A001353 -id:A001353 - cp's OEIS frontend

A077412 Chebyshev U(n,x) polynomial evaluated at x=8.

1, 16, 255, 4064, 64769, 1032240, 16451071, 262184896, 4178507265, 66593931344, 1061324394239, 16914596376480, 269572217629441, 4296240885694576, 68470281953483775, 1091228270370045824, 17391182043967249409

Offset: 0

Wolfdieter Lang, Nov 08 2002

For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 16'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,...,15}. - Milan Janjic, Jan 23 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..800
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.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (16,-1).

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), this sequence (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:=8;; 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

I:=[1, 16, 255]; [n le 3 select I[n] else 16*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012

seq( simplify(ChebyshevU(n, 8)), n=0..20); # G. C. Greubel, Dec 22 2019

Table[GegenbauerC[n, 1, 8], {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
CoefficientList[Series[1/(1-16x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Dec 24 2012 *)
LinearRecurrence[{16,-1}, {1,16}, 30] (* G. C. Greubel, Jan 18 2018 *)
ChebyshevU[Range[21] -1, 8] (* G. C. Greubel, Dec 22 2019 *)

vector( 21, n, polchebyshev(n-1, 2, 8) ) \\ G. C. Greubel, Jan 18 2018

[lucas_number1(n,16,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n,8) for n in (0..20)] # G. C. Greubel, Dec 22 2019

a(n) = 16*a(n-1) - a(n-2), n>=1, a(-1)=0, a(0)=1.
a(n) = S(n, 16) with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. See A049310.
G.f.: 1/(1 - 16*x + x^2).
a(n) = (((8 + 3*sqrt(7))^(n+1) - (8 - 3*sqrt(7))^(n+1)))/(6*sqrt(7)).
a(n) = sqrt((A001081(n+1)^2-1)/63).
a(n) = Sum_{k=0..n} A101950(n,k)*15^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/7*(7 + 3*sqrt(7)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/16*(7 + 3*sqrt(7)). - Peter Bala, Dec 23 2012

A061278 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

Original entry on oeis.org

0, 1, 5, 20, 76, 285, 1065, 3976, 14840, 55385, 206701, 771420, 2878980, 10744501, 40099025, 149651600, 558507376, 2084377905, 7779004245, 29031639076, 108347552060, 404358569165, 1509086724601, 5631988329240, 21018866592360, 78443478040201, 292755045568445

Offset: 0

Henry Bottomley, Jun 04 2001

Indices m of triangular numbers T(m) which are one-third of another triangular number: 3*T(m) = T(k); the k's are given by A001571. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002
On the previous comment: for m=0 this is actually one third of the same triangular number. - Zak Seidov, Apr 07 2011
Also numbers n such that the n-th centered 24-gonal number 12*n*(n+1)+1 is a perfect square A001834(n)^2, where A001834(n) is defined by the recursion: a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2) + 1. - Alexander Adamchuk, Apr 21 2007
Also numbers n such that RootMeanSquare(5,...,6*n-1) is an integer. - Ctibor O. Zizka, Dec 17 2008 (Corrected by Robert K. Moniot, Jul 22 2020)
Also numbers n such that n*(n+1) = Sum_{i=1..x} n+i for some x. (This does not apply to the first term.). - Gil Broussard, Dec 23 2008
From John P. McSorley, May 26 2020: (Start)
Consecutive terms (a(n-1), a(n)) = (u,v) give all points on the hyperbola u^2 - u + v^2 - v - 4*u*v = 0 in quadrant I with both coordinates an integer.
Also related to the block sizes of small multi-set designs. (End)
If a(n) white balls and a(n+1) black balls are mixed in a bag, and a pair of balls is drawn without replacement, the probability that one ball of each color is drawn is exactly 1/3. These are the only integers for which the probability is 1/3. For example, if there are 20 white balls and 76 black balls, the probability of drawing one of each is (20/96)*(76/95) + (76/96)*(20/95) = 1/3. - Elliott Line, May 13 2022

			a(2)=5 and T(5)=15 which is 1/3 of 45=T(9).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.
Brian Lawrence and Will Sawin, The Shafarevich conjecture for hypersurfaces in abelian varieties, arXiv:2004.09046 [math.NT], 2020.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
L. A. Medina and A. Straub, On multiple and infinite log-concavity, 2013.
S. Morier-Genoud, V. Ovsienko and S. Tabachnikov, 2-frieze patterns and the cluster structure of the space of polygons, Annales de l'institut Fourier, 62 no. 3 (2012), 937-987. - From _N. J. A. Sloane_, Dec 26 2012
Robert Phillips, A triangular number result, 2009.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Eric Weisstein, Centered Polygonal Number.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001075, A001353, A001571, A001834, A001835, A079935, A101265. Also cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

I:=[0, 1]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2) + 1: n in [1..30]]; // Vincenzo Librandi, Dec 23 2012

f:= gfun:-rectoproc({a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3),a(1)=1,a(0)=0,a(-1)=0},a(n),remember):
map(f, [$0..50]); # Robert Israel, Jun 05 2015

CoefficientList[Series[x/(1 - 5*x + 5*x^2 - x^3), {x, 0, nn}], x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{5,-5,1},{0,1,5},30] (* Harvey P. Dale, Dec 23 2012 *)

M = [1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=1, 30, print1(([1, 0, 0]*M^i)[3], ",")) \\ Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

a(n) = 4*a(n-1) - a(n-2) + 1.
a(n) = A001075(n) - a(n-1) - 1.
a(n) = (A001835(n+1) - 1)/2 = (A001353(n+1) - A001353(n) - 1)/2.
a(n) = a(n-1) + A001353(n), i.e., partial sum of A001353.
From Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002: (Start)
a(n+2) = 4*a(n+1) - a(n) + 1 for a(0)=0, a(1)=1.
G.f.: x/((1 - x)*(1 - 4*x + x^2)).
a(n) = (1/12)*((3 - sqrt(3))*(2 - sqrt(3))^n + (3 + sqrt(3))*(2 + sqrt(3))^n-6). (End)
a(n) = (1/12)*(A003500(n) + A003500(n+1)-6). - Mario Catalani (mario.catalani(AT)unito.it), Apr 11 2003
a(n+1) = Sum_{k=0..n} U(k, 2) = Sum_{k=0..n} S(k, 4), where U(n,x) and S(n,x) are Chebyshev polynomials. - Paul Barry, Nov 14 2003
G.f.: x/(1 - 5*x + 5*x^2 - x^3).
a(n) = A079935(n+1) + A001571(n) for n>0, a(0)=0. - Gerry Martens, Jun 05 2015
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Bruno Berselli, Nov 29 2016
From John P. McSorley, May 25 2020: (Start)
a(n)^2 - a(n) + a(n-1)^2 - a(n-1) - 4*a(n)*a(n-1) = 0.
a(n) = A001834(n-1) + a(n-2). (End)
(T(a(n)-1) + T(a(n+1)-1))/T(a(n) + a(n+1) - 1) = 2/3 where T(i) is the i-th triangular number. - Robert K. Moniot, Oct 11 2020
E.g.f.: exp(x)*(exp(x)*(3*cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) - 3)/6. - Stefano Spezia, Feb 05 2021
a(n) = A101265(n) - 1. - Jon E. Schoenfield, Jan 01 2022

More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, -6, -32, -4, 312, 664, -1792, -10224, -2528, 97184, 219648, -532544, -3261568, -1197696, 30220288, 72417536, -157367808, -1038910976, -504143872, 9380822016, 23803082752, -46202054656, -330434936832, -198849327104, 2906650714112, 7801794699264

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

For the difference equation a(n) = c*a(n-1) - d*a(n-2), with a(0) = 0, a(1) = 1, the solution is a(n) = d^((n-1)/2) * ChebyshevU(n-1, c/(2*sqrt(d))) and has the alternate form a(n) = ( ((c + sqrt(c^2 - 4*d))/2)^n - ((c - sqrt(c^2 - 4*d))/2)^n )/sqrt(c^2 - 4*d). In the case c^2 = 4*d then the solution is a(n) = n*d^((n-1)/2). The generating function is x/(1 - c*x + d^2) and the exponential generating function takes the form (2/sqrt(c^2 - 4*d))*exp(c*x/2)*sinh(sqrt(c^2 - 4*d)*x/2) for c^2 > 4*d, (2/sqrt(4*d - c^2))*exp(c*x/2)*sin(sqrt(4*d - c^2)*x/2) for 4*d > c^2, and x*exp(sqrt(d)*x) if c^2 = 4*d. - G. C. Greubel, Jun 10 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..190
Index entries for linear recurrences with constant coefficients, signature (2,-10).

Sequences of the form a(n) = c*a(n-1) - d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A128834,A107920,A106852,A106853,A106854,A145934,A145976,A145978,A146078,A146080
.A000027,A009545,A088137,A088138,A045873,A088139,A089181,A090591,A127357,A190958
.A001906,A000225,A057083,A049072,A190959,A190960,A190961,A190962,A190963,A190964
.A001353,A007070,A003462,A001787,A099456,A190965,A168175,A190966,A190967,A190968
.A004254,A107839,A116415,A002450,A030191,A001047,A099450,A190969,A190970,A190971
.A001109,A154244,A138395,A084326,A003463,A030192,A081179,A006516,A027471,A106392
.A004187,A186446,A190972,A190973,A190974,A003464,A030240,A152268,A099459,A016127
.A001090,A190975,A190976,A099156,A190977,A190978,A023000,A057084,A154245,A154235
.A018913,A190979,A190980,A190981,A190982,A190983,A190984,A023001,A057085,A178869
A004189,A190869,A190985,A190986,A190987,A190988,A190989,A190990,A002452,A057086

I:=[0,1]; [n le 2 select I[n] else 2*Self(n-1)-10*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 17 2011

Mathematica
```
LinearRecurrence[{2,-10}, {0,1}, 50]
```

a(n)=([0,1; -10,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,2,10) for n in (0..50)] # G. C. Greubel, Jun 10 2022

G.f.: x / ( 1 - 2*x + 10*x^2 ). - R. J. Mathar, Jun 01 2011
E.g.f.: (1/3)*exp(x)*sin(3*x). - Franck Maminirina Ramaharo, Nov 13 2018
a(n) = 10^((n-1)/2) * ChebyshevU(n-1, 1/sqrt(10)). - G. C. Greubel, Jun 10 2022
a(n) = (1/3)*10^(n/2)*sin(n*arctan(3)) = Sum_{k=0..floor(n/2)} (-1)^k*3^(2*k)*binomial(n,2*k+1). - Gerry Martens, Oct 15 2022

A029547 Expansion of g.f. 1/(1 - 34*x + x^2).

Original entry on oeis.org

1, 34, 1155, 39236, 1332869, 45278310, 1538129671, 52251130504, 1775000307465, 60297759323306, 2048348816684939, 69583562007964620, 2363792759454112141, 80299370259431848174, 2727814796061228725775, 92665403695822344828176, 3147895910861898495432209

Offset: 0

N. J. A. Sloane

Chebyshev sequence U(n,17)=S(n,34) with Diophantine property.
b(n)^2 - 2*(12*a(n))^2 = 1 with the companion sequence b(n)=A056771(n+1). - Wolfdieter Lang, Dec 11 2002
More generally, for t(m) = m + sqrt(m^2-1) and u(n) = (t(m)^(n+1) - 1/t(m)^(n+1))/(t(m) - 1/t(m)), we can verify that ((u(n+1) - u(n-1))/2)^2 - (m^2-1)*u(n)^2 = 1. - Bruno Berselli, Nov 21 2011
a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,33}. - Milan Janjic, Jan 26 2015

Indranil Ghosh, Table of n, a(n) for n = 0..651 (terms 0..200 from Vincenzo Librandi)
Z. Cerin and G. M. Gianella, On sums of squares of Pell-Lucas numbers, INTEGERS 6 (2006) #A15.
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 (34,-1).

A091761 is an essentially identical sequence.
Cf. A200441, A200724.
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), A078987 (m=19), A097316 (m=33).

m:=17;; 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

I:=[1,34]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

with (combinat):seq(fibonacci(4*n+4,2)/12, n=0..15); # Zerinvary Lajos, Apr 21 2008

Table[GegenbauerC[n, 1, 17], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
LinearRecurrence[{34,-1},{1,34},20] (* Vincenzo Librandi, Nov 22 2011 *)
ChebyshevU[Range[21] -1, 17] (* G. C. Greubel, Dec 22 2019 *)

A029547(n, x=[0,1],A=[17,72*4;1,17]) = vector(n,i,(x*=A)[1]) \\ M. F. Hasler, May 26 2007

vector( 21, n, polchebyshev(n-1, 2, 17) ) \\ G. C. Greubel, Dec 22 2019

[lucas_number1(n,34,1) for n in range(1, 16)] # Zerinvary Lajos, Nov 07 2009

[chebyshev_U(n,17) for n in (0..20)] # G. C. Greubel, Dec 22 2019

a(n) = 34*a(n-1) - a(n-2), a(-1)=0, a(0)=1.
a(n) = S(n, 34) with S(n, x):= U(n, x/2) Chebyshev's polynomials of the 2nd kind. See A049310. - Wolfdieter Lang, Dec 11 2002
a(n) = (ap^(n+1) - am^(n+1))/(ap - am) with ap = 17+12*sqrt(2) and am = 17-12*sqrt(2).
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*binomial(n-k, k)*34^(n-2*k).
a(n) = sqrt((A056771(n+1)^2 -1)/2)/12.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) with a(-1)=0, a(0)=1, a(1)=34. Also a(n) = (sqrt(2)/48)*((17+12*sqrt(2))^n-(17-12*sqrt(2))^n) = (sqrt(2)/48)*((3+2*sqrt(2))^(2n+2)-(3-2*sqrt(2))^(2n+2)) = (sqrt(2)/48)*((1+sqrt(2))^(4n+4)-(1-sqrt(2))^(4n+4)). - Antonio Alberto Olivares, Mar 19 2008
a(n) = Sum_{k=0..n} A101950(n,k)*33^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 0} (1 + 1/a(n)) = 1/4*(4 + 3*sqrt(2)).
Product {n >= 1} (1 - 1/a(n)) = 2/17*(4 + 3*sqrt(2)). (End)
E.g.f.: exp(17*x)*(24*cosh(12*sqrt(2)*x) + 17*sqrt(2)*sinh(12*sqrt(2)*x))/24. - Stefano Spezia, Apr 16 2023

A052530 a(n) = 4*a(n-1) - a(n-2), with a(0) = 0, a(1) = 2.

Original entry on oeis.org

0, 2, 8, 30, 112, 418, 1560, 5822, 21728, 81090, 302632, 1129438, 4215120, 15731042, 58709048, 219105150, 817711552, 3051741058, 11389252680, 42505269662, 158631825968, 592022034210, 2209456310872, 8245803209278, 30773756526240

Offset: 0

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

a(n-1) and a(n+1) are the solutions for c if b = a(n) in (b^2 + c^2)/(b*c + 1) = 4 and there are no other pairs of solutions apart from consecutive pairs of terms in this sequence. Cf. A061167. - Henry Bottomley, Apr 18 2001
a(n)^2 for n >= 1 gives solutions to A007913(3*x+4) = A007913(x). - Benoit Cloitre, Apr 07 2002
For all terms k of the sequence, 3*k^2 + 4 is a perfect square. Limit_{n->oo} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 06 2002
a(n) = the number of compositions of the integer 2*n into even parts, where each part 2*i comes in 2*i colors. (Dedrickson, Theorem 3.2.6) An example is given below. Cf. A052529, A095263. - Peter Bala, Sep 17 2013
Except for an initial 1, this is the p-INVERT of (1, 1, 1, 1, 1, ...) for p(S) = 1 - 2*S - 2*S^2; see A291000.  - Clark Kimberling, Aug 24 2017
a(n+1) is the number of spanning trees of the graph P_n, where P_n is a 2 X n grid with two additional vertices, u and v, where u is adjacent to (1,1) and (2,1), and v is adjacent to (1,n) and (2,n). - Kevin Long, May 04 2018
a(n) is also the output of Tesler's formula for the number of perfect matchings of an m X n Mobius band where m is even and n is odd, specialized to m=2. (The twist is on the length-n side.) - Sarah-Marie Belcastro, Feb 15 2022
In general, values of x and y which satisfy (x^2 + y^2)/(x*y + 1) = k^2 are any two adjacent terms of a second-order recurrence with initial terms 0 and k and signature (k^2,-1). This can also be expressed as a first-order recurrence a(n+1) = (k^2*a(n) + sqrt((k^4-4)*a(n)^2 + 4*k^2))/2, n > 1. - Gary Detlefs, Feb 27 2024

			Colored compositions. a(2) = 8: There are two compositions of 4 into even parts, namely 4 and 2 + 2. Using primes to indicate the coloring of parts, the 8 colored compositions are 4, 4', 4'', 4''', 2 + 2, 2 + 2', 2' + 2 and 2' + 2'. - _Peter Bala_, Sep 17 2013

T. D. Noe, Table of n, a(n) for n = 0..200
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.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, (an + b)-color compositions, arXiv:1707.07798 [math.CO], 2017.
Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.
C. R. Dedrickson III, Compositions, Bijections, and Enumerations Thesis, 2012, Jack N. Averitt College of Graduate Studies, Georgia Southern University.
J.-P. Ehrmann et al., Problem POLYA002, Integer pairs (x,y) for which (x^2+y^2)/(1+pxy) is an integer.
F. Goebel and A. A. Jagers, On a conjecture of Tutte concerning minimal tree numbers, J. Combin. Theory Ser. B 26 (1979), no. 3, 346-348. MR0535948 (80m:05064). [From _N. J. A. Sloane_, Feb 20 2012]
A. F. Horadam, Basic Properties of a Certain Generalized Sequence of Numbers, Fibonacci Quarterly, Vol. 3, No. 3, 1965, pp. 161-176.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 460
N. J. A. Sloane, Transforms
G. Tesler, Matchings in graphs on non-orientable surfaces, Journal of Combinatorial Theory B, 78(2000), 198-231.
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001353, A007913, A003699, A217233.

Cf. A052529, A095263.

a052530 n = a052530_list !! n
a052530_list =
   0 : 2 : zipWith (-) (map (* 4) $ tail a052530_list) a052530_list
-- Reinhard Zumkeller, Sep 29 2011

I:=[0,2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Feb 25 2019

spec := [S,{S=Sequence(Prod(Union(Z,Z),Sequence(Z),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec, size=n), n=0..20);
s := sqrt(3): a := n -> ((2-s)^n-(s+2)^n)/(s*(s-2)*(s+2)):
seq(simplify(a(n)), n=0..24); # Peter Luschny, Apr 28 2020

p=1; c=2; a[0]=0; a[1]=c; a[n_]:=a[n]=p*c^2*a[n-1]-a[n-2]; Table[a[n], {n, 0, 20}]
NestList[2 # + Sqrt[4 + 3 #^2]&, 0, 200] (* Zak Seidov, Mar 31 2011 *)
LinearRecurrence[{4, -1}, {0, 2}, 25] (* T. D. Noe, Jan 09 2012 *)
CoefficientList[Series[2x/(1-4x+x^2),{x,0,30}],x] (* Harvey P. Dale, May 31 2023 *)

{ polya002(p,c,m) = local(v,w,j,a); w=0; print1(w,", "); v=c; print1(v,", "); j=1; while(j<=m,a=p*c^2*v-w; print1(a,", "); w=v; v=a; j++) };
polya002(1,2,25)

my(x='x+O('x^30)); concat([0], Vec(2*x/(1-4*x+x^2))) \\ G. C. Greubel, Feb 25 2019

first(n) = n = max(n, 2); my(res = vector(n)); res[1] = 0; res[2] = 2; for(i = 3, n, res[i] = 4 * res[i-1] - res[i-2]); res \\ David A. Corneth, Apr 28 2020

(2*x/(1-4*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

G.f.: 2*x/(1 - 4*x + x^2).
Invert transform of even numbers: a(n) = 2*Sum_{k=1..n} k*a(n-k). - Vladeta Jovovic, Apr 27 2001
From Gregory V. Richardson, Oct 06 2002: (Start)
a(n) = Sum_{alpha} -(1/3)*(-1 + 2*alpha)*alpha^(-1 - n), alpha = root of (1 - 4*Z + Z^2).
a(n) = (((2+sqrt(3))^(n+1) - (2-sqrt(3))^(n+1)) - ((2+sqrt(3))^n - (2-sqrt(3))^n) + ((2+sqrt(3))^(n-1) - (2-sqrt(3))^(n-1)))/(3*sqrt(3)). (End)
a(n) = A071954(n) - 2. - N. J. A. Sloane, Feb 20 2005
a(n) = (2*sinh(2n*arcsinh(1/sqrt(2))))/sqrt(3). - Herbert Kociemba, Apr 24 2008
a(n) = 2*A001353(n). - R. J. Mathar, Oct 26 2009
a(n) = ((3 - 2*sqrt(3))/3)*(2 - sqrt(3))^(n - 1) + ((3 + 2*sqrt(3))/3)*(2 + sqrt(3))^(n - 1). - Vincenzo Librandi, Nov 20 2010
a(n) = floor((2 + sqrt(3))^n/sqrt(3)). - Zak Seidov, Mar 31 2011
a(n) = ((2 + sqrt(3))^n - (2 - sqrt(3))^n)/sqrt(3). (See Horadam for construction.) - Johannes Boot, Jan 08 2012
a(n) = A217233(n) + A217233(n-1) with A217233(-1) = -1. - Bruno Berselli, Oct 01 2012
a(n) = A001835(n+1) - A001835(n). - Kevin Long, May 04 2018
E.g.f.: (exp((2 + sqrt(3))*x) - exp((2 - sqrt(3))*x))/sqrt(3). - Franck Maminirina Ramaharo, Nov 12 2018
a(n+1) = 2*a(n) + sqrt(3*a(n)^2 + 4), n > 1. - Gary Detlefs, Feb 27 2024

More terms from James Sellers, Jun 06 2000
Edited by N. J. A. Sloane, Nov 11 2006
a(0) changed to 0 and entry revised accordingly by Max Alekseyev, Nov 15 2007
Signs in definition corrected by John W. Layman, Nov 20 2007

A004191 Expansion of 1/(1 - 12*x + x^2).

Original entry on oeis.org

1, 12, 143, 1704, 20305, 241956, 2883167, 34356048, 409389409, 4878316860, 58130412911, 692686638072, 8254109243953, 98356624289364, 1172025382228415, 13965947962451616, 166419350167190977, 1983066254043840108, 23630375698358890319, 281581442126262843720

Offset: 0

N. J. A. Sloane

Chebyshev's polynomials U(n,x) evaluated at x=6.
a(n) give all (nontrivial, integer) solutions of Pell equation b(n)^2 - 35*a(n)^2 = +1 with b(n)=A023038(n+1), n>=0.
For positive n, a(n) equals the permanent of the tridiagonal matrix of order n with 12'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,...,11}. - Milan Janjic, Jan 26 2015
a(n) = -a(-2-n) for all n in Z. - Michael Somos, Jun 29 2019

			G.f. = 1 + 12*x + 143*x^2 + 1704*x^3 + 20305*x^4 + 241956*x^5 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..900
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.
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 12.
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
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (12,-1).

Cf. A023038, A049310, A077417, A101950.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), this sequence (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:=8;; 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 23 2019

I:=[1, 12]; [n le 2 select I[n] else 12*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Jun 13 2012

seq( simplify(ChebyshevU(n, 6)), n=0..20); # G. C. Greubel, Dec 23 2019

Table[GegenbauerC[n, 1, 6], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
CoefficientList[Series[1/(1-12*x+x^2), {x,0,30}], x] (* T. D. Noe, Aug 01 2011 *)
LinearRecurrence[{12,-1},{1,12},30] (* Harvey P. Dale, Feb 17 2016 *)
a[n_]:= ChebyshevU[n, 6]; (* Michael Somos, Jun 29 2019 *)

Vec(1/(1-12*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

{a(n) = polchebyshev(n, 2, 6)}; \\ Michael Somos, Jun 29 2019

[lucas_number1(n,12,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n, 6) for n in (0..20)] # G. C. Greubel, Dec 23 2019

a(n) = S(n, 12) with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310.
a(n) = ((6+sqrt(35))^(n+1) - (6-sqrt(35))^(n+1))/(2*sqrt(35)).
a(n) = sqrt((A023038(n)^2 - 1)/35).
[A077417(n), a(n)] = the 2 X 2 matrix [1,10; 1,11]^(n+1) * [1,0]. - Gary W. Adamson, Mar 19 2008
a(n) = 12*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=12. - Philippe Deléham, Nov 17 2008
a(n) = b such that (-1)^(n+1)*Integral_{x=0..Pi/2} (sin((n+1)*x))/(6+cos(x)) dx = c + b*(log(2)+log(3)-log(7)). - Francesco Daddi, Aug 01 2011
a(n) = Sum_{k=0..n} A101950(n,k)*11^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012 (Start):
Product_{n>=0} (1 + 1/a(n)) = 1/5*(5 + sqrt(35)).
Product_{n>=1} (1 - 1/a(n)) = 1/12*(5 + sqrt(35)). (End)
E.g.f.: exp(6*x)*(35*cosh(sqrt(35)*x) + 6*sqrt(35)*sinh(sqrt(35)*x))/35. - Stefano Spezia, Dec 14 2022

Chebyshev comments and a(n) formulas from Wolfdieter Lang, Nov 08 2002

A005246 a(n) = (1 + a(n-1)*a(n-2))/a(n-3), a(0) = a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 1, 2, 3, 7, 11, 26, 41, 97, 153, 362, 571, 1351, 2131, 5042, 7953, 18817, 29681, 70226, 110771, 262087, 413403, 978122, 1542841, 3650401, 5757961, 13623482, 21489003, 50843527, 80198051, 189750626, 299303201, 708158977, 1117014753

Offset: 0

N. J. A. Sloane

For n >= 4 we have the linear recurrence a(n) = 4*a(n-2) - a(n-4). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 04 2001
Integer solutions to the equation floor(sqrt(3)*x^2) = x*floor(sqrt(3)*x). - Benoit Cloitre, Mar 18 2004
For n > 2, a(n) is the smallest integer > a(n-1) such that sqrt(3)*a(n) is closer to and greater than an integer than sqrt(3)*a(n-1). I.e., a(n) is the smallest integer > a(n-1) such that frac(sqrt(3)*a(n)) < frac(sqrt(3)*a(n-1)). - Benoit Cloitre, Jan 20 2003
The lower principal and intermediate convergents to 3^(1/2), beginning with 1/1, 3/2, 5/3, 12/7, 19/11, form a strictly increasing sequence; essentially, numerators=A143643 and denominators=A005246. - Clark Kimberling, Aug 27 2008
This sequence is a particular case of the following situation: a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n) where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z. The g.f. is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4); so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n). The general form of a(n) is given by: a(2*m) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (b-Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p) and a(2*m+1) = a*Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (a*b+q-a*Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p). - Richard Choulet, Feb 24 2010
a(n) for n > 1 are the integer square roots of (floor(m^2/3)+1), where the values of m are given by A143643.  Also see A082630. - Richard R. Forberg, Nov 14 2013
The a(n) = (1 + a(n-1)*a(n-2))/a(n-3) recursion has the Laurent property. If a(0), a(1), a(2) are variables, then a(n) is a Laurent polynomial (a rational function with a monomial denominator). - Michael Somos, Feb 27 2019

			G.f. = 1 + x + x^2 + 2*x^3 + 3*x^4 + 7*x^5 + 11*x^6 + 26*x^7 + 41*x^8 + ...
From _Richard Choulet_, Feb 24 2010: (Start)
a(4) = 4^2 - 4^0 - 3*4^1 = 3.
a(7) = 4^3 - 4*binomial(2,1) - 2*(4^2-1) = 26. (End)

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..500
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ..., [Annotated scanned copy]
Peter Cameron's Blog, The ADE affair, 3, Posted 23/06/2011.
T. Crilly, Double sequences of positive integers, Math. Gaz., 69 (1985), 263-271.
Enrica Duchi, Andrea Frosini, Renzo Pinzani and Simone Rinaldi, A Note on Rational Succession Rules, J. Integer Seqs., Vol. 6, 2003.
R. K. Guy, Letter to N. J. A. Sloane, Feb 1986
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Valentin Ovsienko, Serge Tabachnikov, Dual numbers, weighted quivers, and extended Somos and Gale-Robinson sequences, arXiv:1705.01623 [math.CO], 2017. See p. 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
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).

Bisections are A001835 and A001075.
Cf. A101265. Row sums of A211956.
Cf. A001353.

a005246 n = a005246_list !! n
a005246_list = 1 : 1 : 1 : map (+ 1) (zipWith div
   (zipWith (*) (drop 2 a005246_list) (tail a005246_list)) a005246_list)
-- Reinhard Zumkeller, Mar 07 2012

A005246:=-(-1-z+2*z**2+z**3)/(1-4*z**2+z**4); # Conjectured by Simon Plouffe in his 1992 dissertation. Gives sequence except for one of the leading 1's.
for q from 1 to 10 do :a:=1:b:=1:Q:=(a*b^2+q*b+a+q)/(a*b): for m from 0 to 15 do U(m):=sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(b-Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)):od: for m from 0 to 15 do V(m):=a*sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(a*b+q-a*Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)):od:for m from 0 to 15 do W(2*m):=U(m):od:for m from 0 to 14 do W(2*m+1):=V(m):od:seq(W(m),m=0..30):od; # Richard Choulet, Feb 24 2010

RecurrenceTable[{a[0]==a[1]==a[2]==1,a[n]==(1+a[n-1]a[n-2])/a[n-3]},a,{n,40}] (* Harvey P. Dale, May 28 2013 *)
a[n_] := Cosh[(n-1)*ArcSinh[1/Sqrt[2]]]*If[EvenQ[n], Sqrt[2/3], 1]; Table[a[n] // FunctionExpand, {n, 0, 34}] (* Jean-François Alcover, Dec 10 2014, after Peter Bala *)
a[ n_] := With[{m = If[ n < 0, 2 - n, n]}, SeriesCoefficient[ (1 + x - 3 x^2 - 2 x^3) / (1 - 4 x^2 + x^4), {x, 0, m}]]; (* Michael Somos, Feb 10 2017 *)

{a(n) = if( n<0, n = 2 - n); polcoeff((1 + x - 3*x^2 - 2*x^3) / (1 - 4*x^2 + x^4) + x * O(x^n), n)}; /* Michael Somos, Nov 15 2006 */

{a(n) = real( (2 + quadgen(12))^(n\2) * if( n%2, 1, 1 - 1 / quadgen(12)) )}; /* Michael Somos, May 24 2012 */

G.f.: (1 + x - 3*x^2 - 2*x^3)/(1 - 4*x^2 + x^4).
Limit_{n->oo} a(2n+1)/a(2n) = (3+sqrt(3))/3 = 1.5773502...; lim_{n->oo} a(2n)/a(2n-1) = (3+sqrt(3))/2 = 2.3660254.... - Benoit Cloitre, Aug 07 2002
A101265(n) = a(n)*a(n+1). - Franklin T. Adams-Watters, Apr 24 2006
a(n) = a(2-n) for all n in Z. - Michael Somos, Nov 15 2006
a(2*n + 1) = A001075(n). a(2*n) = A001835(n). a(2*n + 1) - a(2*n) = a(2*n + 2) - a(2*n + 1) = A001353(n). - Michael Somos, May 24 2012
For n > 2: a(n) = a(n-1) + Sum_{k=1..floor((n-1)/2)} a(2*k). - Reinhard Zumkeller, Dec 16 2007
From Richard Choulet, Feb 24 2010: (Start)
a(2*m) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*4^(m-2*p) - 3*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*4^(m-1-2*p).
a(2*m+1) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*4^(m-2*p) - 2*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*4^(m-1-2*p). (End)
From Tim Monahan, Jul 01 2011: (Start)
Closed form without extra leading 1: ((sqrt(6)+3)*(sqrt(2+sqrt(3))^n+(sqrt(2-sqrt(3))^n))+(3-sqrt(6))*((-sqrt(2+sqrt(3)))^n+(-sqrt(2-sqrt(3)))^n))/12.
Closed form with extra leading 1: ((6+3*sqrt(6)-2*sqrt(3)-3*sqrt(2))*(sqrt(2+sqrt(3))^n)+(6+3*sqrt(6)+2*sqrt(3)+3*sqrt(2))*(sqrt(2-sqrt(3))^n)+(6-3*sqrt(6)-2*sqrt(3)+3*sqrt(2))*((-sqrt(2+sqrt(3)))^n)+(6-3*sqrt(6)+2*sqrt(3)-3*sqrt(2))*((-sqrt(2-sqrt(3)))^n))/24. (End)
a(2*n+2) = Sum_{k = 0..n} 2^k*binomial(n+k,2*k); a(2*n+1) = Sum_{k = 0..n} n/(n+k)*2^k*binomial(n+k,2*k) for n >= 1. Row sums of A211956. - Peter Bala, May 01 2012
a(n) = ((sqrt(2)+sqrt(3)+(-1)^n*(sqrt(2)-sqrt(3)))*sqrt(2+(2-sqrt(3))^n*(2+ sqrt(3))-(-2+sqrt(3))*(2+ sqrt(3))^n))/(4*sqrt(3)). - Gerry Martens, Jun 06 2015
0 = a(n) - 2*a(n+1) + a(n+2) if n is even, 0 = a(n) - 3*a(n+1) + a(n+2) if n is odd for all n in Z. - Michael Somos, Feb 10 2017

More terms from Michael Somos, Aug 01 2001

A077421 Chebyshev sequence U(n,11)=S(n,22) with Diophantine property.

Original entry on oeis.org

1, 22, 483, 10604, 232805, 5111106, 112211527, 2463542488, 54085723209, 1187422368110, 26069206375211, 572335117886532, 12565303387128493, 275864339398940314, 6056450163389558415, 132966039255171344816

Offset: 0

Wolfdieter Lang, Nov 29 2002

b(n)^2 - 30*(2*a(n))^2 = 1 with the companion sequence b(n)=A077422(n+1).
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 22's along the main diagonal, and i's along the subdiagonal and the superdiagonal (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,...,21}. - Milan Janjic, Jan 25 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..700
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (22,-1).
Index entries for sequences related to Chebyshev polynomials.

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), this sequence (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:=11;; 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 23 2019

I:=[1, 22]; [n le 2 select I[n] else 22*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012

seq( simplify(ChebyshevU(n, 11)), n=0..20); # G. C. Greubel, Dec 23 2019

Table[GegenbauerC[n, 1, 11], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
CoefficientList[Series[1/(1-22x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Dec 24 2012 *)
ChebyshevU[Range[21] -1, 11] (* G. C. Greubel, Dec 23 2019 *)

vector( 21, n, polchebyshev(n-1, 2, 11) ) \\ G. C. Greubel, Dec 23 2019

[lucas_number1(n,22,1) for n in range(1,20)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n,11) for n in (0..20)] # G. C. Greubel, Dec 23 2019

a(n) = 22*a(n-1) - a(n-1), a(-1)=0, a(0)=1.
a(n) = S(n, 22) with S(n, x) := U(n, x/2) Chebyshev's polynomials of the 2nd kind. See A049310.
a(n) = (ap^(n+1) - am^(n+1))/(ap - am) with ap := 11+2*sqrt(30) and am := 11-2*sqrt(30).
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)*22^(n-2*k).
a(n) = sqrt((A077422(n+1)^2-1)/30)/2.
G.f.: 1/(1-22*x+x^2). - Philippe Deléham, Nov 18 2008
a(n) = Sum_{k, 0<=k<=n} A101950(n,k)*21^k. - Philippe Deléham, Feb 10 2012
Product {n >= 0} (1 + 1/a(n)) = 1/5*(5 + sqrt(30)). - Peter Bala, Dec 23 2012
Product {n >= 1} (1 - 1/a(n)) = 1/11*(5 + sqrt(30)). - Peter Bala, Dec 23 2012

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)

A075843 Numbers k such that 99*k^2 + 1 is a square.

Original entry on oeis.org

0, 1, 20, 399, 7960, 158801, 3168060, 63202399, 1260879920, 25154396001, 501827040100, 10011386405999, 199725901079880, 3984506635191601, 79490406802752140, 1585823629419851199, 31636982181594271840

Offset: 0

Gregory V. Richardson, Oct 14 2002

From Wolfdieter Lang, Nov 08 2002: (Start)
Chebyshev's polynomials U(n,x) evaluated at x=10.
The a(n) give all (unsigned, integer) solutions of Pell equation b(n)^2 - 99*a(n)^2 = +1 with b(n)= A001085(n). (End)
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 20's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imagianry 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,...,19}. - Milan Janjic, Jan 25 2015

A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Vincenzo Librandi, Table of n, a(n) for n = 0..700
Tanya Khovanova, Recursive Sequences
J. J. O'Connor and E. F. Robertson, Pell's Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (20,-1).

Cf. A001084.
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), this sequence (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:=10;; 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 22 2019

I:=[0,1]; [n le 2 select I[n] else 20*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 24 2012

seq( simplify(ChebyshevU(n-1, 10)), n=0..20); # G. C. Greubel, Dec 22 2019

Table[GegenbauerC[n-1, 1, 10], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
CoefficientList[Series[x/(1-20x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Dec 24 2012 *)
ChebyshevU[Range[22] -2, 10] (* G. C. Greubel, Dec 22 2019 *)
LinearRecurrence[{20,-1},{0,1},20] (* Harvey P. Dale, Dec 03 2023 *)

vector( 22, n, polchebyshev(n-2, 2, 10) ) \\ G. C. Greubel, Dec 22 2019

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

[chebyshev_U(n-1,10) for n in (0..20)] # G. C. Greubel, Dec 22 2019

a(n) = ((10+3*sqrt(11))^n - (10-3*sqrt(11))^n) / (6*sqrt(11)).
a(n) = 20*a(n-1) - a(n-2), n>=1, a(0)=0, a(1)=1.
a(n) = S(n-1, 20), with S(n, x) := U(n, x/2), Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
G.f.: x/(1 - 20*x + x^2).
a(n) = sqrt((A001085(n)^2 - 1)/99).
Lim_{n->inf.} a(n)/a(n-1) = 10 + 3*sqrt(11).
a(n+1) = Sum_{k=0..n} A101950(n,k)*19^k. - Philippe Deléham, Feb 10 2012
Product_{n>=1} (1 + 1/a(n)) = 1/3*(3 + sqrt(11)). - Peter Bala, Dec 23 2012
Product_{n>=2} (1 - 1/a(n)) = 3/20*(3 + sqrt(11)). - Peter Bala, Dec 23 2012

cp's OEIS Frontend

A077412 Chebyshev U(n,x) polynomial evaluated at x=8.

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A061278 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

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A190958 a(n) = 2*a(n-1) - 10*a(n-2), with a(0) = 0, a(1) = 1.

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A029547 Expansion of g.f. 1/(1 - 34*x + x^2).

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A052530 a(n) = 4*a(n-1) - a(n-2), with a(0) = 0, a(1) = 2.

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A061278 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

A190958 a(n) = 2a(n-1) - 10a(n-2), with a(0) = 0, a(1) = 1.