A001091 -id:A001091 - cp's OEIS frontend

A001090 a(n) = 8*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

0, 1, 8, 63, 496, 3905, 30744, 242047, 1905632, 15003009, 118118440, 929944511, 7321437648, 57641556673, 453811015736, 3572846569215, 28128961537984, 221458845734657, 1743541804339272, 13726875588979519, 108071462907496880, 850844827670995521, 6698687158460467288

Offset: 0

N. J. A. Sloane

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 15*y^2 = 1; the corresponding values of x are in A001091. - Vincenzo Librandi, Nov 12 2010 [edited by Jon E. Schoenfield, May 02 2014]
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 8'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,...,7}. - Milan Janjic, Jan 25 2015
From Klaus Purath, Jul 25 2024: (Start)
For any three consecutive terms (x, y, z) y^2 - x*z = 1 always applies.
a(n) = (t(i+2n) - t(i))/(t(i+n+1) - t(i+n-1)) where (t) is any recurrence t(k) = 9t(k-1) - 9t(k-2) + t(k-3) or t(k) = 8t(k-1) - t(k-2) without regard to initial values.
In particular, if the recurrence (t) of the form (9,-9,1) has the initial values t(0) = 1, t(1) = 2, t(2) = 9, a(n) = t(n) - 1 applies. (End)

			G.f. = x + 8*x^2 + 63*x^3 + 496*x^4 + 3905*x^5 + 30744*x^6 + 242047*x^7 + ...

Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..100 from T. D. Noe)
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
H. Brocard, Notes élémentaires sur le problème de Peel, Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
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., 5.5 (1967), 424-434. Case a=0,b=1; p=8, q=-1.
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
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq.(44), lhs, m=10.
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.
Shobhna Singh and Felix Flicker, Exact Solution to the Quantum and Classical Dimer Models on the Spectre Aperiodic Monotiling, arXiv:2309.14447 [cond-mat.str-el], 2023.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A001091, A001906, A004187, A004254, A049310, A070997.
Equals one-third A136325.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), this sequence (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), A078987 (m=19), A097316 (m=33).
Cf. A323182.

m:=4;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

I:=[0,1]; [n le 2 select I[n] else 8*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 20 2017

A001090:=1/(1-8*z+z**2); # Simon Plouffe in his 1992 dissertation
seq( simplify(ChebyshevU(n-1, 4)), n=0..20); # G. C. Greubel, Dec 23 2019

Table[GegenbauerC[n-1, 1, 4], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
LinearRecurrence[{8,-1},{0,1},30] (* Harvey P. Dale, Aug 29 2012 *)
a[n_]:= ChebyshevU[n-1, 4]; (* Michael Somos, May 28 2014 *)
CoefficientList[Series[x/(1-8*x+x^2), {x,0,20}], x] (* G. C. Greubel, Dec 20 2017 *)

{a(n) = subst(poltchebi(n+1) - 4 * poltchebi(n), x, 4) / 15}; /* Michael Somos, Apr 05 2008 */

{a(n) = polchebyshev(n-1, 2, 4)}; /* Michael Somos, May 28 2014 */

my(x='x+O('x^30)); concat([0], Vec(x/(1-8*x-x^2))) \\ G. C. Greubel, Dec 20 2017

[lucas_number1(n,8,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,4) for n in (0..20)] # G. C. Greubel, Dec 23 2019

15*a(n)^2 - A001091(n)^2 = -1.
a(n) = sqrt((A001091(n)^2 - 1)/15).
a(n) = S(2*n-1, sqrt(10))/sqrt(10) = S(n-1, 8); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310, with S(-1, x) := 0.
From Barry E. Williams, Aug 18 2000: (Start)
a(n) = ((4+sqrt(15))^n - (4-sqrt(15))^n)/(2*sqrt(15)).
G.f.: x/(1-8*x+x^2). (End)
Limit_{n->infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 13 2002
[A070997(n-1), a(n)] = [1,6; 1,7]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(-n) = -a(n). - Michael Somos, Apr 05 2008
a(n+1) = Sum_{k=0..n} A101950(n,k)*7^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = (1/3)*(3 + sqrt(15)).
Product_{n >= 2} (1 - 1/a(n)) = (1/8)*(3 + sqrt(15)).
(End)
a(n) = A136325(n)/3. - Greg Dresden, Sep 12 2019
E.g.f.: exp(4*x)*sinh(sqrt(15)*x)/sqrt(15). - Stefano Spezia, Dec 12 2022
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*6^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*10^k. - Peter Bala, Jul 17 2023

More terms from Wolfdieter Lang, Aug 02 2000

A070997 a(n) = 8*a(n-1) - a(n-2), a(0)=1, a(-1)=1.

Original entry on oeis.org

1, 7, 55, 433, 3409, 26839, 211303, 1663585, 13097377, 103115431, 811826071, 6391493137, 50320119025, 396169459063, 3119035553479, 24556114968769, 193329884196673, 1522082958604615, 11983333784640247, 94344587318517361

Offset: 0

Joe Keane (jgk(AT)jgk.org), May 18 2002

A Pellian sequence.
In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,8), where L is defined as in A108299; see also A057080 for L(n,-8). - Reinhard Zumkeller, Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7} which do not end in 0. - Tanya Khovanova, Jan 10 2007
Hankel transform of A158197. - Paul Barry, Mar 13 2009
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(6)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Values of x (or y) in the solutions to x^2 - 8xy + y^2 + 6 = 0. - Colin Barker, Feb 05 2014
From Klaus Purath, May 06 2025: (Start)
Nonnegative solutions to the Diophantine equation 3*b(n)^2 - 5*a(n)^2 = -2. The corresponding b(n) are A057080(n). Note that (b(n)*b(n+2) - b(n+1)^2)/2 = -5 and (a(n)*a(n+2) - a(n+1)^2)/2 = 3.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A001090(n+1) = Lucas U(8,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n+1) + t(i))/(t(i+n+1) + t(i+n)) as long as t(i+n+1) + t(i+n) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 7*t(i+2) - 7*t(i+1) + t(i) or t(i+2) = 8*t(i+1) - t(i) regardless of initial values and including this sequence itself. (End)

			1 + 7*x + 55*x^2 + 433*x^3 + 3409*x^4 + 26839*x^5 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Fink, R. K. Guy and M. Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

a(n) = sqrt((3*A057080(n)^2+2)/5) (cf. Richardson comment).
Cf. A057080, A001090, A001091.
Row 8 of array A094954.
Cf. A001090.
Cf. similar sequences listed in A238379.
Cf. A041023.

I:=[1, 7]; [n le 2 select I[n] else 8*Self(n-1) - Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 26 2013

CoefficientList[Series[(1 - x)/(1 - 8*x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jan 26 2013 *)
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 A041023 *)
a[15, 20] (* Gerry Martens, Jun 07 2015 *)
LinearRecurrence[{8,-1},{1,7},20] (* Harvey P. Dale, Dec 04 2021 *)

{a(n) = subst( 9*poltchebi(n) - poltchebi(n-1), x, 4) / 5} /* Michael Somos, Jun 07 2005 */

{a(n) = if( n<0, n=-1-n); polcoeff( (1 - x) / (1 - 8*x + x^2) + x * O(x^n), n)} /* Michael Somos, Jun 07 2005 */

[lucas_number1(n,8,1)-lucas_number1(n-1,8,1) for n in range(1, 21)] # Zerinvary Lajos, Nov 10 2009

For all members x of the sequence, 15*x^2 - 6 is a square. Lim_{n->infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 12 2002
a(n) = (5+sqrt(15))/10 * (4+sqrt(15))^n + (5-sqrt(15))/10 * (4-sqrt(15))^n.
a(n) ~ 1/10*sqrt(10)*(1/2*(sqrt(10)+sqrt(6)))^(2*n+1)
a(n) = U(n, 4)-U(n-1, 4) = T(2*n+1, sqrt(5/2))/sqrt(5/2), with Chebyshev's U and T polynomials and U(-1, x) := 0. U(n, 4)=A001090(n+1), n>=-1.
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 6) = a(n) - Benoit Cloitre, Nov 10 2002
a(n)*a(n+3) = 48 + a(n+1)a(n+2). - Ralf Stephan, May 29 2004
a(n) = (-1)^n*U(2n, i*sqrt(6)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
G.f.: (1-x)/(1-8*x+x^2).
a(n) = a(-1-n).
a(n) = Jacobi_P(n,-1/2,1/2,4)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006
[a(n), A001090(n+1)] = [1,6; 1,7]^(n+1) * [1,0]. - Gary W. Adamson, Mar 21 2008
For n>0, a(n) is the numerator of the continued fraction [2,3,2,3,...,2,3] with n repetitions of 2,3. For the denominators see A136325. - Greg Dresden, Sep 12 2019
From Peter Bala, Apr 30 2025: (Start)
a(n) = (1/sqrt(5)) * sqrt(1 - T(2*n+1, -4)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(n) divides a(3*n+1); a(n) divides a(5*n+2); in general, for k >= 0, a(n) divides a((2*k+1)*n + k).
The aerated sequence [b(n)]n>=1 = [1, 0, 7, 0, 55, 0, 433, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -10, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/6 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A291033(n-1) - 1/A291033(n).) (End)
In addition to the first formula above: In general, the following applies to all recurrences (a(n)) of the form (8,-1) with a(0) = 1 and arbitrary a(1): 15*a(n)^2 + y = b^2 where y = x^2 + 8*x + 1 and x = a(1) - 8. Also y = a(k+1)^2 - a(k)*a(k+1) for any k >=0. - Klaus Purath, May 06 2025

A008310 Triangle of coefficients of Chebyshev polynomials T_n(x).

Original entry on oeis.org

1, 1, -1, 2, -3, 4, 1, -8, 8, 5, -20, 16, -1, 18, -48, 32, -7, 56, -112, 64, 1, -32, 160, -256, 128, 9, -120, 432, -576, 256, -1, 50, -400, 1120, -1280, 512, -11, 220, -1232, 2816, -2816, 1024, 1, -72, 840, -3584, 6912, -6144, 2048, 13, -364, 2912, -9984, 16640, -13312, 4096

Offset: 0

N. J. A. Sloane

The row length sequence of this irregular array is A008619(n), n >= 0. Even or odd powers appear in increasing order starting with 1 or x for even or odd row numbers n, respectively. This is the standard triangle A053120 with 0 deleted. - Wolfdieter Lang, Aug 02 2014

Let T* denote the triangle obtained by replacing each number in this triangle by its absolute value. Then T* gives the coefficients for cos(nx) as a polynomial in cos x. - Clark Kimberling, Aug 04 2024

			Rows are: (1), (1), (-1,2), (-3,4), (1,-8,8), (5,-20,16) etc., since if c = cos(x): cos(0x) = 1, cos(1x) = 1c; cos(2x) = -1+2c^2; cos(3x) = -3c+4c^3, cos(4x) = 1-8c^2+8c^4, cos(5x) = 5c-20c^3+16c^5, etc.
From _Wolfdieter Lang_, Aug 02 2014: (Start)
This irregular triangle a(n,k) begins:
  n\k   0    1     2      3      4      5      6      7 ...
  0:    1
  1:    1
  2:   -1    2
  3:   -3    4
  4:    1   -8     8
  5:    5  -20    16
  6:   -1   18   -48     32
  7:   -7   56  -112     64
  8:    1  -32   160   -256    128
  9:    9 -120   432   -576    256
 10:   -1   50  -400   1120  -1280    512
 11:  -11  220 -1232   2816  -2816   1024
 12:    1  -72   840  -3584   6912  -6144   2048
 13:   13 -364  2912  -9984  16640 -13312   4096
 14:   -1   98 -1568   9408 -26880  39424 -28672   8192
 15:  -15  560 -6048  28800 -70400  92160 -61440  16384
  ...
T(4,x) = 1 - 8*x^2 + 8*x^4, T(5,x) = 5*x - 20*x^3 +16*x^5.
(End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
E. A. Guilleman, Synthesis of Passive Networks, Wiley, 1957, p. 593.
Yaroslav Zolotaryuk, J. Chris Eilbeck, "Analytical approach to the Davydov-Scott theory with on-site potential", Physical Review B 63, p543402, Jan. 2001. The authors write, "Since the algebra of these is 'hyperbolic', contrary to the usual Chebyshev polynomials defined on the interval 0 <= x <= 1, we call the set of functions (21) the hyperbolic Chebyshev polynomials." (This refers to the triangle T* described in Comments.)

R. J. Mathar, Table of n, a(n) for n = 0..2600
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Renato Ferreira Pinto Jr. and Nathaniel Harms, Testing Support Size More Efficiently Than Learning Histograms, arXiv:2410.18915 [cs.DS], 2024. See p. 40.
D. Foata and G.-N. Han, Nombres de Fibonacci et polynomes orthogonaux
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
I. Rivin, Growth in free groups (and other stories), arXiv:math/9911076 [math.CO], 1999.
Eric Weisstein's World of Mathematics, Chebyshev Polynomial of the First Kind.
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

A039991 is a row reversed version, but has zeros which enable the triangle to be seen. Columns/diagonals are A011782, A001792, A001793, A001794, A006974, A006975, A006976 etc.
Reflection of A028297. Cf. A008312, A053112.
Row sums are one. Polynomial evaluations include A001075 (x=2), A001541 (x=3), A001091, A001079, A023038, A011943, A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203.
Cf. A053120.

A008310 := proc(n,m) local x ; coeftayl(simplify(ChebyshevT(n,x),'ChebyshevT'),x=0,m) ; end: i := 0 : for n from 0 to 100 do for m from n mod 2 to n by 2 do printf("%d %d ",i,A008310(n,m)) ; i := i+1 ; od ; od ; # R. J. Mathar, Apr 20 2007
# second Maple program:
b:= proc(n) b(n):= `if`(n<2, 1, expand(2*b(n-1)-x*b(n-2))) end:
T:= n-> (p-> (d-> seq(coeff(p, x, d-i), i=0..d))(degree(p)))(b(n)):
seq(T(n), n=0..15);  # Alois P. Heinz, Sep 04 2019

Flatten[{1, Table[CoefficientList[ChebyshevT[n, x], x], {n, 1, 13}]}]//DeleteCases[#, 0, Infinity]& (* or *) Flatten[{1, Table[Table[((-1)^k*2^(n-2 k-1)*n*Binomial[n-k, k])/(n-k), {k, Floor[n/2], 0, -1}], {n, 1, 13}]}] (* Eugeniy Sokol, Sep 04 2019 *)

a(n,m) = 2^(m-1) * n * (-1)^((n-m)/2) * ((n+m)/2-1)! / (((n-m)/2)! * m!) if n>0. - R. J. Mathar, Apr 20 2007
From Paul Weisenhorn, Oct 02 2019: (Start)
T_n(x) = 2*x*T_(n-1)(x) - T_(n-2)(x), T_0(x) = 1, T_1(x) = x.
T_n(x) = ((x+sqrt(x^2-1))^n + (x-sqrt(x^2-1))^n)/2. (End)
From Peter Bala, Aug 15 2022: (Start)
T(n,x) = [z^n] ( z*x + sqrt(1 + z^2*(x^2 - 1)) )^n.
Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x).
exp( Sum_{n >= 1} T(n,x)*t^n/n ) = Sum_{n >= 0} P(n,x)*t^n, where P(n,x) denotes the n-th Legendre polynomial. (End)

Additional comments and more terms from Henry Bottomley, Dec 13 2000

Edited: Corrected Cf. A039991 statement. Cf. A053120 added. - Wolfdieter Lang, Aug 06 2014

A057080 Even-indexed Chebyshev U-polynomials evaluated at sqrt(10)/2.

Original entry on oeis.org

1, 9, 71, 559, 4401, 34649, 272791, 2147679, 16908641, 133121449, 1048062951, 8251382159, 64962994321, 511452572409, 4026657584951, 31701808107199, 249587807272641, 1965000650073929, 15470417393318791, 121798338496476399

Offset: 0

Wolfdieter Lang, Aug 04 2000

a(n) = L(n,-8)*(-1)^n, where L is defined as in A108299; see also A070997 for L(n,+8). - Reinhard Zumkeller, Jun 01 2005
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim n->infinity a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 71, 34649, 16908641, 8251382159, 31701808107199,... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 9, 0, 71, 0, 559, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -6, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
From Klaus Purath, May 06 2025: (Start)
Nonnegative solutions to the Diophantine equation 3*a(n)^2 - 5*b(n)^2 = -2. The corresponding b(n) are A070997(n). Note that (a(n)*a(n+2) - a(n+1)^2)/2 = -5 and (b(n)*b(n+2) - b(n+1)^2)/2 = 3.
(a(n) + b(n))/2 = (a(n+1) - b(n+1))/2 = A001090(n+1) = Lucas U(8,1). Also a(n)*b(n+1) - a(n+1)*b(n) = -2.
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)) as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0 where (t) is a sequence satisfying t(i+3) = 9*t(i+2) - 9*t(i+1) + t(i) or t(i+2) = 8*t(i+1) - t(i), regardless of the initial values and including this sequence itself. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
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), rhs, m=10.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A033890, A100047.

a:=[1,9];; for n in [3..30] do a[n]:=8*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

I:=[1,9]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

A057080 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,9]);
    else
        8*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1+x)/(1-8x+x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)

Vec((1+x)/(1-8*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,8,1)-lucas_number2(n-1,8,1))/6 for n in range(1, 21)] # Zerinvary Lajos, Nov 10 2009

For all elements x of the sequence, 15*x^2 + 10 is a square. Lim. n-> Inf. a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 13 2002
a(n) = 8*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 8) + S(n-1, 8) = S(2*n, sqrt(10)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 8) = A001090(n).
G.f.: (1+x)/(1-8*x+x^2).
a(n) = ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)). - Gregory V. Richardson, Oct 13 2002
a(n) = sqrt((5*A070997(n)^2 - 2)/3) (cf. Richardson comment).
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i) then a(n) = (-1)^n*q(n,-10). - Benoit Cloitre, Nov 10 2002
a(n) = Jacobi_P(n,1/2,-1/2,4)/Jacobi_P(n,-1/2,1/2,1); - Paul Barry, Feb 03 2006
a(n+1) = 4*a(n) + sqrt(5*(3*a(n)^2 + 2)). - Richard Choulet, Aug 30 2007
In addition to the first formula above: In general, the following applies to all recurrences (a(n)) of the form (8,-1) with a(0) = 1 and arbritrary a(1): 15*a(n)^2 + y = b^2 where y = x^2 + 8*x + 1 and x = a(1) - 8. Also y = a(k+1)^2 - a(k)*a(k+1) for any k >=0. - Klaus Purath, May 06 2025
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 4), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 8*a(n)*a(n+1) + a(n+1)^2 = 10.
More generally, for arbitrary x, a(n+x)^2 - 8*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 10 with a(n) := ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)) as given above.
a(n+1/2) = sqrt(10) * A001090(n+1).
a(n+3/4) + a(n+1/4) = sqrt(10)*sqrt(sqrt(10) + 2) * A001090(n+1).
a(n+3/4) - a(n+1/4) = sqrt((sqrt(40) - 4)/3) * A001091(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/10 (telescoping series: for n >= 1, 10/(a(n) - 1/a(n)) = 1/A001090(n) + 1/A001090(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5/3) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 5/3 * (1 - 2/(1 + A001091(k+1)))). (End)

A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

Original entry on oeis.org

1, 17, 577, 19601, 665857, 22619537, 768398401, 26102926097, 886731088897, 30122754096401, 1023286908188737, 34761632124320657, 1180872205318713601, 40114893348711941777, 1362725501650887306817, 46292552162781456490001

Offset: 0

Henry Bottomley, Aug 16 2000

The sequence satisfies the Pell equation a(n)^2 - 18 * A202299(n+1)^2 = 1. - Vincenzo Librandi, Dec 19 2011
Also numbers n such that n - 1 and 2*n + 2 are squares. - Arkadiusz Wesolowski, Mar 15 2015
And they, n - 1 and 2*n + 2, are the squares of A005319 and A003499. - Michel Marcus, Mar 15 2015
This sequence {a(n)} gives all the nonnegative integer solutions of the Pell equation a(n)^2 - 32*(3*A091761(n))^2 = +1. - Wolfdieter Lang, Mar 09 2019

			G.f. = 1 + 17*x + 577*x^2 + 19601*x^3 + 665857*x^4 + 22619537*x^5 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..600 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A001075, A001541, A001091, A001079, A023038, A011943, A001081, A023039, A001085 and note relationship with square triangular number sequences A001110 and A001109. A091761.

Row 3 of array A188644.

I:=[1, 17]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 18 2011

LinearRecurrence[{34,-1},{1,17},30] (* Vincenzo Librandi, Dec 18 2011 *)
a[ n_] := ChebyshevT[ 2 n, 3]; (* Michael Somos, May 28 2014 *)

makelist(expand(((17+sqrt(288))^n+(17-sqrt(288))^n))/2, n, 0, 15); /* Vincenzo Librandi, Dec 18 2011 */

{a(n) = polchebyshev( n, 1, 17)}; /* Michael Somos, Apr 05 2019 */

[lucas_number2(n,34,1)/2 for n in range(0,15)] # Zerinvary Lajos, Jun 27 2008

a(n) = (r^n + 1/r^n)/2 with r = 17 + sqrt(17^2-1).
a(n) = 16*A001110(n) + 1 = A001541(2n) = (4*A001109(n))^2 + 1 = 3*A001109(2n-1) - A001109(2n-2) = A001109(2n) - 3*A001109(2n-1).
a(n) = T(n, 17) = T(2*n, 3) with T(n, x) Chebyshev's polynomials of the first kind. See A053120. T(n, 3)= A001541(n).
G.f.: (1-17*x)/(1-34*x+x^2).
G.f.: (1 - 17*x / (1 - 288*x / (17 - x))). - Michael Somos, Apr 05 2019
a(n) = cosh(2n*arcsinh(sqrt(8))). - Herbert Kociemba, Apr 24 2008
a(n) = (a^n + b^n)/2 where a = 17 + 12*sqrt(2) and b = 17 - 12*sqrt(2); sqrt(a(n)-1)/4 = A001109(n). - James R. Buddenhagen, Dec 09 2011
a(-n) = a(n). - Michael Somos, May 28 2014
a(n) = sqrt(1 + 32*9*A091761(n)^2), n >= 0. See one of the Pell comments above. - Wolfdieter Lang, Mar 09 2019

More terms from James Sellers, Sep 07 2000

Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A237606 Positive integers k such that x^2 - 8xy + y^2 + k = 0 has integer solutions.

Original entry on oeis.org

6, 11, 14, 15, 24, 35, 44, 51, 54, 56, 59, 60, 71, 86, 96, 99, 110, 119, 126, 131, 134, 135, 140, 150, 159, 176, 179, 191, 204, 206, 215, 216, 224, 231, 236, 239, 240, 251, 254, 275, 284, 294, 311, 315, 326, 335, 339, 344, 350, 359, 366, 371, 374, 375, 384

Offset: 1

Colin Barker, Feb 10 2014

nonn

From Klaus Purath, Feb 17 2024: (Start)
Positive numbers of the form 15x^2 - y^2. The reduced form is -x^2 + 6xy + 6y^2.
Even powers of terms as well as products of an even number of terms belong to A243188. This can be proved with respect to the forms [a,0,-c] and [a, 0, +c] by the following identities: (au^2 - cv^2)(ax^2 - cy^2) = (aux + cvy)^2 - ac(uy + vx)^2 and (au^2 + cv^2)(ax^2 + cy^2) = (aux - cvy)^2 + ac(uy + vx)^2 for all a, c, u, v, x, y in R. This can be verified by expanding both sides of the equations. Generalization (conjecture): This multiplication rule applies to all sequences represented by any binary quadratic form [a, b, c].
Odd powers of terms as well as products of an odd number of terms belong to the sequence. This can be proved with respect to the forms [a,0,-c] and [a, 0, +c] by the following identities: (as^2 - ct^2)(au^2 - cv^2)(ax^2 - cy^2) = a[s(aux + cvy) + ct(uy + vx)]^2 - c[as(uy + vx) + t(aux + cvy)]^2 and (as^2 + ct^2)(au^2 + cv^2)(ax^2 + cy^2) = a[s(aux - cvy) - ct(uy + vx)]^2 + c[as(uy + vx) + t(aux - cvy)]^2 for all a, c, s, t, u, v, x, y in R. This can be verified by expanding both sides of the equations. Generalization (conjecture): This multiplication rule applies to all sequences represented by any binary quadratic form [a, b, c].
If we denote any term of this sequence by B and correspondingly of A243189 by C and of A243190 by D, then B*C = D, C*D = B and B*D = C. This can be proved by the following identities, where the sequence (B) is represented by [kn, 0, -1], (C) by [n, 0, -k] and (D) by [k, 0, -n].
Proof of B*C = D: (knu^2 - v^2)(nx^2 - ky^2) =  k(nux + vy)^2 - n(kuy + vx)^2 for k, n, u, v, x, y in R.
Proof of C*D = B: (nu^2 - kv^2)(kx^2 - ny^2) = kn(ux + vy)^2 - (nuy + kvx)^2 for k, n, u, v, x, y in R.
Proof of B*D = C: (knu^2 - v^2)(kx^2 - ny^2) = n(kux + vy)^2 - k(nuy + vx)^2 for k, n, u, v, x, y in R. This can be verified by expanding both sides of the equations.
Generalization (conjecture): If there are three sequences of a given positive discriminant that are represented by the forms [a1, b1, c1], [a2, b2, c2] and [a1*a2, b3, c3] for a1, a2 != 1, then the BCD rules apply to these sequences. (End)

			6 is in the sequence because x^2 - 8xy + y^2 + 6 = 0 has integer solutions, for example (x, y) = (1, 7).

N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)

Cf. A070997 (k = 6), A199336 (k = 14), A001091 (k = 15), A077248 (k = 35).
Cf. A031363, A084917, A237351, A237599, A237609, A237610.
For primes see A141302.
Cf. A378710, A378711 (subsequence of properly represented numbers and fundamental solutions).

A322836 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{n}(x), evaluated at x=k.

Original entry on oeis.org

1, 1, 0, 1, 1, -1, 1, 2, 1, 0, 1, 3, 7, 1, 1, 1, 4, 17, 26, 1, 0, 1, 5, 31, 99, 97, 1, -1, 1, 6, 49, 244, 577, 362, 1, 0, 1, 7, 71, 485, 1921, 3363, 1351, 1, 1, 1, 8, 97, 846, 4801, 15124, 19601, 5042, 1, 0, 1, 9, 127, 1351, 10081, 47525, 119071, 114243, 18817, 1, -1

Offset: 0

Seiichi Manyama, Dec 28 2018

			Square array begins:
   1, 1,    1,     1,      1,      1,       1, ...
   0, 1,    2,     3,      4,      5,       6, ...
  -1, 1,    7,    17,     31,     49,      71, ...
   0, 1,   26,    99,    244,    485,     846, ...
   1, 1,   97,   577,   1921,   4801,   10081, ...
   0, 1,  362,  3363,  15124,  47525,  120126, ...
  -1, 1, 1351, 19601, 119071, 470449, 1431431, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Mirror of A101124.
Columns 0-20 give A056594, A000012, A001075, A001541, A001091, A001079, A023038, A011943(n+1), A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203, A322888, A056771, A322889, A078986, A322890.
Rows 0-10 give A000012, A001477, A056220, A144129, A144130, A243131, A243132, A243133, A243134, A243135, A243136.
Main diagonal gives A115066.
Cf. A323182 (Chebyshev polynomial of the second kind).

Table[ChebyshevT[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 28 2018 *)

T(n,k) = polchebyshev(n,1,k);
matrix(7, 7, n, k, T(n-1,k-1)) \\ Michel Marcus, Dec 28 2018

T(n, k) = round(cos(n*acos(k)));\\ Seiichi Manyama, Mar 05 2021

T(n, k) = if(n==0, 1, n*sum(j=0, n, (2*k-2)^j*binomial(n+j, 2*j)/(n+j))); \\ Seiichi Manyama, Mar 05 2021

A(0,k) = 1, A(1,k) = k and A(n,k) = 2 * k * A(n-1,k) - A(n-2,k) for n > 1.
A(n,k) = cos(n*arccos(k)). - Seiichi Manyama, Mar 05 2021
A(n,k) = n * Sum_{j=0..n} (2*k-2)^j * binomial(n+j,2*j)/(n+j) for n > 0. - Seiichi Manyama, Mar 05 2021

A005828 a(n) = 2*a(n-1)^2 - 1, a(0) = 4, a(1) = 31.

Original entry on oeis.org

4, 31, 1921, 7380481, 108942999582721, 23737154316161495960243527681, 1126904990058528673830897031906808442930637286502826475521

Offset: 0

Jeffrey Shallit

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004

The next term has 115 digits. - Harvey P. Dale, May 25 2018

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

G. C. Greubel, Table of n, a(n) for n = 0..10
Anonymous, Fermat's rule for 3-fold perfect numbers [Broken link]

Cf. A001091, A001601, A002812, A084764 (essentially the same).

[n le 2 select 2^(3*n-1)-n+1 else 2*Self(n-1)^2 - 1: n in [1..10]]; // G. C. Greubel, May 17 2023

NestList[2#^2-1&,4,10] (* Harvey P. Dale, May 25 2018 *)

PARI
```
a(n)=if(n<1,4*(n==0),2*a(n-1)^2-1)
```

a(n)=if(n<0,0,subst(poltchebi(2^n),x,4))

[chebyshev_T(2^n, 4) for n in range(11)] # G. C. Greubel, May 17 2023

a(n) = A001091(2^n).
From Peter Bala, Nov 11 2012, (Start)
a(n) = (1/2)*((4 + sqrt(15))^(2^n) + (4 - sqrt(15))^(2^n)).
2*sqrt(15)/9 = Product_{n>=0} (1 - 1/(2*a(n))).
sqrt(5/3) = Product_{n>=0} (1 + 1/a(n)).
See A002812 for general properties of the recurrence a(n+1) = 2*a(n)^2 - 1.
(End)
a(n) = T(2^n,4), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Feb 01 2017
a(n) = cos(2^n*arccos(4)). - Peter Luschny, Oct 12 2022

A101124 Number triangle associated to Chebyshev polynomials of first kind.

Original entry on oeis.org

1, 0, 1, -1, 1, 1, 0, 1, 2, 1, 1, 1, 7, 3, 1, 0, 1, 26, 17, 4, 1, -1, 1, 97, 99, 31, 5, 1, 0, 1, 362, 577, 244, 49, 6, 1, 1, 1, 1351, 3363, 1921, 485, 71, 7, 1, 0, 1, 5042, 19601, 15124, 4801, 846, 97, 8, 1, -1, 1, 18817, 114243, 119071, 47525, 10081, 1351, 127, 9, 1, 0, 1, 70226, 665857, 937444, 470449, 120126, 18817, 2024, 161

Offset: 0

Paul Barry, Dec 02 2004

			As a number triangle, rows begin:
  {1},
  {0,1},
  {-1,1,1},
  {0,1,2,1},
  ...
As a square array, rows begin
   1, 1,  1,   1,    1, ...
   0, 1,  2,   3,    4, ...
  -1, 1,  7,  17,   31, ...
   0, 1, 26,  99,  244, ...
   1, 1, 97, 577, 1921, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Index entries for sequences related to Chebyshev polynomials.

Columns include A001075, A001541, A001091, A001079, A023038, A011943.
Row sums are A101125.
Diagonal sums are A101126.
Main diagonal gives A115066.
Mirror of A322836.
Cf. A053120.

T[n_, k_] := SeriesCoefficient[x^k (1 - k x)/(1 - 2 k x + x^2), {x, 0, n}];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 12 2017 *)

Number triangle S(n, k)=T(n-k, k), k

Columns have g.f. x^k(1-kx)/(1-2kx+x^2).

Also, square array if(n=0, 1, T(n, k)) read by antidiagonals.

A080871 a(n)a(n+3) - a(n+1)a(n+2) = 3, given a(0)=a(1)=1, a(2)=4.

Original entry on oeis.org

1, 1, 4, 7, 31, 55, 244, 433, 1921, 3409, 15124, 26839, 119071, 211303, 937444, 1663585, 7380481, 13097377, 58106404, 103115431, 457470751, 811826071, 3601659604, 6391493137, 28355806081, 50320119025, 223244789044

Offset: 0

Paul D. Hanna, Feb 22 2003

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (0, 8, 0, -1).

Cf. A001519, A079496, A080872, A080873, A080874, A080875.

Bisections are A001091 and A070997.

RecurrenceTable[{a[0]==a[1]==1,a[2]==4,a[n]==(3+a[n+1]a[n+2])/a[n+3]},a,{n,30}] (* Harvey P. Dale, Jun 08 2017 *)

a(n) = (3 + a(n-1)*a(n-2))/a(n-3) for n>2.
G.f.: (-x^3 - 4*x^2 + x + 1)/(x^4 - 8*x^2 + 1)
a(n+4) = 8*a(n+2)-a(n). [Richard Choulet, Dec 04 2008]
a(n) = (0.25 + sqrt(10)/20)*(sqrt(4 + sqrt(15)))^n + (0.25 + sqrt(10)/20)*(sqrt(4 - sqrt(15)))^n + ( - 1/20*10^(1/2) + 1/4)*( - sqrt(4 + sqrt(15)))^n + ( - 1/20*10^(1/2) + 1/4)*( - (sqrt(4 - sqrt(15))))^n. [Richard Choulet, Dec 06 2008]

Showing 1-10 of 22 results. Next

cp's OEIS Frontend

A001090 a(n) = 8*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

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A070997 a(n) = 8*a(n-1) - a(n-2), a(0)=1, a(-1)=1.

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A008310 Triangle of coefficients of Chebyshev polynomials T_n(x).

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A057080 Even-indexed Chebyshev U-polynomials evaluated at sqrt(10)/2.

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A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

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A080871 a(n)a(n+3) - a(n+1)a(n+2) = 3, given a(0)=a(1)=1, a(2)=4.