A018913 -id:A018913 - cp's OEIS frontend

A056918 a(n) = 9*a(n-1)-a(n-2); a(0)=2, a(1)=9.

2, 9, 79, 702, 6239, 55449, 492802, 4379769, 38925119, 345946302, 3074591599, 27325378089, 242853811202, 2158358922729, 19182376493359, 170483029517502, 1515164889164159, 13466000972959929, 119678843867475202

Offset: 0

Barry E. Williams, Aug 21 2000

All nonnegative integer solutions of Pell equation a(n)^2 - 77*b(n)^2 = +4 together with b(n)=A018913(n), n>=0. - Wolfdieter Lang, Aug 31 2004

Except for the first term, positive values of x (or y) satisfying x^2 - 9xy + y^2 + 77 = 0. - Colin Barker, Feb 13 2014

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A018913. a(n)=sqrt(77*A018913(n)^2 + 4). A005248.

a056918 n = a056918_list !! n
a056918_list = 2 : 9 :
   zipWith (-) (map (* 9) $ tail a056918_list) a056918_list
-- Reinhard Zumkeller, Jan 06 2013

a[0] = 2; a[1] = 9; a[n_] := 9a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)

[lucas_number2(n,9,1) for n in range(23)] # Zerinvary Lajos, Jun 25 2008

a(n) = 9*S(n-1, 9) - 2*S(n-2, 9) = S(n, 9) - S(n-2, 9) = 2*T(n, 9/2), with S(n, x) := U(n, x/2) (see A049310), S(-1, x) := 0, S(-2, x) := -1. S(n-1, 9)=A018913(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind.
a(n) = {9*[((9+sqrt(77))/2)^n - ((9-sqrt(77))/2)^n] - 2*[((9+sqrt(77))/2)^(n-1) - ((9-sqrt(77))/2)^(n-1)]}/sqrt(77).
G.f.: (2-9*x)/(1-9*x+x^2).
a(n) = ap^n + am^n, with ap := (9+sqrt(77))/2 and am := (9-sqrt(77))/2.
G.f.: (2-9*x)/(1-9*x+x^2). - Philippe Deléham, Nov 03 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 = 1/2*(9 - sqrt(77)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.11095 50589 89701 91909 ... = 2 + 1/(9 + 1/(79 + 1/(702 + ...))).
Also F(-alpha) = 0.88873 23915 40314 47623 ... has the continued fraction representation 1 - 1/(9 - 1/(79 - 1/(702 - ...))) and the simple continued fraction expansion 1/(1 + 1/((9-2) + 1/(1 + 1/((79-2) + 1/(1 + 1/((702-2) + 1/(1 + ...))))))). F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((9^2-4) + 1/(1 + 1/((79^2-4) + 1/(1 + 1/((702^2-4) + 1/(1 + ...))))))). Cf. A005248.
(End)

More terms from James Sellers, Sep 07 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A078364 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 15, 224, 3345, 49951, 745920, 11138849, 166336815, 2483913376, 37092363825, 553901543999, 8271430796160, 123517560398401, 1844491975179855, 27543862067299424, 411313439034311505

Offset: 0

Wolfdieter Lang, Nov 29 2002

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 221*a^2 = +4 with companion sequence b(n)=A078365(n+1), n>=0.
This is the m=17 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..16 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362 and A007655. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 15's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,14}. - Milan Janjic, Jan 23 2015

Harvey P. Dale, Table of n, a(n) for n = 0..850
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=15, q=-1.
Tanya Khovanova, Recursive Sequences
W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=17.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (15,-1).

a(n) = sqrt((A078365(n+1)^2 - 4)/221), n>=0, (Pell equation d=221, +4).

Cf. A077428, A078355 (Pell +4 equations).

LinearRecurrence[{15,-1},{1,15},30] (* Harvey P. Dale, Oct 16 2011 *)

[lucas_number1(n,15,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

a(n) = 15*a(n-1) - a(n-2), n>= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(17))/sqrt(17) = S(n, 15); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (15+sqrt(221))/2 and am = (15-sqrt(221))/2.
G.f.: 1/(1 - 15*x + x^2). - Philippe Deléham, Nov 17 2008
a(n) = Sum_{k=0..n} A101950(n,k)*14^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/13*(13 + sqrt(221)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/30*(13 + sqrt(221)). - Peter Bala, Dec 23 2012
For n>=1, a(n) = U(n-1,15/2), where U(k,x) is Chebyshev polynomial of the second kind. - Milan Janjic, Jan 23 2015

A078366 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 17, 288, 4879, 82655, 1400256, 23721697, 401868593, 6808044384, 115334885935, 1953885016511, 33100710394752, 560758191694273, 9499788548407889, 160935647131239840, 2726406212682669391, 46187969968474139807, 782469083251377707328, 13255786445304946884769

Offset: 0

Wolfdieter Lang, Nov 29 2002

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 285*a^2 = +4 with companion sequence b(n)=A078367(n+1), n >= 0.
This is the m=19 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..18 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362, A007655, A078364 and A077412. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 17's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n >= 2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,16}. - Milan Janjic, Jan 23 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..800
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=17, q=-1.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=19.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (17,-1).

Cf. A077428, A078355 (Pell +4 equations). Cf. A078367.

a:=[1,17,288];; for n in [4..20] do a[n]:=17*a[n-1]-a[n-2]; od; a; # G. C. Greubel, May 25 2019

I:=[1, 17, 288]; [n le 3 select I[n] else 17*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012

CoefficientList[Series[1/(1 - 17 x + x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Dec 24 2012 *)
LinearRecurrence[{17,-1},{1,17},20] (* Harvey P. Dale, Aug 02 2018 *)

my(x='x+O('x^20)); Vec(1/(1-17*x+x^2)) \\ G. C. Greubel, May 25 2019

[lucas_number1(n,17,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

a(n) = 17*a(n-1) - a(n-2), n >= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(19))/sqrt(19) = S(n, 17), where S(n, x) = U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (17+sqrt(285))/2 and am = (17-sqrt(285))/2.
G.f.: 1/(1-17*x+x^2).
a(n) = Sum_{k=0..n} A101950(n,k)*16^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = (1/15)*(15 + sqrt(285)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = (1/34)*(15 + sqrt(285)). - Peter Bala, Dec 23 2012
For n >= 1, a(n) = U(n-1,13/2), where U(k,x) represents Chebyshev polynomial of the second order. - Milan Janjic, Jan 23 2015
a(n) = sqrt((A078367(n+1)^2 - 4)/285), n>=0, (Pell equation d=285, +4).
E.g.f.: exp(17*x/2)*(sqrt(285)*cosh(sqrt(285)*x/2) + 17*sinh(sqrt(285)*x/2))/sqrt(285). - Stefano Spezia, Aug 19 2023

A078368 A Chebyshev S-sequence with Diophantine property.

Original entry on oeis.org

1, 19, 360, 6821, 129239, 2448720, 46396441, 879083659, 16656193080, 315588584861, 5979526919279, 113295422881440, 2146633507828081, 40672741225852099, 770635449783361800, 14601400804658022101

Offset: 0

Wolfdieter Lang, Nov 29 2002

a(n) gives the general (positive integer) solution of the Pell equation b^2 - 357*a^2 =+4 with companion sequence b(n)=A078369(n+1), n>=0.
This is the m=21 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..20 (nonnegative) sequences are: A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190, A004191, A078362, A007655, A078364, A077412, A078366 and A049660. The m=1..3 (signed) sequences are A049347, A056594, A010892.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 19's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=2, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,18}. Milan Janjic, Jan 25 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..200
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=19, q=-1.
Tanya Khovanova, Recursive Sequences
W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=21.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (19,-1).

a(n) = sqrt((A078369(n+1)^2 - 4)/357), n>=0, (Pell equation d=357, +4).

Cf. A077428, A078355 (Pell +4 equations).

Join[{a=1,b=19},Table[c=19*b-a;a=b;b=c,{n,40}]] (* Vladimir Joseph Stephan Orlovsky, Feb 14 2011 *)
LinearRecurrence[{19,-1},{1,19},20] (* Harvey P. Dale, Feb 10 2019 *)

[lucas_number1(n,19,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

a(n) = 19*a(n-1)-a(n-2), n >= 1; a(-1)=0, a(0)=1.
a(n) = (ap^(n+1)-am^(n+1))/(ap-am) with ap = (19+sqrt(357))/2 and am = (19-sqrt(357))/2.
a(n) = S(2*n+1, sqrt(21))/sqrt(21) = S(n, 19); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
G.f.: 1/(1-19*x+x^2).
a(n) = Sum_{k=0..n} A101950(n,k)*18^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/17*(17 + sqrt(357)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/38*(17 + sqrt(357)). - Peter Bala, Dec 23 2012

A179943 Triangle read by rows, antidiagonals of an array (r,k), r=(0,1,2,...), generated from 2 X 2 matrices of the form [1,r; 1,(r+1)].

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 8, 4, 1, 5, 15, 21, 5, 1, 6, 24, 56, 55, 6, 1, 7, 35, 115, 209, 144, 7, 1, 8, 48, 204, 551, 780, 377, 8, 1, 9, 63, 329, 1189, 2640, 2911, 987, 9, 1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10, 1, 11, 99, 711, 3905, 15456, 40391, 60605, 40545, 6765, 11

Offset: 0

Gary W. Adamson, Aug 07 2010

Row sums = A179944: (1, 3, 7, 17, 47, 148, 518,...)
Row 1 = A001906, row 2 = A001353, row 3 = A004254, row 4 = A001109, row 5 = A004187, row 6 = A001090, row 7 = A018913, row 9 = A004189.
Let S_m(x) be the m-th Chebyshev S-polynomial, described by Wolfdieter Lang in his draft [Lang], defined by S_0(x)=1, S_1(x)=x and S_m(x)=x*S_{m-1}(x)-S_{m-2}(x) (m>1). Let A = (A(r,c)) denote the rectangular array (not the triangle). Then A(r,c) = S_c(r+2), r,c=0,1,2,.... - L. Edson Jeffery, Aug 14 2011
As to the array, (n+1)-th row is the INVERT transform of n-th row. - Gary W. Adamson, Jun 30 2013
If the array sequences are labeled (2,3,4,...) for the n-th sequence, convergence tends to (n + sqrt(n^2 - 4))/2. - Gary W. Adamson, Aug 20 2013

			First few rows of the array:
  1, 2,  3,   4,    5,    6,     7,...
  1, 3,  8,  21,   55,  144,   377,...
  1, 4, 15,  56,  209,  780,  2911,...
  1, 5, 24, 115,  551, 2640, 12649,...
  1, 6, 35, 204, 1189, 6930, 40391,...
Taking antidiagonals, we obtain triangle A179943:
  1;
  1, 2;
  1, 3, 3;
  1, 4, 8, 4;
  1, 5, 15, 21, 5;
  1, 6, 24, 56, 55, 6;
  1, 7, 35, 115, 209, 144, 7;
  1, 8, 48, 204, 551, 780, 377, 8;
  1, 9, 63, 329, 1189, 2640, 2911, 987, 9;
  1, 10, 80, 496, 2255, 6930, 12649, 10864, 2584, 10;
  1, 11, 99, 711, 3905, 15456, 40391, 60605, 40545, 6765, 11;
  1, 12, 120, 980, 6319, 30744, 105937, 235416, 290376, 151316, 17711, 12;
  ...
Examples: Row 1 of the array: (1, 3, 8, 21, 55, 144,...); 144 = term (1,5) of the array = term (2,1) of M^6; where M = the 2 X 2 matrix [1,1; 1,2] and M^6 = [89,144; 144,233].
Term (1,5) of the array = 144 = (r+2)*(term (1,4)) - (term (1,3)) = 3*55 - 21.

Alois P. Heinz, Rows n = 0..140, flattened
Robert G. Donnelly, Molly W. Dunkum, Sasha V. Malone, and Alexandra Nance, Symmetric Fibonaccian distributive lattices and representations of the special linear Lie algebras, arXiv:2012.14991 [math.CO], 2020.
Wolfdieter Lang, Chebyshev S-polynomials: ten applications.

Cf. A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.

invtr:= proc(b) local a;
          a:= proc(n) option remember; local i;
          `if`(n<1, 1, add(a(n-i) *b(i-1), i=1..n+1)) end
        end:
A:= proc(n) A(n):= `if`(n=0, k->k+1, invtr(A(n-1))) end:
seq(seq(A(d-k)(k), k=0..d), d=0..10);  # Alois P. Heinz, Jul 17 2013
# using observation by Gary W. Adamson

a[, 0] = 0; a[, 1] = 1; a[r_, k_] := a[r, k] = (r+1)*a[r, k-1] - a[r, k-2]; Table[a[r-k+2, k], {r, 0, 10}, {k, 1, r+1}] // Flatten (* Jean-François Alcover, Feb 23 2015 *)

Antidiagonals of an array, (r,k), a(k) = (r+2)*a(k-1) - a*(k-2), r=0,1,2,... where (r,k) = term (2,1) in the 2 X 2 matrix [1,r; 1,r+1]^(k+1).

G.f. for row r of array: 1/(1 - (r+2)*x + x^2). - L. Edson Jeffery, Oct 26 2012

A073134 Table by antidiagonals of T(n,k)=n*T(n,k-1)-T(n,k-2) starting with T(n,1)=1.

Original entry on oeis.org

1, 1, 1, 0, 2, 1, -1, 3, 3, 1, -1, 4, 8, 4, 1, 0, 5, 21, 15, 5, 1, 1, 6, 55, 56, 24, 6, 1, 1, 7, 144, 209, 115, 35, 7, 1, 0, 8, 377, 780, 551, 204, 48, 8, 1, -1, 9, 987, 2911, 2640, 1189, 329, 63, 9, 1, -1, 10, 2584, 10864, 12649, 6930, 2255, 496, 80, 10, 1, 0, 11, 6765, 40545, 60605, 40391, 15456, 3905, 711, 99, 11, 1, 1, 12

Offset: 1

Henry Bottomley, Jul 16 2002

			Rows start:
  1, 1,  0, -1,  -1,   0,    1, ...;
  1, 2,  3,  4,   5,   6,    7, ...;
  1, 3,  8, 21,  55, 144,  377, ...;
  1, 4, 15, 56, 209, 780, 2911, ...;
  ...

Shmuel T. Klein, Combinatorial Representation of Generalized Fibonacci Numbers, Fib. Quarterly 29 (2) (1991) 124-131, variable U_n^m. [From _R. J. Mathar_, Feb 19 2010]

Rows include A010892, A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189, A004190. Columns include (with some gaps) A000012, A000027, A005563, A057722.

Cf. A094954.

T(n,k) = sum(j=0,k-1,A049310(k-1,j)*n^j) \\ Jason Yuen, Aug 20 2024

T(n, k) = A073133(n, k)-2*A073135(n, k-2).

T(n, k) = Sum_{j=0..k-1} A049310(k-1, j)*n^j.

A092499 Chebyshev polynomials S(n-1,21) with Diophantine property.

Original entry on oeis.org

0, 1, 21, 440, 9219, 193159, 4047120, 84796361, 1776676461, 37225409320, 779956919259, 16341869895119, 342399310878240, 7174043658547921, 150312517518628101, 3149388824232642200, 65986852791366858099

Offset: 0

Rainer Rosenthal, Apr 05 2004

Sequence R_21: Starts with 0,1,21 and satisfies A*C=B^2-1 for successive A,B,C.
The natural numbers a(n)=n satisfy the recurrence a(n-1)*a(n+1)=a(n)^2-1. Let R_r denote the sequence starting with 0,1,r and with this recurrence. We see that R_2 = "the natural numbers" and we find R_3 = A001906. These R_r form a "family" of sequences, which coincides with the m-family (r=m-2, n -> n+1) provided by Wolfdieter Lang (see A078368). This sequence R_21 is strongly related to A041833, which gives the denominators in the continued fraction of sqrt(437).
All positive integer solutions of Pell equation b(n)^2 - 437*a(n)^2 = +4 together with b(n)=A097777(n), n>=0.
For n>=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 21's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,20}. - Milan Janjic, Jan 25 2015

			a(3)=440 because a(1)*440 = a(2)^2-1.

Indranil Ghosh, Table of n, a(n) for n = 0..756
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (21,-1).

Cf. R_3=A001906, R_4=A001353, R_5=A004254, R_6=A001109, R_7=A004187, R_8=A001090, R_9=A018913, R_10=A004189, R_11=A004190, R_12=A004191, R_13=A078362, R_14=A007655, R_15=A078364, R_16=A077412, R_17=A078366, R_18=A049660, R_19=A078368, R_20=A075843, R_21=this, sequence, R_22=A077421. See also A041219 and A041917.

LinearRecurrence[{21,-1},{0,1},30] (* Harvey P. Dale, Apr 23 2015 *)

[lucas_number1(n,21,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008

a(0)=0, a(1)=1, a(2)=21 and a(n-1)*a(n+1) = a(n)^2-1
a(n) = S(n-1, 21)=U(n-1, 21/2) with S(n, x)=U(n, x/2) Chebyshev's polynomials of the 2nd kind, A049310. S(-1, x)= 0 = U(-1, x).
a(n) = S(2*n-1, sqrt(23))/sqrt(23), n>=1.
a(n) = 21*a(n-1)-a(n-2), n >= 1; a(0)=0, a(1)=1.
a(n) = (ap^n-am^n)/(ap-am) with ap := (21+sqrt(437))/2 and am := (21-sqrt(437))/2.
G.f.: x/(1-21*x+x^2).
a(n+1) = Sum_{k, 0<=k<=n} A101950(n,k)*20^k. - Philippe Deléham, Feb 10 2012
Product {n >= 1} (1 + 1/a(n)) = 1/19*(19 + sqrt(437)). - Peter Bala, Dec 23 2012
Product {n >= 2} (1 - 1/a(n)) = 1/42*(19 + sqrt(437)). - Peter Bala, Dec 23 2012

Extension, Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

Corrected by T. D. Noe, Nov 07 2006

A221367 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)*(9 - sqrt(77)).

Original entry on oeis.org

1, 7, 1, 77, 1, 700, 1, 6237, 1, 55447, 1, 492800, 1, 4379767, 1, 38925117, 1, 345946300, 1, 3074591597, 1, 27325378087, 1, 242853811200, 1, 2158358922727, 1, 19182376493357, 1, 170483029517500, 1, 1515164889164157, 1, 13466000972959927, 1

Offset: 0

Peter Bala, Jan 15 2013

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 9. See also A221364 (N = 3), A221365 (N = 5) and A221366 (N = 7).

If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(9 - sqrt(77)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

			F(1/2*(9 - sqrt(77))) = 1.12519 81018 34502 81936 ... = 1 + 1/(7 + 1/(1 + 1/(77 + 1/(1 + 1/(700 + 1/(1 + 1/(6237 + ...))))))).
F((1/2*(9 - sqrt(77)))^2) = 1.01282 05391 65421 74656 ... = 1 + 1/(77 + 1/(1 + 1/(6237 + 1/(1 + 1/(492800 + 1/(1 + 1/(38925117 + ...))))))).
F((1/2*(9 - sqrt(77)))^3) = 1.00142 65335 27667 24640 ... = 1 + 1/(700 + 1/(1 + 1/(492800 + 1/(1 + 1/(345946300 + 1/(1 + 1/(242853811200 + ...))))))).

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,10,0,-10,0,1).

Cf. A018193, A057081, A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221366 (N = 7).

a(2*n-1) = (1/2*(9 + sqrt(77)))^n + (1/2*(9 - sqrt(77)))^n - 2; a(2*n) = 1.
a(4*n-1) = 77*(A018913(n))^2; a(4*n+1) = 7*(A057081(n))^2.
a(n) = 10*a(n-2)-10*a(n-4)+a(n-6). G.f.: -(x^4+7*x^3-9*x^2+7*x+1) / ((x-1)*(x+1)*(x^4-9*x^2+1)). - Colin Barker, Jan 20 2013

A291034 p-INVERT of the positive integers, where p(S) = 1 - 7*S.

Original entry on oeis.org

7, 63, 560, 4977, 44233, 393120, 3493847, 31051503, 275969680, 2452675617, 21798110873, 193730322240, 1721774789287, 15302242781343, 135998410242800, 1208683449403857, 10742152634391913, 95470690260123360, 848494059706718327, 7540975847100341583

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Clark Kimberling, Aug 19 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A290890 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9, -1)

Cf. A000027, A290890.

z = 60; s = x/(1 - x)^2; p = 1 - 7 s;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291034 *)

G.f.: 7/(1 - 9 x + x^2).
a(n) = 9*a(n-1) - a(n-2).
a(n) = 7*A018913(n) for n >= 1.

cp's OEIS Frontend

A056918 a(n) = 9*a(n-1)-a(n-2); a(0)=2, a(1)=9.

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A078364 A Chebyshev S-sequence with Diophantine property.

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A078366 A Chebyshev S-sequence with Diophantine property.

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A078368 A Chebyshev S-sequence with Diophantine property.

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A179943 Triangle read by rows, antidiagonals of an array (r,k), r=(0,1,2,...), generated from 2 X 2 matrices of the form [1,r; 1,(r+1)].

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A073134 Table by antidiagonals of T(n,k)=n*T(n,k-1)-T(n,k-2) starting with T(n,1)=1.

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A092499 Chebyshev polynomials S(n-1,21) with Diophantine property.

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A221367 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4n+3))/(1 - x^(4n+1)) when x = (1/2)*(9 - sqrt(77)).