A001109 -id:A001109 - cp's OEIS frontend

A029549 a(n + 3) = 35a(n + 2) - 35a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

0, 6, 210, 7140, 242556, 8239770, 279909630, 9508687656, 323015470680, 10973017315470, 372759573255306, 12662852473364940, 430164224521152660, 14612920781245825506, 496409142337836914550, 16863297918705209269200, 572855720093639278238256

Offset: 0

Don N. Page

Triangular numbers that are twice other triangular numbers. - Don N. Page
Triangular numbers that are also pronic numbers. These will be shown to have a Pythagorean connection in a paper in preparation. - Stuart M. Ellerstein (ellerstein(AT)aol.com), Mar 09 2002
In other words, triangular numbers which are products of two consecutive numbers. E.g., a(2) = 210: 210 is a triangular number which is the product of two consecutive numbers: 14 * 15. - Shyam Sunder Gupta, Oct 26 2002
Coefficients of the series giving the best rational approximations to sqrt(8). The partial sums of the series 3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(8) = 2 sqrt(2), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [2; 1, 4, 1], [2; 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1], [2; 1, 4, 1, 4, 1, 4, 1, 4, 1] and so forth. - Gene Ward Smith, Sep 30 2006
This sequence satisfy the same recurrence as A165518. - Ant King, Dec 13 2010
Intersection of A000217 and A002378.
This is the sequence of areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + 1,z(n)) with x(0) = 0, y(0) = 1, z(0) = 1, a(0) = 0 and x(1) = 3, y(1) = 4, z(1) = 5, a(1) = 6. - George F. Johnson, Aug 20 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..100
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Shyam Sunder Gupta Fascinating Triangular Numbers
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
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, 2021.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A123478, A123479, A123480, A123482, A075528, A082405 (first differences).
Cf. A000129, A001108, A002203, A009111, A245031.
Cf. A000217, A001109, A001652, A002315, A002378, A011900, A029546, A046090, A046729, A053141, A084159, A096979, A157259, A165518.

List([0..20], n-> (Lucas(2,-1, 4*n+2)[2] -6)/32 ); # G. C. Greubel, Jan 13 2020

a029549 n = a029549_list !! n
a029549_list = [0,6,210] ++
   zipWith (+) a029549_list
               (map (* 35) $ tail delta)
   where delta = zipWith (-) (tail a029549_list) a029549_list
-- Reinhard Zumkeller, Sep 19 2011

(makelist(binom(n,2),n,1,999999),intersection(%%,2*%%)) /* Bill Gosper, Feb 07 2010 */

R:=PowerSeriesRing(Integers(), 25); [0] cat Coefficients(R!(6/(1-35*x+35*x^2-x^3))); // G. C. Greubel, Jul 15 2018

A029549 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[0,6]) ;
    else
        34*procname(n-1)-procname(n-2)+6 ;
    end if;
end proc: # R. J. Mathar, Feb 05 2016

Table[Floor[(Sqrt[2] + 1)^(4n + 2)/32], {n, 0, 20} ] (* Original program from author, corrected by Ray Chandler, Jul 09 2015 *)
CoefficientList[Series[6/(1 - 35x + 35x^2 - x^3), {x, 0, 14}], x]
Intersection[#, 2#] &@ Table[Binomial[n, 2], {n, 999999}] (* Bill Gosper, Feb 07 2010 *)
LinearRecurrence[{35, -35, 1}, {0, 6, 210}, 20] (* Harvey P. Dale, Jun 06 2011 *)
(LucasL[4Range[20] - 2, 2] -6)/32 (* G. C. Greubel, Jan 13 2020 *)

concat(0,Vec(6/(1-35*x+35*x^2-x^3)+O(x^25))) \\ Charles R Greathouse IV, Jun 13 2013

[(lucas_number2(4*n+2, 2, -1) -6)/32 for n in (0..20)] # G. C. Greubel, Jan 13 2020

val triNums = (0 to 39999).map(n => (n * n + n)/2)
triNums.filter( % 2 == 0).filter(n => (triNums.contains(n/2))) // _Alonso del Arte, Jan 12 2020

G.f.: 6*x/(1 - 35*x + 35*x^2 - x^3) = 6*x /( (1-x)*(1 - 34*x + x^2) ).
a(n) = 6*A029546(n-1) = 2*A075528(n).
a(n) = -3/16 + ((3+2*sqrt(2))/32) *(17 + 12*sqrt(2))^n + ((3-2*sqrt(2))/32) *(17 - 12*sqrt(2))^n. - Gene Ward Smith, Sep 30 2006
From Bill Gosper, Feb 07 2010: (Start)
a(n) = (cosh((4*n + 2)*log(1 + sqrt(2))) - 3)/16.
a(n) = binomial(A001652(n) + 1, 2) = 2*binomial(A053141(n) + 1, 2). (End)
a(n) = binomial(A046090(n), 2) = A000217(A001652(n)). - Mitch Harris, Apr 19 2007, R. J. Mathar, Jun 26 2009
a(n) = ceiling((3 + 2*sqrt(2))^(2n + 1) - 6)/32 = floor((1/32) (1+sqrt(2))^(4n+2)). - Ant King, Dec 13 2010
Sum_{n >= 1} 1/a(n) = 3 - 2*sqrt(2) = A157259 - 4. - Ant King, Dec 13 2010
a(n) = a(n - 1) + A001109(2n). - Charlie Marion, Feb 10 2011
a(n+2) = 34*a(n + 1) - a(n) + 6. - Charlie Marion, Feb 11 2011
From George F. Johnson, Aug 20 2012: (Start)
a(n) = ((3 + 2*sqrt(2))^(2*n + 1) + (3 - 2*sqrt(2))^(2*n + 1) - 6)/32.
8*a(n) + 1 = (A002315(n))^2, 4*a(n) + 1 = (A000129(2*n + 1))^2, 32*a(n)^2 + 12*a(n) + 1 are perfect squares.
a(n + 1) = 17*a(n) + 3 + 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
a(n - 1) = 17*a(n) + 3 - 3*sqrt((8*a(n) + 1)*(4*a(n) + 1)).
a(n - 1)*a(n + 1) = a(n)*(a(n) - 6), a(n) = A096979(2*n).
a(n) = (1/2)*A084159(n)*A046729(n) = (1/2)*A001652(n)*A046090(n).
Limit_{n->infinity} a(n)/a(n - 1) = 17 + 12*sqrt(2).
Limit_{n->infinity} a(n)/a(n - 2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Limit_{n->infinity} a(n)/a(n - r) = (17 + 12*sqrt(2))^r.
Limit_{n->infinity} a(n - r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r. (End)
a(n) = 3 * T( b(n) ) + (2*b(n) + 1)*sqrt( T( b(n) ) ) where b(n) = A001108(n) (indices of the square triangular numbers), T(n) = A000217(n) (the n-th triangular number). - Dimitri Papadopoulos, Jul 07 2017
a(n) = (Pell(2*n + 1)^2 - 1)/4 = (Q(4*n + 2) - 6)/32, where Q(n) are the Pell-Lucas numbers (A002203). - G. C. Greubel, Jan 13 2020
a(n) = A002378(A011900(n)-1) = A002378(A053141(n)). - Pontus von Brömssen, Sep 11 2024

Additional comments from Christian G. Bower, Sep 19 2002; T. D. Noe, Nov 07 2006; and others

Edited by N. J. A. Sloane, Apr 18 2007, following suggestions from Andrew S. Plewe and Tanya Khovanova

A046090 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values.

Original entry on oeis.org

1, 4, 21, 120, 697, 4060, 23661, 137904, 803761, 4684660, 27304197, 159140520, 927538921, 5406093004, 31509019101, 183648021600, 1070379110497, 6238626641380, 36361380737781, 211929657785304, 1235216565974041, 7199369738058940, 41961001862379597, 244566641436218640

Offset: 0

Eric W. Weisstein

Solution to a*(a-1) = 2b*(b-1) in natural numbers: a = a(n), b = b(n) = A011900(n).
n such that n^2 = (1/2)*(n+floor(sqrt(2)*n*floor(sqrt(2)*n))). - Benoit Cloitre, Apr 15 2003
Place a(n) balls in an urn, of which b(n) = A011900(n) are red; draw 2 balls without replacement; 2*Probability(2 red balls) = Probability(2 balls); this is equivalent to the Pell equation A(n)^2-2*B(n)^2 = -1 with a(n) = (A(n)+1)/2; b(n) = (B(n)+1)/2; and the fundamental solution (7;5) and the solution (3;2) for the unit form. - Paul Weisenhorn, Aug 03 2010
Find base x in which repdigit yy has a square that is repdigit zzzz, corresponding to Diophantine equation zzzz_x = (yy_x)^2; then, solution z = a(n) with x = A002315(n) and y = A001653(n+1) for n >= 1 (see Maurice Protat reference). - Bernard Schott, Dec 21 2022

			For n=4: a(4)=697; b(4)=493; 2*binomial(493,2)=485112=binomial(697,2). - _Paul Weisenhorn_, Aug 03 2010

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
Maurice Protat, Des Olympiades à l'Agrégation, De zzzz_x = (yy_x)^2 à Pell-Fermat, Problème 23, pp. 52-54, Ellipses, Paris, 1997.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 94-104.
L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Ron Knott, Pythagorean Triples and Online Calculators
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Eric Weisstein's World of Mathematics, Pythagorean Triple
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Other 2 sides are A001652 and A001653.
Cf. A001009, A001541, A001542, A011900, A046729, A053141, A084159, A084703, A156035.
See comments in A301383.
Cf. A001109, A001653, A002024, A002315, A029549.

a046090 n = a046090_list !! n
a046090_list = 1 : 4 : map (subtract 2)
   (zipWith (-) (map (* 6) (tail a046090_list)) a046090_list)
-- Reinhard Zumkeller, Jan 10 2012

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-3*x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018

Digits:=100: seq(round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^(n-1))/2)/2, n=0..20); # Paul Weisenhorn, Aug 03 2010

Join[{1},#+1&/@With[{c=3+2Sqrt[2]},NestList[Floor[c #]+3&,3,20]]] (* Harvey P. Dale, Aug 19 2011 *)
LinearRecurrence[{7,-7,1},{1,4,21},25] (* Harvey P. Dale, Apr 13 2012 *)
a[n_] := (2-ChebyshevT[n, 3]+ChebyshevT[n+1, 3])/4; Array[a, 21, 0] (* Jean-François Alcover, Jul 10 2016, adapted from PARI *)

a(n)=(2-subst(poltchebi(abs(n))-poltchebi(abs(n+1)),x,3))/4

x='x+O('x^30); Vec((1-3*x)/((1-6*x+x^2)*(1-x))) \\ G. C. Greubel, Jul 15 2018

a(n) = (-1+sqrt(1+8*b(n)*(b(n)+1)))/2 with b(n) = A011900(n). [corrected by Michel Marcus, Dec 23 2022]
a(n) = 6*a(n-1) - a(n-2) - 2, n >= 2, a(0) = 1, a(1) = 4.
a(n) = (A(n+1) - 3*A(n) + 2)/4 with A(n) = A001653(n).
A001652(n) = -a(-1-n).
From Barry E. Williams, May 03 2000: (Start)
G.f.: (1-3*x)/((1-6*x+x^2)*(1-x)).
a(n) = partial sums of A001541(n). (End)
From Charlie Marion, Jul 01 2003: (Start)
A001652(n)*A001652(n+1) + a(n)*a(n+1) = A001542(n+1)^2 = A084703(n+1).
Let a(n) = A001652(n), b(n) = this sequence and c(n) = A001653(n). Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. (End)
a(n) = 1/2 + ((1-2^(1/2))/4)*(3 - 2^(3/2))^n + ((1+2^(1/2))/4)*(3 + 2^(3/2))^n. - Antonio Alberto Olivares, Oct 13 2003
2*a(n) = 2*A084159(n) + 1 + (-1)^(n+1) = 2*A046729(n) + 1 - (-1)^(n+1). - Lekraj Beedassy, Jul 16 2004
a(n) = A001109(n+1) - A053141(n). - Manuel Valdivia, Apr 03 2010
From Paul Weisenhorn, Aug 03 2010: (Start)
a(n+1) = round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^n)/2);
b(n+1) = round((2+(10+7*sqrt(2))*(3+2*sqrt(2))^n)/4) = A011900(n+1).
(End)
a(n)*(a(n)-1)/2 = b(n)*b(n+1) and 2*a(n) - 1 = b(n) + b(n+1), where b(n) = A001109. - Kenneth J Ramsey, Apr 24 2011
T(a(n)) = A011900(n)^2 + A001109(n), where T(n) is the n-th triangular number. See also A001653. - Charlie Marion, Apr 25 2011
a(0)=1, a(1)=4, a(2)=21, a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - Harvey P. Dale, Apr 13 2012
Limit_{n->oo} a(n+1)/a(n) = 3 + 2*sqrt(2) = A156035. - Ilya Gutkovskiy, Jul 10 2016
a(n) = A001652(n)+1. - Dimitri Papadopoulos, Jul 06 2017
a(n) = (A002315(n) + 1)/2. - Bernard Schott, Dec 21 2022
E.g.f.: (exp(x) + exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/2. - Stefano Spezia, Mar 16 2024
a(n) = A002024(A029549(n))+1. - Pontus von Brömssen, Sep 11 2024

Additional comments from Wolfdieter Lang

Comment moved to A001653 by Claude Morin, Sep 22 2023

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

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

A015519 a(n) = 2a(n-1) + 7a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 11, 36, 149, 550, 2143, 8136, 31273, 119498, 457907, 1752300, 6709949, 25685998, 98341639, 376485264, 1441362001, 5518120850, 21125775707, 80878397364, 309637224677, 1185423230902, 4538307034543, 17374576685400

Offset: 0

Olivier Gérard

The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 8 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(8). - Cino Hilliard, Sep 25 2005

Pisano period lengths: 1, 2, 8, 4, 24, 8, 3, 8, 24, 24, 15, 8, 168, 6, 24, 16, 16, 24, 120, 24, ... . - R. J. Mathar, Aug 10 2012

John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,7).

The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.

[ n eq 1 select 0 else n eq 2 select 1 else 2*Self(n-1)+7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 23 2011

LinearRecurrence[{2,7},{0,1},30] (* Harvey P. Dale, Oct 09 2017 *)

a(n)=([0,1; 7,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, May 10 2016

[lucas_number1(n,2,-7) for n in range(0, 25)] # Zerinvary Lajos, Apr 22 2009

From Mario Catalani (mario.catalani(AT)unito.it), Apr 23 2003: (Start)
a(n) = a(n-1) + A083100(n-2), n>1.
A083100(n)/a(n+1) converges to sqrt(8). (End)
From Paul Barry, Jul 17 2003: (Start)
G.f.: x/ ( 1-2*x-7*x^2 ).
a(n) = ((1+2*sqrt(2))^n-(1-2*sqrt(2))^n)*sqrt(2)/8. (End)
E.g.f.: exp(x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - Paul Barry, Nov 20 2003
Second binomial transform is A000129(2n)/2 (A001109). - Paul Barry, Apr 21 2004
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-k-1, k)*(7/2)^k*2^(n-k-1). - Paul Barry, Jul 17 2004
a(n) = Sum_{k=0..n} binomial(n, 2*k+1)*8^k. - Paul Barry, Sep 29 2004
G.f.: G(0)*x/(2*(1-x)), where G(k)= 1 + 1/(1 - x*(8*k-1)/(x*(8*k+7) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013

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

A002965 Interleave denominators (A000129) and numerators (A001333) of convergents to sqrt(2).

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 5, 7, 12, 17, 29, 41, 70, 99, 169, 239, 408, 577, 985, 1393, 2378, 3363, 5741, 8119, 13860, 19601, 33461, 47321, 80782, 114243, 195025, 275807, 470832, 665857, 1136689, 1607521, 2744210, 3880899, 6625109, 9369319, 15994428, 22619537

Offset: 0

N. J. A. Sloane

Denominators of Farey fraction approximations to sqrt(2). The fractions are 1/0, 0/1, 1/1, 2/1, 3/2, 4/3, 7/5, 10/7, 17/12, .... See A082766(n+2) or A119016 for the numerators. "Add" (meaning here to add the numerators and add the denominators, not to add the fractions) 1/0 to 1/1 to make the fraction bigger: 2/1. Now 2/1 is too big, so add 1/1 to make the fraction smaller: 3/2, 4/3. Now 4/3 is too small, so add 3/2 to make the fraction bigger: 7/5, 10/7, ... Because the continued fraction for sqrt(2) is all 2's, it will always take exactly two terms here to switch from a number that's bigger than sqrt(2) to one that's less. A097545/A097546 gives the similar sequence for Pi. A119014/A119015 gives the similar sequence for e. - Joshua Zucker, May 09 2006
The principal and intermediate convergents to 2^(1/2) begin with 1/1, 3/2 4/3, 7/5, 10/7; essentially, numerators=A143607, denominators=A002965. - Clark Kimberling, Aug 27 2008
(a(2n)*a(2n+1))^2 is a triangular square. - Hugh Darwen, Feb 23 2012
a(2n) are the interleaved values of m such that 2*m^2+1 and 2*m^2-1 are squares, respectively; a(2n+1) are the interleaved values of their corresponding integer square roots. - Richard R. Forberg, Aug 19 2013
Coefficients of (sqrt(2)+1)^n are a(2n)*sqrt(2)+a(2n+1). - John Molokach, Nov 29 2015
Apart from the first two terms, this is the sequence of denominators of the convergents of the continued fraction expansion sqrt(2) = 1/(1 - 1/(2 + 1/(1 - 1/(2 + 1/(1 - ....))))). - Peter Bala, Feb 02 2017
Limit_{n->infinity} a(2n+1)/a(2n) = sqrt(2); lim_{n->infinity} a(2n)/a(2n-1) = (2+sqrt(2))/2. - Ctibor O. Zizka, Oct 28 2018

			The convergents are rational numbers given by the recurrence relation p/q -> (p + 2*q)/(p + q). Starting with 1/1, the next three convergents are (1 + 2*1)/(1 + 1) = 3/2, (3 + 2*2)/(3 + 2) = 7/5, and (7 + 2*5)/(7 + 5) = 17/12. The sequence puts the denominator first, so a(2) through a(9) are 1, 1, 2, 3, 5, 7, 12, 17. - _Michael B. Porter_, Jul 18 2016

C. Brezinski, History of Continued Fractions and Padé Approximants. Springer-Verlag, Berlin, 1991, p. 24.
Jay Kappraff, Musical Proportions at the Basis of Systems of Architectural Proportion both Ancient and Modern, in Volume I of K. Williams and M.J. Ostwald (eds.), Architecture and Mathematics from Antiquity to the Future, DOI 10.1007/978-3-319-00143-2_27, Springer International Publishing Switzerland 2015. See Eq. 32.7.
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).
Guelena Strehler, Chess Fractal, April 2016, p. 24.

T. D. Noe, Table of n, a(n) for n = 0..500
Damanvir Singh Binner, Proofs of Chappelon and Alfonsín Conjectures On Square Frobenius Numbers and its Relationship to Simultaneous Pell's Equations, arXiv:2112.15474 [math.NT], 2021.
Jonathan Chappelon and Jorge Luis Ramírez Alfonsín, The Square Frobenius Number, arXiv:2006.14219 [math.NT], 2020.
H. S. M. Coxeter, The role of intermediate convergents in Tait's explanation for phyllotaxis, J. Algebra 20 (1972), 167-175.
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Pierre Lamothe, En marge du problème des cercles tangents
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
Dave Rusin, Farey fractions on sci.math [Broken link]
Dave Rusin, Farey fractions on sci.math [Cached copy]
K. Williams, The sacred cult revisited: the pavement of the baptistery of San Giovanni, Florence, Math. Intellig., 16 (No. 2, 1994), 18-24.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).

Cf. A000129(n) = a(2n), A001333(n) = a(2n+1).

Cf. A001109, A155046.

a:=[0,1];; for n in [3..45] do a[n]:=a[n-1]+a[n-2-((n-1) mod 2)]; od; a; # Muniru A Asiru, Oct 28 2018

import Data.List (transpose)
a002965 n = a002965_list !! n
a002965_list = concat $ transpose [a000129_list, a001333_list]
-- Reinhard Zumkeller, Jan 01 2014

a=new Array(); a[0]=0; a[1]=1;
for (i=2;i<50;i+=2) {a[i]=a[i-1]+a[i-2];a[i+1]=a[i]+a[i-2];}
document.write(a); // Jon Perry, Sep 12 2012

I:=[0,1,1,1]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Nov 30 2015

A002965 := proc(n) option remember; if n <= 0 then 0; elif n <= 3 then 1; else 2*A002965(n-2)+A002965(n-4); fi; end;
A002965:=-(1+2*z+z**2+z**3)/(-1+2*z**2+z**4); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence except for two leading terms

LinearRecurrence[{0, 2, 0, 1}, {0, 1, 1, 1}, 42] (* Vladimir Joseph Stephan Orlovsky, Feb 13 2012 *)
With[{c=Convergents[Sqrt[2],20]},Join[{0,1},Riffle[Denominator[c], Numerator[c]]]] (* Harvey P. Dale, Oct 03 2012 *)

PARI
```
a(n)=if(n<4,n>0,2*a(n-2)+a(n-4))
```

x='x+O('x^100); concat(0, Vec((x+x^2-x^3)/(1-2*x^2-x^4))) \\ Altug Alkan, Dec 04 2015

a(n) = 2*a(n-2) + a(n-4) if n>3; a(0)=0, a(1)=a(2)=a(3)=1.
a(2*n) = a(2*n-1) + a(2*n-2) and a(2*n+1) = 2*a(2*n) - a(2*n-1).
G.f.: (x+x^2-x^3)/(1-2*x^2-x^4).
a(0)=0, a(1)=1, a(n) = a(n-1) + a(2*[(n-2)/2]). - Franklin T. Adams-Watters, Jan 31 2006
For n > 0, a(2*n) = a(2*n-1) + a(2*n-2) and a(2*n+1) = a(2*n) + a(2*n-2). - Jon Perry, Sep 12 2012
a(n) = (((sqrt(2) - 2)*(-1)^n + 2 + sqrt(2))*(1 + sqrt(2))^(floor(n/2)) - ((2 + sqrt(2))*(-1)^n -2 + sqrt(2))*(1 - sqrt(2))^(floor(n/2)))/8. - Ilya Gutkovskiy, Jul 18 2016
a(n) = a(n-1) + a(n-2-(n mod 2)); a(0)=0, a(1)=1. - Ctibor O. Zizka, Oct 28 2018

Thanks to Michael Somos for several comments which improved this entry.

A038761 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=9.

Original entry on oeis.org

1, 9, 53, 309, 1801, 10497, 61181, 356589, 2078353, 12113529, 70602821, 411503397, 2398417561, 13979001969, 81475594253, 474874563549, 2767771787041, 16131756158697, 94022765165141, 548004834832149, 3194006243827753, 18616032628134369, 108502189524978461

Offset: 0

Barry E. Williams, May 02 2000

Bisection of A048654. - Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 24 2004

This gives part of the (increasingly sorted) positive solutions y to the Pell equation x^2 - 2*y^2 = +7. For the x solutions see A038762. For the other part of solutions see A101386 and A253811. - Wolfdieter Lang, Feb 05 2015

			A038762(3)^2 - 2*a(4)^2 = 2547^2 - 2*1801^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
M. J. DeLeon, Pell's Equation and Pell Number Triples, Fib. Quart., 14(1976), pp. 456-460.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001109, A001541, A001542, A001652, A001653, A038762, A046090, A048654, A055997, A101386, A124124, A253811.

I:=[1, 9]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

a[0]:=1: a[1]:=9: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..19); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{6,-1},{1,9},40] (* Vincenzo Librandi, Nov 16 2011 *)

a(n)=([0,1; -1,6]^n*[1;9])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

a(n) = (9*((3+2*sqrt(2))^n -(3-2*sqrt(2))^n)-((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = sqrt(2*(A038762(n))^2-14)/2.
For n>1, a(n)-4a(n-1)=A001541(n)-A001542(n-2); e.g. 309-4*53=97=99-2. - Charlie Marion, Nov 12 2003
For n>0, a(n)=A046090(n)+A001653(n)+A001652(n-1)=A055997(n+1)+A001652(n-1); e.g., 309=120+169+20. - Charlie Marion, Oct 11 2006
G.f.: (1+3*x)/(1-6*x+x^2). - Philippe Deléham, Nov 03 2008
a(n) = third binomial transform of 1,6,8,48,64,384. - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009
a(n)^2 + 2^2 = A124124(2*n+1)^2 + (A124124(2*n+1)+1)^2. - Hermann Stamm-Wilbrandt, Aug 31 2014
a(n) = irrational part of z(n) = (3 + sqrt(2))*(3 + 2*sqrt(2))^n, n >= 0. z(n) gives only part of the general positive solutions to the Pell equation x^2 - 2*y^2 = 7. See the Nagell reference in A038762 on how to find z(n), and a comment above. - Wolfdieter Lang, Feb 05 2015
a(n) = S(n, 6) + 3*S(n-1, 6), n >= 0, with the Chebyshev S-polynomials evaluated at x=6. See S(n-1, 6) = A001109(n). - Wolfdieter Lang, Mar 30 2015
E.g.f.: exp(3*x)*(2*cosh(2*sqrt(2)*x) + 3*sqrt(2)*sinh(2*sqrt(2)*x))/2. - Stefano Spezia, Mar 16 2024

Edited: Replaced the unspecific Pell comment. Moved a formula from the comment section to the formula section. - Wolfdieter Lang, Feb 05 2015

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)

cp's OEIS Frontend

A029549 a(n + 3) = 35*a(n + 2) - 35*a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

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A046090 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values.

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

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A029549 a(n + 3) = 35a(n + 2) - 35a(n + 1) + a(n), with a(0) = 0, a(1) = 6, a(2) = 210.

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

A015519 a(n) = 2a(n-1) + 7a(n-2), with a(0) = 0, a(1) = 1.