A099369 -id:A099369 - cp's OEIS frontend

A041025 Denominators of continued fraction convergents to sqrt(17).

1, 8, 65, 528, 4289, 34840, 283009, 2298912, 18674305, 151693352, 1232221121, 10009462320, 81307919681, 660472819768, 5365090477825, 43581196642368, 354014663616769, 2875698505576520, 23359602708228929, 189752520171407952, 1541379764079492545

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A041024(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 17*a^2 = +1, a(2*n) with b(2*n) := A041024(2*n), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 17*a^2 = -1 (cf. Emerson reference).
Bisection: a(2*n) = T(2*n+1,sqrt(17))/sqrt(17) = A078988(n), n >= 0 and a(2*n+1) = 8*S(n-1,66), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003
Sqrt(17) = 8/2 + 8/65 + 8/(65*4289) + 8/(4289*283009) + ... . - Gary W. Adamson, Dec 26 2007
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 8's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
De Moivre's formula: a(n) = (r^n - s^n)/(r-s), for r > s, gives sequences with integers if r and s are conjugates. With r=4+sqrt(17) and s=4-sqrt(17), a(n+1)/a(n) converges to r=4+sqrt(17). - Sture Sjöstedt, Nov 11 2011
a(n) equals the number of words of length n on alphabet {0,1,...,8} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 21 2023: (Start)
Also called the 8-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 8 kinds of squares available. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle, Chaos, Solitons & Fractals 2007; 33(1): 38-49.
S. Falcón and Á. Plaza, On k-Fibonacci sequences and polynomials and their derivatives, Chaos, Solitons & Fractals (2007).
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,1).

Cf. A041024, A040012.
Cf. A000045, A000129, A006190, A001076, A052918, A005668, A054413, A243399.
Row n=8 of A073133, A172236 and A352361.
Cf. A099369 (squares).

I:=[1, 8]; [n le 2 select I[n] else 8*Self(n-1)+Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 23 2013

CoefficientList[Series[1/(-z^2 - 8 z + 1), {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 23 2011 *)
Denominator[Convergents[Sqrt[17],30]] (* Harvey P. Dale, Aug 15 2011 *)
LinearRecurrence[{8,1}, {1,8}, 50] (* Sture Sjöstedt, Nov 11 2011 *)

Vec(1/(1-8*x-x^2)+O(x^99)) \\ Charles R Greathouse IV, Dec 09 2014

[lucas_number1(n,8,-1) for n in range(1, 20)] # Zerinvary Lajos, Apr 25 2009

G.f.: 1/(1 - 8*x - x^2).
a(n) = ((-i)^n)*S(n, 8*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind and i^2 = -1. See A049310.
a(n) = F(n, 8), the n-th Fibonacci polynomial evaluated at x=8. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = ((4 + sqrt(17))^n - (4 - sqrt(17))^n)/(2*sqrt(17));
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*8^(n-1-2i). (End)
Let T be the 2 X 2 matrix [0, 1; 1, 8]. Then T^n * [1, 0] = [a(n-2), a(n-1)]. - Gary W. Adamson, Dec 26 2007
a(n) = 8*a(n-1) + a(n-2), n > 1; a(0)=1, a(1)=8. - Philippe Deléham, Nov 20 2008
a(p-1) == 68^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009 [Corrected by Jason Yuen, Apr 05 2025. See A087475 for more info about this congruence.]
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = sqrt(17) - 4. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 8*x - x^2) = Sum_{n >= 0} x^n *( Product_{k = 1..n} (m*k + 8 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

A099372 a(n) = A099371(n)^2.

Original entry on oeis.org

0, 1, 81, 6724, 558009, 46308025, 3843008064, 318923361289, 26466795978921, 2196425142889156, 182276820063821025, 15126779640154255921, 1255340433312739420416, 104178129185317217638609, 8645529381948016324584129, 717474760572500037722844100, 59541759598135555114671476169

Offset: 0

Wolfdieter Lang, Oct 18 2004

See the comment in A099279. This is example a=9.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 9 kinds of half-square available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 9 kinds of (1/4,1/4)-fence available. - Michael A. Allen, Mar 21 2024

Stefano Spezia, Table of n, a(n) for n = 0..500
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (82,82,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A099371, A099373.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, A099367, A099369, this sequence, A099374.

LinearRecurrence[{82,82,-1},{0,1,81},17] (* Stefano Spezia, Apr 06 2023 *)

a(n) = A099371(n)^2.
a(n) = 82*a(n-1) + 82*a(n-2) - a(n-3), n>=3; a(0)=0, a(1)=1, a(2)=81.
a(n) = 83*a(n-1) - a(n-2) - 2*(-1)^n, n>=2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 83/2)-(-1)^n)/85 with twice the Chebyshev polynomials of the first kind: 2*T(n, 83/2) = A099373(n).
G.f.: x*(1-x)/((1-83*x+x^2)*(1+x)) = x*(1-x)/(1-82*x-82*x^2+x^3).
E.g.f.: 2*exp(-x)*(exp(85*x/2)*cosh(9*sqrt(85)*x/2) - 1)/85. - Stefano Spezia, Apr 06 2023
a(n) = (1 - (-1)^n)/2 + 81*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Mar 21 2024
Product_{n>=2} (1 + (-1)^n/a(n)) = (9 + sqrt(85))/18 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099279 Squares of A001076.

Original entry on oeis.org

0, 1, 16, 289, 5184, 93025, 1669264, 29953729, 537497856, 9645007681, 173072640400, 3105662519521, 55728852710976, 1000013686278049, 17944517500293904, 322001301319012225, 5778078906241926144, 103683419011035658369, 1860523463292399924496, 33385738920252162982561

Offset: 0

Wolfdieter Lang, Oct 18 2004

For the generalized Fibonacci sequences U(n-1;a) = (ap(a)^n - am(a)^n)/(ap(a) - am(a)) with ap(a) = (a + sqrt(a^2+4))/2, am(a) = (a - sqrt(a^2+4))/2, a from the integers, one has for the squared sequences U(n-1;a)^2 = (2*T(n,(a^2+2)/2) - 2*(-1)^n)/(a^2+4). Here T(n,x) are Chebyshev's polynomials of the first kind (see A053120). Therefore the o.g.f. for the squared sequence is x*(1-x)/((1+x)*(1-(a^2+2)*x+x^2)) = x*(1-x)/(1 - (a^2+1)*x - (a^2+1)*x^2 + x^3). For this example a=4.
Unsigned member r=-16 of the family of Chebyshev sequences S_r(n) defined in A092184.
(-1)^(n+1)*a(n) = S_{-16}(n), n >= 0, defined in A092184.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 4 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 4 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Mar 12 2023

G. C. Greubel, Table of n, a(n) for n = 0..750
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (17,17,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001076, A001654, A023039, A053120, A092184.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, this sequence, A099365, A099366, A099367, A099369, A099372, A099374.

[Fibonacci(3*n)^2/4: n in [0..30]]; // G. C. Greubel, Aug 18 2022

with (combinat):seq(fibonacci(n,4)^2,n=0..16); # Zerinvary Lajos, Apr 09 2008
nmax:=48: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: for n from 1 to nmax/3 do a(n):=a(n-1)+A001654(3*n-2) od: seq(a(n),n=0..nmax/3); # Johannes W. Meijer, Sep 22 2010

LinearRecurrence[{17,17,-1},{0,1,16},30] (* Harvey P. Dale, Mar 26 2012 *)
Fibonacci[3*Range[0, 30]]^2/4 (* G. C. Greubel, Aug 18 2022 *)

numlib::fibonacci(3*n)^2/4 $ n = 0..35; // Zerinvary Lajos, May 13 2008

my(x='x+O('x^99)); concat([0], Vec(x*(1-x)/((1-18*x+x^2)*(1+x)))) \\ Altug Alkan, Dec 17 2017

[(fibonacci(3*n))^2/4 for n in range(0, 17)] # Zerinvary Lajos, May 15 2009

a(n) = A001076(n)^2.
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3), n >= 3, a(0)=0, a(1)=1, a(2)=16.
a(n) = 18*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2, a(0)=0, a(1)=1.
a(n) = (T(n, 9) - (-1)^n)/10 with Chebyshev's T(n, x) polynomials of the first kind. T(n, 9) = A023039(n).
G.f.: x*(1-x)/((1+x)*(1-18*x+x^2)) = x*(1-x)/(1-17*x-17*x^2+x^3).
a(n) = a(n-1) + A001654(3*n-2) with a(0)=0, where A001654 are the golden rectangle numbers. - Johannes W. Meijer, Sep 22 2010
a(n+1) = (1 + (-1)^n)/2 + 16*Sum_{r=1..n} ( r*a(n+1-r) ). - Michael A. Allen, Mar 12 2023
E.g.f.: exp(-x)*(exp(10*x)*cosh(4*sqrt(5)*x) - 1)/10. - Stefano Spezia, Apr 06 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (2 + sqrt(5))/4 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099365 Squares of A052918(n-1) (generalized Fibonacci).

Original entry on oeis.org

0, 1, 25, 676, 18225, 491401, 13249600, 357247801, 9632441025, 259718659876, 7002771375625, 188815108482001, 5091005157638400, 137268324147754801, 3701153746831741225, 99793882840309258276, 2690733682941518232225

Offset: 0

Wolfdieter Lang, Oct 18 2004

See the comment in A099279. This is example a=5.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 5 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 5 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Mar 30 2023

G. C. Greubel, Table of n, a(n) for n = 0..650
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (26,26,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, this sequence, A099366, A099367, A099369, A099372, A099374.

[(2/29)*(Evaluate(ChebyshevFirst(n), 27/2) -(-1)^n): n in [0..30]]; // G. C. Greubel, Aug 21 2022

with (combinat):seq(fibonacci(n,5)^2,n=0..16); # Zerinvary Lajos, Apr 09 2008

LinearRecurrence[{26,26,-1},{0,1,25},30] (* Harvey P. Dale, Sep 25 2019 *)

def A099365(n): return (2/29)*(chebyshev_T(n, 27/2) - (-1)^n)
[A099365(n) for n in (0..30)] # G. C. Greubel, Aug 21 2022

a(n) = A052918(n-1)^2, n >= 1, a(0) = 0.
a(n) = 26*a(n-1) + 26*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=25.
a(n) = 27*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 27/2) - (-1)^n)/29 with twice the Chebyshev's T(n, x) polynomials of the first kind. 2*T(n, 27/2) = A090248(n).
G.f.: x*(1-x)/((1-27*x+x^2)*(1+x)) = x*(1-x)/(1-26*x-26*x^2+x^3).
a(n) = (1 - (-1)^n)/2 + 25*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Mar 30 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (5 + sqrt(29))/10 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099374 a(n) = A041041(n-1)^2, n >= 1.

Original entry on oeis.org

0, 1, 100, 10201, 1040400, 106110601, 10822240900, 1103762461201, 112572948801600, 11481337015302001, 1170983802612002500, 119428866529408953001, 12180573402197101203600

Offset: 0

Wolfdieter Lang, Oct 18 2004

See the comment in A099279. This is example a=10.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 10 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 10 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Mar 21 2024

Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (101,101,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, A099367, A099369, A099372, this sequence.

LinearRecurrence[{101,101,-1},{0,1,100},20] (* Harvey P. Dale, Nov 10 2021 *)

a(n) = A041041(n-1)^2, n >= 1, a(0)=0.
a(n) = 101*a(n-1) + 101*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=100.
a(n) = 102*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = (T(n, 51) - (-1)^n)/52 with the Chebyshev polynomials of the first kind: T(n, 51) = (n).
G.f.: x*(1-x)/((1-102*x+x^2)*(1+x)) = x*(1-x)/(1-101*x-101*x^2+x^3).
a(n) = (1 - (-1)^n)/2 + 100*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Mar 21 2024
Product_{n>=2} (1 + (-1)^n/a(n)) = (5 + sqrt(26))/10 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099367 a(n) = A054413(n-1)^2, n >= 1.

Original entry on oeis.org

0, 1, 49, 2500, 127449, 6497401, 331240000, 16886742601, 860892632649, 43888637522500, 2237459621014849, 114066552034234801, 5815156694124960000, 296458924848338725201, 15113590010571150025249, 770496631614280312562500

Offset: 0

Wolfdieter Lang, Oct 18 2004

See the comment in A099279. This is example a=7.

Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (50,50,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A054413.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, this sequence, A099369, A099372, A099374.

LinearRecurrence[{50,50,-1},{0,1,49},20] (* Harvey P. Dale, Jul 27 2023 *)

a(n) = A054413(n-1)^2, n >= 1. a(0)=0.
a(n) = 50*a(n-1) + 50*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=49.
a(n) = 51*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 51/2) - (-1)^n)/53 with twice the Chebyshev polynomials of the first kind: 2*T(n, 51/2) = A099368(n).
G.f.: x*(1-x)/((1-51*x+x^2)*(1+x)) = x*(1-x)/(1-50*x-50*x^2+x^3).
a(n+1) = (1 + (-1)^n)/2 + 49*Sum_{k=1..n} k*a(n+1-k). - Michael A. Allen, Feb 21 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (7 + sqrt(53))/14 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099366 Squares of A005668.

Original entry on oeis.org

0, 1, 36, 1369, 51984, 1974025, 74960964, 2846542609, 108093658176, 4104712468081, 155870980128900, 5918992532430121, 224765845252215696, 8535183127051766329, 324112192982714904804, 12307728150216114616225

Offset: 0

Wolfdieter Lang, Oct 18 2004

See the comment in A099279. This is example a=6.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 6 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 6 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Apr 21 2023

Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (37,37,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, this sequence, A099367, A099369, A099372, A099374.

with (combinat):seq(fibonacci(n,6)^2,n=0..15); # Zerinvary Lajos, Apr 09 2008

LinearRecurrence[{37,37,-1},{0,1,36},20] (* Harvey P. Dale, Sep 23 2018 *)

a(n) = A005668(n)^2.
a(n) = 37*a(n-1) + 37*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=36.
a(n) = 38*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = (T(n, 19) - (-1)^n)/20 with the Chebyshev polynomials of the first kind: T(n, 19) = A078986(n).
G.f.: x*(1-x)/((1 - 38*x + x^2)*(1+x)) = x*(1-x)/(1 - 37*x - 37*x^2 + x^3).
a(n) = (1 - (-1)^n)/2 + 36*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Apr 21 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (3 + sqrt(10))/6 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099370 Chebyshev polynomial of the first kind, T(n,x), evaluated at x=33.

Original entry on oeis.org

1, 33, 2177, 143649, 9478657, 625447713, 41270070401, 2723199198753, 179689877047297, 11856808685922849, 782369683393860737, 51624542295308885793, 3406437421806992601601, 224773245296966202819873

Offset: 0

Wolfdieter Lang, Oct 18 2004

Used in A099369.
Solutions of the Pell equation x^2 - 17y^2 = 1 (x values). After initial term this sequence bisects A041024. See 8*A097316(n-1) with A097316(-1) = 0 for corresponding y values. a(n+1)/a(n) apparently converges to (4+sqrt(17))^2. (See related comments in A088317, which this sequence also bisects.). - Rick L. Shepherd, Jul 31 2006
From a(n) = T(n, 33) (see the formula section) and the de Moivre-Binet formula for T(n,x=33) follows a(n+1)/a(n) = 33  + 8*sqrt(17), which is the conjectured value (4+sqrt(17))^2 given in the previous comment by Rick L. Shepherd. - Wolfdieter Lang, Jun 28 2013
Also numbers k such that 17*(k-1)*(k+1) is a square. - Bruno Berselli, May 31 2025

			a(1)^2 - 17*A121470(1)^2 = 33^2 - 17*8^2 = 1089 - 1088 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
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
Eric Weisstein's World of Mathematics, Pell Equation
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (66,-1).

Cf. A121470, A041024, A040012.

Row 4 of array A188645.

LinearRecurrence[{66, -1},{1, 33},14] (* Ray Chandler, Aug 11 2015 *)

\\ Program uses fact that continued fraction for sqrt(17) = [4,8,8,...].
print1("1, "); forstep(n=2,40,2,v=vector(n,i,if(i>1,8,4)); print1(contfracpnqn(v)[1,1],", ")) \\ Rick L. Shepherd, Jul 31 2006

vector(20,n,polchebyshev(n-1,1,33)) \\ Joerg Arndt, Jan 01 2021

a(n) = 66*a(n-1) - a(n-2), a(-1):= 33, a(0)=1.
a(n) = T(n, 33) = (S(n, 66)-S(n-2, 66))/2 = S(n, 66)-33*S(n-1, 66) with T(n, x), resp. S(n, x), Chebyshev polynomials of the first, resp.second, kind. See A053120 and A049310. S(n, 66) = A097316(n).
a(n) = ((33+8*sqrt(17))^n + (33-8*sqrt(17))^n)/2.
a(n) = Sum_{k=0..floor(n/2)} ((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*33)^(n-2*k), for n>=1, a(0)=1.
G.f.: (1-33*x)/(1-66*x+x^2).

A-number for y values in Pell equation corrected by Wolfdieter Lang, Jun 28 2013

cp's OEIS Frontend

A041025 Denominators of continued fraction convergents to sqrt(17).

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A099372 a(n) = A099371(n)^2.

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A099279 Squares of A001076.

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A099365 Squares of A052918(n-1) (generalized Fibonacci).

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A099374 a(n) = A041041(n-1)^2, n >= 1.

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A099367 a(n) = A054413(n-1)^2, n >= 1.

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A099366 Squares of A005668.

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