17 2005

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000
Jeffrey Shallit, Proving Properties of phi-Representations with the Walnut Theorem-Prover, arXiv:2305.02672 [math.NT], 2023.
Eric Weisstein's World of Mathematics, Phi Number System
Wikipedia, Golden ratio base

Cf. A005248, A104627, A104628, A065034.

[4,5,6] cat [1 + Fibonacci(2*n-3) + Fibonacci(2*n-5): n in [4..50]]; // G. C. Greubel, Aug 13 2018

Join[{4, 5, 6}, Table[LucasL[2*n-4] + 1, {n, 4, 50}]] (* G. C. Greubel, Aug 13 2018 *)

for(n=1,50, print1(if(n==1,4, if(n==2, 5, if(n==3, 6, 1 + fibonacci(2*n-3) + fibonacci(2*n-5)))), ", ")) \\ G. C. Greubel, Aug 13 2018

{n: A055778(n) = 3}. - R. J. Mathar, Sep 05 2010

a(n) = Lucas(2*n-4) + 1, for n>3. - Ralf Stephan, Nov 13 2010

Terms 5 and 6 added by Jaroslav Krizek, May 25 2010

Edited by R. J. Mathar, Sep 05 2010

A104627 Numbers having 4 1's in their base-phi representation.

Original entry on oeis.org

9, 10, 12, 13, 14, 16, 17, 20, 21, 25, 36, 49, 50, 54, 65, 94, 125, 126, 130, 141, 170, 246, 324, 325, 329, 340, 369, 445, 644, 845, 846, 850, 861, 890, 966, 1165, 1686, 2209, 2210, 2214, 2225, 2254, 2330, 2529, 3050, 4414

Offset: 1

Eric W. Weisstein, Mar 17 2005

Amiram Eldar, Table of n, a(n) for n = 1..250
Jeffrey Shallit, Proving Properties of phi-Representations with the Walnut Theorem-Prover, arXiv:2305.02672 [math.NT], 2023.
Eric Weisstein's World of Mathematics, Phi Number System.

Cf. A005248, A104626, A104628.

q[n_] := Plus @@ RealDigits[n, GoldenRatio, 2*Ceiling[ Log[GoldenRatio, n]] ][[1]] == 4; Select[Range[4500], q] (* Amiram Eldar, Jan 20 2022 *)

There is a 24-state automaton accepting exactly the Zeckendorf representation of members of this sequence. - Jeffrey Shallit, May 03 2023

A104628 Numbers having 5 1's in their base-phi representation.

Original entry on oeis.org

11, 15, 22, 23, 24, 26, 30, 31, 32, 34, 35, 37, 41, 42, 43, 45, 46, 51, 52, 53, 55, 66, 83, 95, 112, 127, 128, 129, 131, 142, 171, 217, 247, 293, 326, 327, 328, 330, 341, 370, 446, 568, 645, 767, 847, 848, 849, 851, 862, 891, 967, 1166, 1487, 1687, 2008, 2211

Offset: 1

Eric W. Weisstein, Mar 17 2005

Amiram Eldar, Table of n, a(n) for n = 1..300
Jeffrey Shallit, Proving Properties of phi-Representations with the Walnut Theorem-Prover, arXiv:2305.02672 [math.NT], 2023.
Eric Weisstein's World of Mathematics, Phi Number System.

Cf. A005248, A104626, A104627.

q[n_] := Plus @@ RealDigits[n, GoldenRatio, 2*Ceiling[ Log[GoldenRatio, n]] ][[1]] == 5; Select[Range[2000], q] (* Amiram Eldar, Jan 20 2022 *)

There is a 46-state automaton accepting precisely the Zeckendorf representation of members of this sequence. - Jeffrey Shallit, May 03 2023

cp's OEIS Frontend

A061171 One half of second column of Lucas bisection triangle (odd part).

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A065034 a(n) = Lucas(2*n) + 1.

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A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

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A173027 Numbers of rows R of the Wythoff array such that R is the n-th multiple of a tail of the Fibonacci sequence.

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A198636 One half of total number of round trips, each of length 2n, on the graph P_6 (o-o-o-o-o-o).

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A237132 Values of x in the solutions to x^2 - 3xy + y^2 + 11 = 0, where 0 < x < y.

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A047946 a(n) = 5*F(n)^2 + 3*(-1)^n where F(n) are the Fibonacci numbers A000045.

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A047946 a(n) = 5F(n)^2 + 3(-1)^n where F(n) are the Fibonacci numbers A000045.