A097315 -id:A097315 - cp's OEIS frontend

A195500 Denominators a(n) of Pythagorean approximations b(n)/a(n) to sqrt(2).

3, 228, 308, 5289, 543900, 706180, 1244791, 51146940, 76205040, 114835995824, 106293119818725, 222582887719576, 3520995103197240, 17847666535865852, 18611596834765355, 106620725307595884, 269840171418387336, 357849299891217865

Offset: 1

Clark Kimberling, Sep 20 2011

For each positive real number r, there is a sequence (a(n),b(n),c(n)) of primitive Pythagorean triples such that the limit of b(n)/a(n) is r and
|r-b(n+1)/a(n+1)| < |r-b(n)/a(n)|.  Peter Shiu showed how to find (a(n),b(n)) from the continued fraction for r, and Peter J. C. Moses incorporated Shiu's method in the Mathematica program shown below.
Examples:
r...........a(n)..........b(n)..........c(n)
sqrt(2).....A195500.......A195501.......A195502
sqrt(3).....A195499.......A195503.......A195531
sqrt(5).....A195532.......A195533.......A195534
sqrt(6).....A195535.......A195536.......A195537
sqrt(8).....A195538.......A195539.......A195540
sqrt(12)....A195680.......A195681.......A195682
e...........A195541.......A195542.......A195543
pi..........A195544.......A195545.......A195546
tau.........A195687.......A195688.......A195689
1...........A046727.......A084159.......A001653
2...........A195614.......A195615.......A007805
3...........A195616.......A195617.......A097315
4...........A195619.......A195620.......A078988
5...........A195622.......A195623.......A097727
1/2.........A195547.......A195548.......A195549
3/2.........A195550.......A195551.......A195552
5/2.........A195553.......A195554.......A195555
1/3.........A195556.......A195557.......A195558
2/3.........A195559.......A195560.......A195561
1/4.........A195562.......A195563.......A195564
5/4.........A195565.......A195566.......A195567
7/4.........A195568.......A195569.......A195570
1/5.........A195571.......A195572.......A195573
2/5.........A195574.......A195575.......A195576
3/5.........A195577.......A195578.......A195579
4/5.........A195580.......A195611.......A195612
sqrt(1/2)...A195625.......A195626.......A195627
sqrt(1/3)...{1}+A195503...{0}+A195499...{1}+A195531
sqrt(2/3)...A195631.......A195632.......A195633
sqrt(3/4)...A195634.......A195635.......A195636

			For r=sqrt(2), the first five fractions b(n)/a(n) can be read from the following five primitive Pythagorean triples (a(n), b(n), c(n)) = (A195500, A195501, A195502):
(3,4,5); |r - b(1)/a(1)| = 0.08...
(228,325,397); |r - b(2)/a(2)| = 0.011...
(308,435,533); |r - b(3)/a(3)| = 0.0018...
(5289,7480,9161); |r - b(4)/a(4)| = 0.000042...
(543900,769189,942061); |r - b(5)/a(5)| = 0.0000003...

Ron Knott, Pythagorean Angles
Peter Shiu, The shapes and sizes of Pythagorean triangles, The Mathematical Gazette 67, no. 439 (March 1983) 33-38.

Cf. A195501, A195502.

Shiu := proc(r,n)
        t := r+sqrt(1+r^2) ;
        cf := numtheory[cfrac](t,n+1) ;
        mn := numtheory[nthconver](cf,n) ;
        (mn-1/mn)/2 ;
end proc:
A195500 := proc(n)
        Shiu(sqrt(2),n) ;
        denom(%) ;
end proc: # R. J. Mathar, Sep 21 2011

r = Sqrt[2]; z = 18;
p[{f_, n_}] := (#1[[2]]/#1[[
      1]] &)[({2 #1[[1]] #1[[2]], #1[[1]]^2 - #1[[
         2]]^2} &)[({Numerator[#1], Denominator[#1]} &)[
     Array[FromContinuedFraction[
        ContinuedFraction[(#1 + Sqrt[1 + #1^2] &)[f], #1]] &, {n}]]]];
{a, b} = ({Denominator[#1], Numerator[#1]} &)[
  p[{r, z}]]  (* A195500, A195501 *)
Sqrt[a^2 + b^2] (* A195502 *)

A085478 Triangle read by rows: T(n, k) = binomial(n + k, 2*k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 15, 7, 1, 1, 15, 35, 28, 9, 1, 1, 21, 70, 84, 45, 11, 1, 1, 28, 126, 210, 165, 66, 13, 1, 1, 36, 210, 462, 495, 286, 91, 15, 1, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1

Offset: 0

Philippe Deléham, Aug 14 2003

Coefficient array for Morgan-Voyce polynomial b(n,x). A053122 (unsigned) is the coefficient array for B(n,x). Reversal of A054142. - Paul Barry, Jan 19 2004
This triangle is formed from even-numbered rows of triangle A011973 read in reverse order. - Philippe Deléham, Feb 16 2004
T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k+1 peaks. T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k peaks at height >= 2. T(n,k) is the number of directed column-convex polyominoes of area n+1, having k+1 columns. - Emeric Deutsch, May 31 2004
Riordan array (1/(1-x), x/(1-x)^2). - Paul Barry, May 09 2005
The triangular matrix a(n,k) = (-1)^(n+k)*T(n,k) is the matrix inverse of A039599. - Philippe Deléham, May 26 2005
The n-th row gives absolute values of coefficients of reciprocal of g.f. of bottom-line of n-wave sequence. - Floor van Lamoen (fvlamoen(AT)planet.nl), Sep 24 2006
Unsigned version of A129818. - Philippe Deléham, Oct 25 2007
T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k >=1 (height(alpha) = |Im(alpha)|) and of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
A085478 is jointly generated with A078812 as a triangular array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n>1, u(n,x) = u(n-1,x)+x*v(n-1)x and v(n,x) = u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012
Per Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016
Subtriangle of the triangle given by (0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 26 2012
T(n,k) is also the number of compositions (ordered partitions) of 2*n+1 into 2*k+1 parts which are all odd. Proof: The o.g.f. of column k, x^k/(1-x)^(2*k+1) for k >= 0, is the o.g.f. of the odd-indexed members of the sequence with o.g.f. (x/(1-x^2))^(2*k+1) (bisection, odd part). Thus T(n,k) is obtained from the sum of the multinomial numbers A048996 for the partitions of 2*n+1 into 2*k+1 parts, all of which are odd. E.g., T(3,1) = 3 + 3 from the numbers for the partitions [1,1,5] and [1,3,3], namely 3!/(2!*1!) and 3!/(1!*2!), respectively. The number triangle with the number of these partitions as entries is A152157. - Wolfdieter Lang, Jul 09 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*A039599(n,k). - R. J. Mathar, Mar 12 2013
T(n,k) = A258993(n+1,k) for k = 0..n-1. - Reinhard Zumkeller, Jun 22 2015
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the algebraic function F(x)*G(x)^n about 0, where F(x) = (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) and G(x) = ((1 + sqrt(1 + 4*x))/2)^2. For example, for n = 4, (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) * ((1 + sqrt(1 + 4*x))/2)^8 = (x^4  + 10*x^3 + 15*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 23 2018
Row n also gives the coefficients of the characteristc polynomial of the tridiagonal n X n matrix M_n given in A332602: Phi(n, x) := Det(M_n - x*1_n) = Sum_{k=0..n} T(n, k)*(-x)^k, for n >= 0, with Phi(0, x) := 1. - Wolfdieter Lang, Mar 25 2020
It appears that the largest root of the n-th degree polynomial is equal to the sum of the distinct diagonals of a (2*n+1)-gon including the edge, 1. The largest root of x^3 - 6*x^2 + 5*x - 1 is 5.048917... = the sum of (1 + 1.80193... + 2.24697...). Alternatively, the largest root of the n-th degree polynomial is equal to the square of sigma(2*n+1). Check: 5.048917... is the square of sigma(7), 2.24697.... Given N = 2*n+1, sigma(N) (N odd) can be defined as 1/(2*sin(Pi/(2*N))). Relating to the 9-gon, the largest root of x^4 - 10*x^3 + 15*x^2 - 7*x + 1 is 8.290859..., = the sum of (1 + 1.879385... + 2.532088... + 2.879385...), and is the square of sigma(9), 2.879385... Refer to A231187 for a further clarification of sigma(7). - Gary W. Adamson, Jun 28 2022
For n >=1, the n-th row is given by the coefficients of the minimal polynomial of -4*sin(Pi/(4*n + 2))^2. - Eric W. Weisstein, Jul 12 2023
Denoting this lower triangular array by L, then L * diag(binomial(2*k,k)^2) * transpose(L) is the LDU factorization of A143007, the square array of crystal ball sequences for the A_n X A_n lattices. - Peter Bala, Feb 06 2024
T(n, k) is the number of occurrences of the periodic substring (01)^k in the periodic string (01)^n (see Proposition 4.7 at page 7 in Fang). - Stefano Spezia, Jun 09 2024

			Triangle begins as:
  1;
  1    1;
  1    3    1;
  1    6    5    1;
  1   10   15    7    1;
  1   15   35   28    9    1;
  1   21   70   84   45   11    1;
  1   28  126  210  165   66   13    1;
  1   36  210  462  495  286   91   15    1;
  1   45  330  924 1287 1001  455  120   17    1;
  1   55  495 1716 3003 3003 1820  680  153   19    1;
...
From _Philippe Deléham_, Mar 26 2012: (Start)
(0, 1, 0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) begins:
  1
  0, 1
  0, 1,  1
  0, 1,  3,   1
  0, 1,  6,   5,   1
  0, 1, 10,  15,   7,   1
  0, 1, 15,  35,  28,   9,  1
  0, 1, 21,  70,  84,  45, 11,  1
  0, 1, 28, 126, 210, 165, 66, 13, 1. (End)

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
J.-P. Allouche and M. Mendès France, Stern-Brocot polynomials and power series, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - _N. J. A. Sloane_, May 10 2012
Peter Bala, A 4-parameter family of embedded Riordan arrays
E. Barcucci, A. Del Lungo, S. Fezzi, and R. Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170, 1997, 211-217.
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, J. Integer Sequ., Vol. 8 (2005), Article 05.4.5.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
Paul Barry, The second production matrix of a Riordan array, arXiv:2011.13985 [math.CO], 2020.
Paul Barry and Aoife Hennessy, Notes on a Family of Riordan Arrays and Associated Integer Hankel Transforms , JIS 12 (2009) 09.5.3.
Eduardo H. M. Brietzke, Recurrence Relations for Sums of Binomial Coefficients and Some Generalizations, Journal of Integer Sequences, Vol. 27 (2024), Article 24.3.4.
E. Czabarka et al, Enumerations of peaks and valleys on non-decreasing Dyck paths, Disc. Math. 341 (2018) 2789-2807, Theorem 3.
Emeric Deutsch and H. Prodinger, A bijection between directed column-convex polyominoes and ordered trees of height at most three, Theoretical Comp. Science, 307, 2003, 319-325.
James East and Nicholas Ham, Lattice paths and submonoids of Z^2, arXiv:1811.05735 [math.CO], 2018.
Wenjie Fang, Maximal number of subword occurrences in a word, arXiv:2406.02971 [math.CO], 2024.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62. - _Abdullahi Umar_, Oct 02 2008
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
Yidong Sun, Numerical Triangles and Several Classical Sequences, Fib. Quart. 43, no. 4, (2005) 359-370.
Eric Weisstein's World of Mathematics, Morgan-Voyce Polynomials

Row sums: A001519. Signed versions: A123970, A129818.
Cf. A000108, A001006, A007318, A098158, A183160, A078812, A091042.
Cf. A258993, A025581, A051162, A054142, A332602.
Cf. A231187, A143007.

Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k, 2*k) ))); # G. C. Greubel, Aug 01 2019

a085478 n k = a085478_tabl !! n !! k
a085478_row n = a085478_tabl !! n
a085478_tabl = zipWith (zipWith a007318) a051162_tabl a025581_tabl
-- Reinhard Zumkeller, Jun 22 2015

[Binomial(n+k, 2*k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019

T := (n,k) -> binomial(n+k,2*k): seq(seq(T(n,k), k=0..n), n=0..11);

(* First program *)
u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A085478 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A078812 *) (*Clark Kimberling, Feb 25 2012 *)
(* Second program *)
Table[Binomial[n + k, 2 k], {n, 0, 12}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 01 2019 *)
CoefficientList[Table[Fibonacci[2 n + 1, Sqrt[x]], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Jul 03 2023 *)
Join[{{1}}, CoefficientList[Table[MinimalPolynomial[-4 Sin[Pi/(4 n + 2)]^2, x], {n, 20}], x]] (* Eric W. Weisstein, Jul 12 2023 *)

PARI
```
T(n,k) = binomial(n+k,n-k)
```

[[binomial(n+k,2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

T(n, k) = (n+k)!/((n-k)!*(2*k)!).
G.f.: (1-z)/((1-z)^2-tz). - Emeric Deutsch, May 31 2004
Row sums are A001519 (Fibonacci(2n+1)). Diagonal sums are A011782. Binomial transform of A026729 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
T(n, 0) = 1, T(n, k) = 0 if n=0} T(n-1-j, k-1)*(j+1). T(0, 0) = 1, T(0, k) = 0 if k>0; T(n, k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j). For the column k, g.f.: Sum_{n>=0} T(n, k)*x^n = (x^k) / (1-x)^(2*k+1). - Philippe Deléham, Feb 15 2004
Sum_{k=0..n} T(n,k)*x^(2*k) = A000012(n), A001519(n+1), A001653(n), A078922(n+1), A007805(n), A097835(n), A097315(n), A097838(n), A078988(n), A097841(n), A097727(n), A097843(n), A097730(n), A098244(n), A097733(n), A098247(n), A097736(n), A098250(n), A097739(n), A098253(n), A097742(n), A098256(n), A097767(n), A098259(n), A097770(n), A098262(n), A097773(n), A098292(n), A097776(n) for x=0,1,2,...,27,28 respectively. - Philippe Deléham, Dec 31 2007
T(2*n,n) = A005809(n). - Philippe Deléham, Sep 17 2009
A183160(n) = Sum_{k=0..n} T(n,k)*T(n,n-k). - Paul D. Hanna, Dec 27 2010
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Feb 06 2012
O.g.f. for column k: x^k/(1-x)^(2*k+1), k >= 0. [See the o.g.f. of the triangle above, and a comment on compositions. - Wolfdieter Lang, Jul 09 2012]
E.g.f.: (2/sqrt(x + 4))*sinh((1/2)*t*sqrt(x + 4))*cosh((1/2)*t*sqrt(x)) = t + (1 + x)*t^3/3! + (1 + 3*x + x^2)*t^5/5! + (1 + 6*x + 5*x^2 + x^3)*t^7/7! + .... Cf. A091042. - Peter Bala, Jul 29 2013
T(n, k) = A065941(n+3*k, 4*k) = A108299(n+3*k, 4*k) = A194005(n+3*k, 4*k). - Johannes W. Meijer, Sep 05 2013
Sum_{k=0..n} (-1)^k*T(n,k)*A000108(k) = A000007(n) for n >= 0. - Werner Schulte, Jul 12 2017
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A001006(n) for n >= 0. - Werner Schulte, Jul 12 2017
From  Peter Bala, Jun 26 2025: (Start)
The n-th row polynomial b(n, x) = (-1)^n * U(2*n, (i/2)*sqrt(x)), where U(n,x) is the n-th Chebyshev polynomial of the second kind.
b(n, x) = (-1)^n * Dir(n, -1 - x/2), where Dir(n, x) is the n-th row polynomial of the triangle A244419.
b(n, -1 - x) is the n-th row polynomial of A098493. (End)

A005668 Denominators of continued fraction convergents to sqrt(10).

Original entry on oeis.org

0, 1, 6, 37, 228, 1405, 8658, 53353, 328776, 2026009, 12484830, 76934989, 474094764, 2921503573, 18003116202, 110940200785, 683644320912, 4212806126257, 25960481078454, 159975692596981, 985814636660340, 6074863512559021, 37434995712014466, 230684837784645817

Offset: 0

N. J. A. Sloane, Simon Plouffe, R. K. Guy

a(2*n+1) with b(2*n+1) := A005667(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 10*a^2 = -1, a(2*n) with b(2*n) := A005667(2*n), n>=1, give all (positive integer) solutions to Pell equation b^2 - 10*a^2 = +1 (cf. Emerson reference).
Bisection: a(2*n)= 6*S(n-1,2*19) = 6*A078987(n-1), n >= 0 and a(2*n+1) = A097315(n), n >= 0, with S(n,x) Chebyshev's polynomials of the second kind. S(-1,x)=0. See A049310.
Sqrt(10) = 6/2 + 6/37 + 6/(37*1405) + 6/(1405*53353) + ... - Gary W. Adamson, Dec 21 2007
a(p) == 40^((p-1)/2) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009
For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 6's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,6} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 6-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) 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 6 kinds of squares available. (End)

			G.f. = x + 6*x^2 + 37*x^3 + 228*x^4 + 1405*x^5 + 8658*x^6 + 53353*x^7 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

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, On the Fibonacci k-numbers, Chaos, Solitons & Fractals 2007; 32(5): 1615-24.
Sergio Falcón and Ángel Plaza, The k-Fibonacci sequence and the Pascal 2-triangle Chaos, Solitons & Fractals 2007; 33(1): 38-49.
R. K. Guy, Letter to N. J. A. Sloane, 1987
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 427
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
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for linear recurrences with constant coefficients, signature (6,1).
Index entries for sequences related to Chebyshev polynomials.

Row n=6 of A073133, A172236, A352361.

Cf. A005667, A000045, A000129, A006190, A001076, A052918, A175185 (Pisano periods), A243399.

[n le 2 select n-1 else 6*Self(n-1)+Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 23 2013

evalf(sqrt(10),200); convert(%,confrac,fractionlist); fractionlist;
A005668:=-z/(-1+6*z+z**2); - Simon Plouffe in his 1992 dissertation.
a := n -> `if`(n<2,n,6^(n-1)*hypergeom([1-n/2,(1-n)/2], [1-n], -1/9)):
seq(simplify(a(n)), n=0..23); # Peter Luschny, Jun 28 2017

LinearRecurrence[{6,1}, {0,1}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
a[ n_] := (-I)^(n - 1) ChebyshevU[ n - 1, 3 I]; (* Michael Somos, May 28 2014 *)
a[ n_] := MatrixPower[ {{0, 1}, {1, 6}}, n + 1][[1, 1]]; (* Michael Somos, May 28 2014 *)
Fibonacci[Range[0,30],6] (* G. C. Greubel, Jun 06 2019 *)
Join[{0},Convergents[Sqrt[10],30]//Denominator] (* Harvey P. Dale, Dec 28 2022 *)

{a(n) = ([0, 1; 1, 6]^(n+1)) [1, 1]}; /* Michael Somos, May 28 2014 */

{a(n) = (-I)^(n-1) * polchebyshev( n-1, 2, 3*I)}; /* Michael Somos, May 28 2014 */

from sage.combinat.sloane_functions import recur_gen3; it = recur_gen3(0,1,6,6,1,0); [next(it) for i in range(1,22)] # Zerinvary Lajos, Jul 09 2008

[lucas_number1(n,6,-1) for n in range(0, 21)]# Zerinvary Lajos, Apr 24 2009

G.f.: x / (1 - 6*x - x^2).
a(n) = 6*a(n-1) + a(n-2).
a(n) = ((-i)^(n-1))*S(n-1, 3*i) with S(n, x) Chebyshev's polynomials of the second kind (see A049310) and i^2=-1.
a(n) = F(n, 6), the n-th Fibonacci polynomial evaluated at x=6. - T. D. Noe, Jan 19 2006
From Sergio Falcon, Sep 24 2007: (Start)
a(n) = ((3+sqrt(10))^n - (3-sqrt(10))^n)/(2*sqrt(10)).
a(n) = Sum_{i=0..floor((n-1)/2)} binomial(n-1-i,i)*6^(n-1-2*i). (End)
Sum_{n>=1}(-1)^(n-1)/(a(n)*a(n+1)) = sqrt(10) - 3. - Vladimir Shevelev, Feb 23 2013
a(n) = [M^(n+1)]{0,0}, where M = [0,1; 1,6]. - _L. Edson Jeffery, Aug 28 2013
a(-n) = -(-1)^n * a(n). - Michael Somos, May 28 2014
a(n) = 6^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1/9) for n >= 2. - Peter Luschny, Jun 28 2017
G.f.: x/(1 - 6*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 6 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

Chebyshev comments from Wolfdieter Lang, Jan 21 2003

A238379 Expansion of (1 - x)/(1 - 36*x + x^2).

Original entry on oeis.org

1, 35, 1259, 45289, 1629145, 58603931, 2108112371, 75833441425, 2727895778929, 98128414600019, 3529895029821755, 126978092658983161, 4567681440693572041, 164309553772309610315, 5910576254362452399299, 212616435603275976764449

Offset: 0

Bruno Berselli, Feb 25 2014

First bisection of A041611.

Bruno Berselli, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (36,-1).

Cf. similar sequences with g.f. (1-x)/(1-k*x+x^2): A122367 (k=3), A079935 (k=4), A004253 (k=5), A001653 (k=6), A049685 (k=7), A070997 (k=8), A070998 (k=9), A138288 (k=10), A078922 (k=11), A077417 (k=12), A085260 (k=13), A001570 (k=14), A160682 (k=15), A157456 (k=16), A161595 (k=17). From 18 to 38, even k only, except k=27 and k=31: A007805 (k=18), A075839 (k=20), A157014 (k=22), A159664 (k=24), A153111 (k=26), A097835 (k=27), A159668 (k=28), A157877 (k=30), A111216 (k=31), A159674 (k=32), A077420 (k=34), this sequence (k=36), A097315 (k=38).

[n le 2 select 35^(n-1) else 36*Self(n-1)-Self(n-2): n in [1..20]];

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1 - x)/(1 - 36*x + x^2))); // Marius A. Burtea, Jan 14 2020

CoefficientList[Series[(1 - x)/(1 - 36 x + x^2), {x, 0, 20}], x] (* or *) LinearRecurrence[{36, -1}, {1, 35}, 20]

a(n)=([0,1; -1,36]^n*[1;35])[1,1] \\ Charles R Greathouse IV, May 10 2016

m = 20; L. = PowerSeriesRing(ZZ, m); f = (1-x)/(1-36*x+x^2)
print(f.coefficients())

G.f.: (1 - x)/(1 - 36*x + x^2).
a(n) = a(-n-1) = 36*a(n-1) - a(n-2).
a(n) = ((19-sqrt(323))/38)*(1+(18+sqrt(323))^(2*n+1))/(18+sqrt(323))^n.
a(n+1) - a(n) = 34*A144128(n+1).
323*a(n+1)^2 - ((a(n+2)-a(n))/2)^2 = 34.
Sum_{n>0} 1/(a(n) - 1/a(n)) = 1/34.
See also Tanya Khovanova in Links field:
a(n) = 35*a(n-1) + 34*Sum_{i=0..n-2} a(i).
a(n+2)*a(n) - a(n+1)^2 = 36-2 = 34 = 34*1,
a(n+3)*a(n) - a(n+1)*a(n+2) = 36*(36-2) = 1224 = 34*36.
Generalizing:
a(n+4)*a(n) - a(n+1)*a(n+3) = 44030 = 34*1295,
a(n+5)*a(n) - a(n+1)*a(n+4) = 1583856 = 34*46584,
a(n+6)*a(n) - a(n+1)*a(n+5) = 56974786 = 34*1675729, etc.,
where 1, 36, 1295, 46584, 1675729, ... is the sequence A144128, which is the second bisection of A041611.
a(n)^2 - 36*a(n)*a(n+1) + a(n+1)^2 + 34 = 0 (see comments by Colin Barker in similar sequences).

A078987 Chebyshev U(n,x) polynomial evaluated at x=19.

Original entry on oeis.org

1, 38, 1443, 54796, 2080805, 79015794, 3000519367, 113940720152, 4326746846409, 164302439443390, 6239165952002411, 236924003736648228, 8996872976040630253, 341644249085807301386, 12973484592284636822415, 492650770257730391950384, 18707755785201470257292177

Offset: 0

Wolfdieter Lang, Jan 10 2003

A078986(n+1)^2 - 10*(6*a(n))^2 = +1, n>=0 (Pell equation +1, see A033313 and A033317).

a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,37}. - Milan Janjic, Jan 26 2015

Colin Barker, Table of n, a(n) for n = 0..632
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A097314, A097315.

Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), this sequence (m=19), A097316 (m=33).

m:=19;; a:=[1,2*m];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 22 2019

m:=19; I:=[1, 2*m]; [n le 2 select I[n] else 2*m*Self(n-1) -Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 22 2019

seq( simplify(ChebyshevU(n, 19)), n=0..20); # G. C. Greubel, Dec 22 2019

lst={};Do[AppendTo[lst, GegenbauerC[n, 1, 19]], {n, 0, 8^2}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
ChebyshevU[Range[21] -1, 19] (* G. C. Greubel, Dec 22 2019 *)

a(n)=subst(polchebyshev(n,2),x,19) \\ Charles R Greathouse IV, Feb 10 2012

Vec(1/(1-38*x+x^2) + O(x^50)) \\ Colin Barker, Jun 15 2015

[lucas_number1(n,38,1) for n in range(1, 16)] # Zerinvary Lajos, Nov 07 2009

[chebyshev_U(n,19) for n in (0..20)] # G. C. Greubel, Dec 22 2019

a(n) = 38*a(n-1) - a(n-2), n>=1, a(-1)=0, a(0)=1.
a(n) = S(n, 38) with S(n, x) = U(n, x/2), Chebyshev's polynomials of the second kind. See A049310.
G.f.: 1/(1-38*x+x^2).
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)*38^(n-2*k).
a(n) = ((19+6*sqrt(10))^(n+1) - (19-6*sqrt(10))^(n+1))/(12*sqrt(10)).
a(n) = Sum_{k=0..n} A101950(n,k)*37^k. - Philippe Deléham, Feb 10 2012
Product_{n>=0} (1 + 1/a(n)) = 1/3*(3 + sqrt(10)). - Peter Bala, Dec 23 2012
Product_{n>=1} (1 - 1/a(n)) = 3/19*(3 + sqrt(10)). - Peter Bala, Dec 23 2012
From Andrea Pinos, Jan 02 2023: (Start)
a(n) = (A097314(n+1) - A097315(n+1))/2.
a(n) = (A097314(n) + A097315(n))/2. (End)

A188647 Array read by antidiagonals of a(n) = a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2+1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 5, 1, 29, 17, 1, 169, 305, 37, 1, 985, 5473, 1405, 65, 1, 5741, 98209, 53353, 4289, 101, 1, 33461, 1762289, 2026009, 283009, 10301, 145, 1, 195025, 31622993, 76934989, 18674305, 1050601, 21169, 197, 1, 1136689, 567451585, 2921503573, 1232221121, 107151001, 3090529, 39005, 257, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=1 results shown here; y=-1 results are A188646.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is (1/sqrt(n^2+1)) * T_{2*k+1}(sqrt(n^2+1)), with T the Chebyshev polynomial of the first kind. - Seiichi Manyama, Jan 02 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   5,     29,       169,          985, ...
   2 | 1,  17,    305,      5473,        98209, ...
   3 | 1,  37,   1405,     53353,      2026009, ...
   4 | 1,  65,   4289,    283009,     18674305, ...
   5 | 1, 101,  10301,   1050601,    107151001, ...
   6 | 1, 145,  21169,   3090529,    451196065, ...
   7 | 1, 197,  39005,   7722793,   1529074009, ...
   8 | 1, 257,  66305,  17106433,   4413393409, ...
   9 | 1, 325, 105949,  34539049,  11259624025, ...
  10 | 1, 401, 161201,  64802401,  26050404001, ...
  11 | 1, 485, 235709, 114554089,  55673051545, ...
  12 | 1, 577, 333505, 192765313, 111418017409, ...
  13 | 1, 677, 459005, 311204713, 210996336409, ...
  14 | 1, 785, 617009, 484968289, 381184458145, ...
  15 | 1, 901, 812701, 733055401, 661215159001, ...
  ...

Row 1 is A001653, row 2 is A007805, row 3 is A097315, row 4 is A078988, row 5 is A097727, row 6 is A097730, row 7 is A097733, row 8 is A097736, row 9 is A097739, row 10 is A097742, row 11 is A097767, row 12 is A097770, row 13 is A097773.
Column 1 is A053755.
A(n,n) gives A323012.
Cf. A188645, A188646 (f(x, y) as above with y=-1).

A(n,k) = 2 * A188645(n,k) - A(n,k-1).

A(n,k) = Sum_{j=0..k} binomial(2*k+1,2*j)*(n^2+1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 02 2019

Edited and extended by Seiichi Manyama, Jan 02 2019

A157014 Expansion of x(1-x)/(1 - 22x + x^2).

Original entry on oeis.org

1, 21, 461, 10121, 222201, 4878301, 107100421, 2351330961, 51622180721, 1133336644901, 24881784007101, 546265911511321, 11992968269241961, 263299036011811821, 5780585823990618101, 126909589091781786401, 2786230374195208682721, 61170158643202809233461

Offset: 1

Paul Weisenhorn, Feb 21 2009

This sequence is part of a solution of a general problem involving 2 equations, three sequences a(n), b(n), c(n) and a constant A:
    A    * c(n)+1 = a(n)^2,
   (A+1) * c(n)+1 = b(n)^2, where solutions are given by the recurrences:
a(1) = 1, a(2) = 4*A+1, a(n) = (4*A+2)*a(n-1)-a(n-2) for n>2, resulting in a(n) terms 1, 4*A+1, 16*A^2+12*A+1, 64*A^3+80*A^2+24*A+1, ...;
b(1) = 1, b(2) = 4*A+3, b(n) = (4*A+2)*b(n-1)-b(n-2) for n>2, resulting in b(n) terms 1, 4*A+3, 16*A^2+20*A+5, 64*A^3+112*A^2+56*A+7, ...;
c(1) = 0, c(2) = 16*A+8, c(3) = (16*A^2+16*A+3)*c(2), c(n) = (16*A^2+16*A+3) * (c(n-1)-c(n-2)) + c(n-3) for n>3, resulting in c(n) terms 0, 16*A+8, 256*A^3+384*A^2+176*A+24, 4096*A^5 + 10240*A^4 + 9472*A^3 + 3968*A^2 + 736*A + 48, ... .
A157014 is the a(n) sequence for A=5.
For other A values the a(n), b(n) and c(n) sequences are in the OEIS:
  A   a-sequence    b-sequence     c-sequence
  1     A001653       A002315(n-1)   A078522
  2     A072256       A054320(n-1)   A045502(n-1)
  3     A001570       A028230        A059989(n-1)
  4     A007805       A049629(n-1)   A157459
  5  -> A157014 <-    A133283        A157460
  6     A153111       A157461        A157874
  7     A157877       A157878        A157879
  8     A077420(n-1)  A046176        A157880
  9     A097315(n-1)  A097314(n-1)   A157881
Positive values of x (or y) satisfying x^2 - 22xy + y^2 + 20 = 0. - Colin Barker, Feb 19 2014
From Klaus Purath, Apr 22 2025: (Start)
Nonnegative solutions to the Diophantine equation 5*b(n)^2 - 6*a(n)^2 = -1. The corresponding b(n) are A133283(n). Note that (b(n+1)^2 - b(n)*b(n+2))/4 = 6 and (a(n)*a(n+2) - a(n+1)^2)/4 = 5.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A077421(n-1) = Lucas U(22,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n-1) + t(i))/(t(i+n) + t(i+n-1)) as long as t(i+n) + t(i+n-1) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 21*t(i+2) - 21*t(i+1) + t(i) or t(i+2) = 22*t(i+1) - t(i) without regard to initial values and including this sequence itself. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (22,-1).

Cf. similar sequences listed in A238379.

a:=[1,21];; for n in [3..20] do a[n]:=22*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

I:=[1,21]; [n le 2 select I[n] else 22*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2014

seq( simplify(ChebyshevU(n-1,11) - ChebyshevU(n-2,11)), n=1..20); # G. C. Greubel, Jan 14 2020

CoefficientList[Series[(1-x)/(1-22x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Feb 21 2014 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041049 *)
a[30, 20] (* Gerry Martens, Jun 07 2015 *)
Table[ChebyshevU[n-1, 11] - ChebyshevU[n-2, 11], {n,20}] (* G. C. Greubel, Jan 14 2020 *)

Vec((1-x)/(1-22*x+x^2)+O(x^20)) \\ Charles R Greathouse IV, Sep 23 2012

[chebyshev_U(n-1,11) - chebyshev_U(n-2,11) for n in (1..20)] # G. C. Greubel, Jan 14 2020

G.f.: x*(1-x)/(1-22*x+x^2).
a(1) = 1, a(2) = 21, a(n) = 22*a(n-1) - a(n-2) for n>2.
5*A157460(n)+1 = a(n)^2 for n>=1.
6*A157460(n)+1 = A133283(n)^2 for n>=1.
a(n) = (6+sqrt(30)-(-6+sqrt(30))*(11+2*sqrt(30))^(2*n))/(12*(11+2*sqrt(30))^n). - Gerry Martens, Jun 07 2015
a(n) = ChebyshevU(n-1, 11) - ChebyshevU(n-2, 11). - G. C. Greubel, Jan 14 2020

Edited by Alois P. Heinz, Sep 09 2011

A221874 Numbers m such that 10*m^2 + 6 is a square.

Original entry on oeis.org

1, 5, 43, 191, 1633, 7253, 62011, 275423, 2354785, 10458821, 89419819, 397159775, 3395598337, 15081612629, 128943316987, 572704120127, 4896450447169, 21747674952197, 185936173675435, 825838944063359, 7060678149219361, 31360132199455445

Offset: 1

Bruno Berselli, Jan 28 2013

The Diophantine equation 10*x^2 + k = y^2, for |k| < 10, has integer solutions with the following k values:
k =  1, the nonnegative x values are in A084070;
k = -1,               "                 A097315;
k =  4,               "                 2*A084070;
k = -4,               "                 2*A097315;
k =  6,               "                 this sequence;
k = -6,               "                 A221875;
k =  9,               "                 A075836;
k = -9,               "                 A052454.
a(n+1)/a(n) tends alternately to (sqrt(2)+sqrt(5))^2/3 and (2*sqrt(2)+sqrt(5))^2/3; a(n+2)/a(n) tends to A176398^2.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,38,0,-1).

Cf. A097315, A084070, A075836, A052454, A129556, A221875.

Subsequence of A031150.

m:=22; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+x)*(1+4*x+x^2)/((1-6*x-x^2)*(1+6*x-x^2))));

A221874:=proc(q)
local n;
for n from 1 to q do if type(sqrt(10*n^2+6),integer) then print(n);
fi; od; end:
A221874(100000000000000000); # Paolo P. Lava, Feb 11 2013

LinearRecurrence[{0, 38, 0, -1}, {1, 5, 43, 191}, 22]

makelist(expand(((-5*(-1)^n+2*sqrt(10))*(3+sqrt(10))^(2*floor(n/2))-(5*(-1)^n+2*sqrt(10))*(3-sqrt(10))^(2*floor(n/2)))/10), n, 1, 22);

G.f.: x*(1+x)*(1+4*x+x^2)/((1-6*x-x^2)*(1+6*x-x^2)).
a(n) = ((-5*(-1)^n+2*t)*(3+t)^(2*floor(n/2)) - (5*(-1)^n+2*t)*(3-t)^(2*floor(n/2)))/10, where t=sqrt(10).
a(n) = 2*A129556(n) + 1.
a(n)*a(n-3) - a(n-1)*a(n-2) = -36 + 12(-1)^n.

A247335 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 10/9 divided by a chord of length 4/3.

Original entry on oeis.org

1, 10, 361, 13690, 519841, 19740250, 749609641, 28465426090, 1080936581761, 41047124680810, 1558709801289001, 59189925324301210, 2247658452522156961, 85351831270517663290, 3241121929827149048041, 123077281502161146162250

Offset: 0

Kival Ngaokrajang, Sep 18 2014

Refer to comment of A240926. Consider a circle C of radius 10/9 (in some length units) with a chord of length 4/3. This has been chosen such that the larger sagitta has length 2. The smaller sagitta has length 2/9. The input, besides the circle C is the circle C_0 with radius R_0 = 1, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle, C_n = 1/R_n, n >= 0, is conjectured to be a(n). If one considers the curvature of touching circles inscribed in the smaller segment, the sequence would be A247512. See an illustration given in the link.
a(n) also seems to be A078986(i)^2 and 10*A097315(j)^2 interleaved; where i = n/2 for n even, j = n/2 - 1/2 for n odd; as following:
   1          = 1^2
   10         = 10*1^2
   361        = 19^2
   13690      = 10*37^2
   519841     = 721^2
   19740250   = 10*1405^2
   749609641  = 27379^2
   ...
A078986; Chebyshev... polynomial: 1, 19, 721, 27379, ...
A097315; Pell equation...       : 1, 37, 1405, 53353, ...

Colin Barker, Table of n, a(n) for n = 0..600
Kival Ngaokrajang, Illustration of initial terms
Wolfdieter Lang, Curvature computation for A247335 and A247512.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (39,-39,1).

Cf. A240926, A078986, A097315, A247512.

I:=[39,-39,1]; [n le 3 select I[n] else Self(n-1) - 10*Self(n-2) + 361*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{39,-39,1}, {1, 10, 361}, 50] (* or *) Table[Round[((19 + 6*Sqrt[10])^(-n)*(1 + (19 + 6*Sqrt[10])^n)^2)]/4, {n, 0, 30}] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.9;print1(1,", ");r1=r;
for (n=1,50,
if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
ac=sqrt(ab^2-r^2);
if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
b=acos(r/ab)-z;
r=r*(1-cos(b))/(1+cos(b));
an=floor(9/(10*r));
print1(if(an>9,an,10),", ")
)
}

Vec(-(10*x^2-29*x+1)/((x-1)*(x^2-38*x+1)) + O(x^20)) \\ Colin Barker, Mar 03 2016

Conjectures from Colin Barker, Sep 18 2014: (Start)
a(n) = 39*a(n-1) - 39*a(n-2) + a(n-3).
G.f.: -(10*x^2-29*x+1) / ((x-1)*(x^2-38*x+1)). (End)
From Wolfdieter Lang, Sep 30 2014 (Start)
See the W. Lang link for proofs of the following statements.
One step nonlinear recurrence: a(n) = -9 + 19*a(n-1) + 60*sqrt(a(n-1)*(a(n-1) - 1)/10), n>=1, with a(0) = 1.
a(n) =  (1 + A078986(n))/2  = (2 + S(n, 38) - S(n-2, 38))/4 =
  (1 + S(n, 38) -19*S(n-1, 38))/2 for n>=0, with Chebyshev's S-polynomials (see A049310).  S(n, 38) = A078987(n).
The G.f. conjectured by Colin Barker above follows from the one for Chebyshev's T(n, 19) = A078986(n):  (1/(1-x) + (1-19*x)/(1-38*x+x^2))/2 = (1-29*x+10*x^2)/((1-x)* (1-38*x+x^2)).
The four term recurrence conjectured by Colin Barker above follows from the expanded g.f. denominator: (1-x)* (1-38*x+x^2) = 1- 39*x + 39*x^2 - x^3.
(End)
a(n) = ((19+6*sqrt(10))^(-n)*(1+(19+6*sqrt(10))^n)^2)/4. - Colin Barker, Mar 03 2016

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