A097781 -id:A097781 - cp's OEIS frontend

A097835 First differences of Chebyshev polynomials S(n,27) = A097781(n) with Diophantine property.

1, 26, 701, 18901, 509626, 13741001, 370497401, 9989688826, 269351100901, 7262490035501, 195817879857626, 5279820266120401, 142359329305393201, 3838422070979496026, 103495036587140999501, 2790527565781827490501

Offset: 0

Wolfdieter Lang, Sep 10 2004

(5*b(n))^2 - 29*a(n)^2 = -4 with b(n)=A097834(n) give all positive solutions of this Pell equation.

			All positive solutions of Pell equation x^2 - 29*y^2 = -4 are (5=5*1,1), (140=5*28,26), (3775=5*755,701), (101785=5*20357,18901), ...

Indranil Ghosh, Table of n, a(n) for n = 0..697
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (27,-1).

Cf. similar sequences listed in A238379.

a:=[1,26];; for n in [3..30] do a[n]:=27*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)/(1-27*x+x^2) )); // G. C. Greubel, Jan 12 2019

LinearRecurrence[{27,-1},{1,26},30] (* Harvey P. Dale, May 31 2013 *)

my(x='x+O('x^30)); Vec((1-x)/(1-27*x+x^2)) \\ G. C. Greubel, Jan 12 2019

((1-x)/(1-27*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 12 2019

a(n) = ((-1)^n)*S(2*n, 5*i) with the imaginary unit i and the S(n, x) = U(n, x/2) Chebyshev polynomials.
G.f.: (1-x)/(1-27*x+x^2).
a(n) = S(n, 27) - S(n-1, 27) = T(2*n+1, sqrt(29)/2)/(sqrt(29)/2), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x) and T(n, x) Chebyshev's polynomials of the first kind, A053120.
a(n) = 27*a(n-1) - a(n-2), a(0)=1, a(1)=26. - Philippe Deléham, Nov 18 2008

A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).

Original entry on oeis.org

1, 5, 26, 135, 701, 3640, 18901, 98145, 509626, 2646275, 13741001, 71351280, 370497401, 1923838285, 9989688826, 51872282415, 269351100901, 1398627786920, 7262490035501, 37711077964425, 195817879857626

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

A087130(n)^2 - 29*a(n-1)^2 = 4*(-1)^n, n >= 1. - Gary W. Adamson, Jul 01 2003, corrected Oct 07 2008, corrected by Jianing Song, Feb 01 2019
a(p-1) == 29^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
For more information about this type of recurrence, follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A015523. - Johannes W. Meijer, Aug 01 2010
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 5's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,5} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 5-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 5 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.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 901
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
Milan Janjić, 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 (5,1).

Row 5 of A073133, A172236, and A352361.

Cf. A087130, A099365 (squares), A100237, A175184 (Pisano periods), A201005 (prime subsequence).

a:=[1,5];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Oct 16 2019

I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 23 2013

R:=PowerSeriesRing(Integers(), 22); Coefficients(R!( 1/(1 - 5*x - x^2) )); // Marius A. Burtea, Oct 16 2019

spec := [S,{S=Sequence(Union(Z,Z,Z,Z,Z,Prod(Z,Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006
with(combinat):a:=n->fibonacci(n,5):seq(a(n),n=1..30); # Zerinvary Lajos, Dec 07 2008

LinearRecurrence[{5, 1}, {1, 5}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
Table[Fibonacci[n+1, 5], {n,0,30}] (* Vladimir Reshetnikov, May 08 2016 *)

Vec(1/(1-5*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011

[lucas_number1(n,5,-1) for n in range(1, 22)] # Zerinvary Lajos, Apr 24 2009

G.f.: 1/(1 - 5*x - x^2).
a(3n) = A041047(5n), a(3n+1) = A041047(5n+3), a(3n+2) = 2*A041047(5n+4). - Henry Bottomley, May 10 2000
a(n) = Sum_{alpha=RootOf(-1+5*z+z^2)} (1/29)*(5+2*alpha)*alpha^(-1-n).
a(n-1) = (((5 + sqrt(29))/2)^n - ((5 - sqrt(29))/2)^n)/sqrt(29). - Gary W. Adamson, Jul 01 2003
a(n) = U(n, 5*i/2)*(-i)^n with i^2 = -1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See triangle A049310.
Let M = {{0, 1}, {1, 5}}, then a(n) is the lower-right term of M^n. - Roger L. Bagula, May 29 2005
a(n) = F(n, 5), the n-th Fibonacci polynomial evaluated at x = 5. - T. D. Noe, Jan 19 2006
a(n) = denominator of n-th convergent to [1, 4, 5, 5, 5, ...], for n > 0. Continued fraction [1, 4, 5, 5, 5, ...] = 0.807417596..., the inradius of a right triangle with legs 2 and 5. n-th convergent = A100237(n)/A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135, 566/701, ... - Gary W. Adamson, Dec 21 2007
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 5*A097781(n), a(2n) = A097835(n).
Limit_{k->oo} a(n+k)/a(k) = (A087130(n) + a(n-1)*sqrt(29))/2.
Limit_{n->oo} A087130(n)/a(n-1) = sqrt(29). (End)
From L. Edson Jeffery, Jan 07 2012: (Start)
Define the 2 X 2 matrix A = {{1, 1}, {5, 4}}. Then:
a(n) is the upper-left term of (1/5)*(A^(n+2) - A^(n+1));
a(n) is the upper-right term of A^(n+1);
a(n) is the lower-left term of (1/5)*A^(n+1);
a(n) is the lower-right term of (Sum_{k=0..n} A^k). (End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(29) - 5)/2. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 5*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 5 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

A090248 a(n) = 27a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 27.

Original entry on oeis.org

2, 27, 727, 19602, 528527, 14250627, 384238402, 10360186227, 279340789727, 7531841136402, 203080369893127, 5475638145978027, 147639149571513602, 3980781400284889227, 107333458658120495527, 2894022602368968490002, 78031276805304028734527

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 24 2004

a(n+1)/a(n) converges to ((27+sqrt(725))/2) = 26.96291201...
Lim a(n)/a(n+1) as n approaches infinity = 0.03708798... = 2/(27+sqrt(725)) = (27-sqrt(725))/2.
Lim a(n+1)/a(n) as n approaches infinity = 26.96291201... = (27+sqrt(725))/2 = 2/(27-sqrt(725)).
Lim a(n)/a(n+1) = 27 - Lim a(n+1)/a(n).
A Chebyshev T-sequence with Diophantine property.
a(n) gives the general (nonnegative integer) solution of the Pell equation a^2 - 29*(5*b)^2 =+4 with companion sequence b(n)=A097781(n-1), n>=0.

			a(4) = 528527 = 27a(3) - a(2) = 27*19602 - 727 = ((27+sqrt(725))/2)^4 + ((27-sqrt(725))/2)^4 = 528526.999998107 + 0.000001892 = 528527.
(x;y) = (2;0), (27;1), (727;27), (19602;728), ... give the nonnegative integer solutions to x^2 - 29*(5*y)^2 = +4.

O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

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 (27, -1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A046213, A078046.
a(n)=sqrt(4 + 29*(5*A097781(n-1))^2), n>=1.
Cf. A077428, A078355 (Pell +4 equations).
Cf. A090733 for 2*T(n, 25/2).
Cf. A087130.

a[0] = 2; a[1] = 27; a[n_] := 27a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
RecurrenceTable[{a[0]==2,a[1]==27,a[n]==27a[n-1]-a[n-2]},a,{n,20}] (* or *) LinearRecurrence[{27,-1},{2,27},20] (* Harvey P. Dale, Jan 03 2018 *)

{a(n) = (-1)^n * subst(2 * poltchebi(2*n), 'x, -5/2 * I)}; /* Michael Somos, Nov 04 2008 */

def aupton(idx):
  alst = [2, 27]
  for n in range(2, idx+1): alst.append(27*alst[-1] - alst[-2])
  return alst
print(aupton(16)) # Michael S. Branicky, Feb 27 2021

[lucas_number2(n,27,1) for n in range(0,16)] # Zerinvary Lajos, Jun 27 2008

a(n) = 27a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 27. a(n) = ((27+sqrt(725))/2)^n + ((27-sqrt(725))/2)^n, (a(n))^2 = a(2n)+2.
a(n) = S(n, 27) - S(n-2, 27) = 2*T(n, 27/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 27)=A097781(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120.
a(n) = ap^n + am^n, with ap := (27+5*sqrt(29))/2 and am := (27-5*sqrt(29))/2.
G.f.: (2-27*x)/(1-27*x+x^2).
a(-n) = a(n). - Michael Somos, Nov 01 2008
A087130(2*n) = a(n). - Michael Somos, Nov 01 2008

More terms from Robert G. Wilson v, Jan 30 2004

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A097834 Chebyshev polynomials S(n,27) + S(n-1,27) with Diophantine property.

Original entry on oeis.org

1, 28, 755, 20357, 548884, 14799511, 399037913, 10759224140, 290100013867, 7821941150269, 210902311043396, 5686540457021423, 153325690028535025, 4134107090313424252, 111467565748433919779, 3005490168117402409781

Offset: 0

Wolfdieter Lang, Sep 10 2004

(5*a(n))^2 - 29*b(n)^2 = -4 with b(n) = A097835(n) give all positive solutions of this Pell equation.
From Klaus Purath, Sep 24 2024: (Start)
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)), where (t) is any sequence satisfying t(i) = 28t(i-1) - 28t(i-2) + t(i-3) or t(i) = 27t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0.
a(n) = (Sum_{i .. i+2n} t(i))/t(i+n), where (t) is any recurrence of the form (27,-1) without regard to initial values and including this sequence itself, as long as t(i+n) != 0 for integer i and n >= 0.
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/28, where (t) is any third order recurrence with constant coefficients (28,-28,1) and initial values t(0) = x, t(1) = x + 1, t(2) = x + 29 for any integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/26, where (t) is any third order recurrence with constant coefficients (26,26,-1) and initial values t(0) = x, t(1) = 1 - x, t(2) = x + 27 for any integer x.
a(n) =  (t(i+4n+2) - t(i))/(t(i+2n+2) - t(i+2n)), where (t) is any recurrence of the form (5,1) without regard to initial values, as long as t(i+2n+2) - t(i+2n) != 0 for nonnegative integer i and n. (End)

			All positive solutions of Pell equation x^2 - 29*y^2 = -4 are
(5=5*1,1), (140=5*28,26), (3775=5*755,701), (101785=5*20357,18901), ...

Indranil Ghosh, Table of n, a(n) for n = 0..697 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (27,-1).
Index entries for sequences related to Chebyshev polynomials.

A087130(2*n + 1) = 5 * a(n). - Michael Somos, Nov 01 2008

Cf. A097781, A097783, A097837, A097840.

a[n_] := -2/5*I*(-1)^n*ChebyshevT[2*n + 1, 5*I/2]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jun 21 2013, from 2nd formula *)

{a(n) = (-1)^n * subst(2 * I / 5 * poltchebi(2*n + 1), 'x, -5/2 * I)}; /* Michael Somos, Nov 04 2008 */

a(n) = S(n, 27) + S(n-1, 27) = S(2*n, sqrt(29)), with S(n, x)=U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x) = 0 = U(-1, x). S(n, 27)=A097781(n).
a(n) = (-2/5)*i*((-1)^n)*T(2*n+1, 5*i/2) with the imaginary unit i and Chebyshev's polynomials of the first kind. See the T-triangle A053120.
G.f.: (1+x)/(1-27*x+x^2).
a(n) = - a(-1-n) for all n in Z. - Michael Somos, Nov 01 2008
From Peter Bala, Aug 26 2022: (Start)
a(n) = (2/5)*(5/2 o 5/2 o ... o 5/2) (2*n+1 terms), where the binary operation o is defined on real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0.
The aerated sequence (b(n))n>=1 = [1, 0, 28, 0, 755, 0, 20357, 0, ...], with o.g.f. x*(1 + x^2)/(1 - 27*x^2 + x^4), is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -25, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials.
b(n) = (1/2)*( (-1)^n - 1 )*F(n,5) + (1/5)*( 1 + (-1)^(n+1) )*F(n+1,5), where F(n,x) is the n-th Fibonacci polynomial - see A168561 (but with row indexing starting at n = 1).
Exp( Sum_{n >= 1} 10*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*x^n.
Exp( Sum_{n >= 1} (-10)*b(n)*x^n/n ) = 1 + Sum_{n >= 1} 10*A052918(n)*(-x)^n.
(End)
a(n) = (a(n-1)*a(n-2) - 783)/a(n-3) for n >= 3. - Klaus Purath, Sep 24 2024

A090251 a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29.

Original entry on oeis.org

2, 29, 839, 24302, 703919, 20389349, 590587202, 17106639509, 495501958559, 14352450158702, 415725552643799, 12041688576511469, 348793243166188802, 10102962363242963789, 292637115290879761079, 8476373381072270107502

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 24 2004

a(n+1)/a(n) converges to ((29+sqrt(837))/2) =28.9654761... Lim a(n)/a(n+1) as n approaches infinity = 0.0345238... =2/(29+sqrt(837)) =(29-sqrt(837))/2. Lim a(n+1)/a(n) as n approaches infinity = 28.9654761... = (29+sqrt(837))/2=2/(29-sqrt(837)). Lim a(n)/a(n+1) = 29 - Lim a(n+1)/a(n).
A Chebyshev T-sequence with a Diophantine property.
a(n) gives the general (nonnegative integer) solution of the Pell equation a^2 - 93*(3*b)^2 =+4 with companion sequence b(n)=A097782(n+1), n>=0.

			a(4) =703919 = 29a(3) - a(2) = 29*24302 - 839= ((29+sqrt(837))/2)^4 + ((29-sqrt(837))/2)^4 = 703918.99999857 + 0.00000142 =703919.
(x,y) = (2;0), (29;1), (839;29), (24302,840), ..., give the
nonnegative integer solutions to x^2 - 93*(3*y)^2 =+4.

O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

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 (29, -1).

Cf. A083148, A007610.
a(n)=sqrt(4 + 93*(3*A097782(n-1))^2), n>=1.
Cf. A077428, A078355 (Pell +4 equations).
Cf. A090248 for 2*T(n, 27/2).

a[0] = 2; a[1] = 29; a[n_] := 29a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{29,-1},{2,29},30] (* Harvey P. Dale, May 28 2013 *)

[lucas_number2(n,29,1) for n in range(0,16)] # Zerinvary Lajos, Jun 27 2008

a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29. a(n) = ((29+sqrt(837))/2)^n + ((29-sqrt(837))/2)^n, (a(n))^2 =a(2n)+2.
a(n) = S(n, 29) - S(n-2, 29) = 2*T(n, 29/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 27)=A097781(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120.
a(n) = ap^n + am^n, with ap := (29+3*sqrt(93))/2 and am := (29-3*sqrt(93))/2.
G.f.: (2-29*x)/(1-29*x+x^2).

More terms from Robert G. Wilson v, Jan 30 2004

Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A097782 Chebyshev polynomials S(n,29) with Diophantine property.

Original entry on oeis.org

1, 29, 840, 24331, 704759, 20413680, 591291961, 17127053189, 496093250520, 14369577211891, 416221645894319, 12056058153723360, 349209464812083121, 10115018421396687149, 292986324755691844200, 8486488399493666794651

Offset: 0

Wolfdieter Lang, Aug 31 2004

All positive integer solutions of Pell equation b(n)^2 - 837*a(n)^2 = +4 together with b(n)=A090251(n+1), n >= 0. Note that D=837=93*3^2 is not squarefree.
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 29'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,...,28}. - Milan Janjic, Jan 26 2015

			(x,y) = (29;1), (839;29), (24302,840), ..., give the positive integer solutions to x^2 - 93*(3*y)^2 =+4.

Indranil Ghosh, Table of n, a(n) for n = 0..682
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (29, -1).

Cf. A097781.

a:=[1,29];; for n in [3..20] do a[n]:=29*a[n-1]-a[n-2]; od; a; # G. C. Greubel, May 25 2019

I:=[1,29]; [n le 2 select I[n] else 29*Self(n-1)-Self(n-2): n in [1..30]]; // G. C. Greubel, May 25 2019

LinearRecurrence[{29,-1},{1,29},20] (* Harvey P. Dale, Dec 14 2011 *)

my(x='x+O('x^30)); Vec(1/(1-29*x+x^2)) \\ G. C. Greubel, May 25 2019

[lucas_number1(n,29,1) for n in range(1,20)] # Zerinvary Lajos, Jun 27 2008

a(n) = S(n, 29) = U(n, 29/2) = S(2*n+1, sqrt(31))/sqrt(31) 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) = 29*a(n-1) - a(n-2), n >= 1; a(0)=1, a(1)=29; a(-1)=0.
a(n) = (ap^(n+1) - am^(n+1))/(ap-am) with ap = (29+3*sqrt(93))/2 and am = (29-3*sqrt(93))/2.
G.f.: 1/(1-29*x+x^2).
a(n) = Sum_{k=0..n} A101950(n,k)*28^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/9*(9 + sqrt(93)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 3/58*(9 + sqrt(93)). - Peter Bala, Dec 23 2012

cp's OEIS Frontend

A097835 First differences of Chebyshev polynomials S(n,27) = A097781(n) with Diophantine property.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Sage

Formula

A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Magma

Maple

Mathematica

PARI

Sage

Formula

A090248 a(n) = 27a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 27.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

Python

Sage

Formula

Extensions

A097834 Chebyshev polynomials S(n,27) + S(n-1,27) with Diophantine property.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A090251 a(n) =29a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 29.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Mathematica

Sage

Formula

Extensions

A097782 Chebyshev polynomials S(n,29) with Diophantine property.

Views

Author

Keywords

Comments

Examples