A003499 -id:A003499 - cp's OEIS frontend

A077445 Numbers k such that (k^2 - 8)/2 is a square.

4, 20, 116, 676, 3940, 22964, 133844, 780100, 4546756, 26500436, 154455860, 900234724, 5246952484, 30581480180, 178241928596, 1038870091396, 6054978619780, 35291001627284, 205691031143924, 1198855185236260

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "(k^2 - 8)/2 is a square" is a version of the generalized Pell Equation "x^2 - D*y^2 = C".
a(n)^2 - 2*A077444(n) = 8.
From Wolfdieter Lang, Jan 18 2013: (Start)
4*(1-z)/(1-6*z+z^2) = Sum_{n>=0} a(n+1)*z^n is the formal power series for tan(4*x)/tan(x) if one lets
  z = (tan(x))^2. For the numerator and denominator of this o.g.f. see A034867 and A034839, respectively. Convergence holds for 0 <= z < 3 - 2*sqrt(2), approximately 0.1715728753. This means for |x| < Pi/8, approximately 0.3926990818.
See also the o.g.f. given by Johannes W. Meijer, Aug 01 2010, in the formula section of A001653 = (this sequence)/4.
(End)
Positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 64 = 0. - Colin Barker, Feb 13 2014

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 = 1..1000
J. J. O'Connor and E. F. Robertson, Pell's Equation
Tanya Khovanova, Recursive Sequences
Klaus Nagel, Length of cycles in bijection of 2 quadratic grids rotated by Pi/6 relative to each other, (2009).
Eric Weisstein's World of Mathematics, Pell Equation
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A001653, A003499, A075870.

CoefficientList[Series[4 (1 - x)/(1 - 6 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 14 2014 *)

a(n)=if(n<1,0,subst(poltchebi(n)+poltchebi(n-1),x,3))

a(n) = (((3+2*sqrt(2))^n + (3-2*sqrt(2))^n) + ((3+2*sqrt(2))^(n-1) + (3-2*sqrt(2))^(n-1))) / 2.
a(n) = 6*a(n-1) - a(n-2) = 4*A001653(n).
G.f.: 4*(x-x^2)/(1-6*x+x^2).
With a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = sqrt(2)*(a^((2*n-1)/2) + b^((2*n-1)/2)). a(n) = sqrt(2*A003499(2*n-1)+4). - Mario Catalani (mario.catalani(AT)unito.it), Mar 24 2003
a(n) = (A003499(n+1) + A003499(n))/2. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
a(n) = (2 + sqrt(2))*(3 + 2*sqrt(2))^n + (2 - sqrt(2))*(3- 2*sqrt(2))^n. - Antonio Alberto Olivares, Feb 23 2006
a(n) = 2*A075870(n). - Bruno Berselli, Nov 27 2013
G.f.: 2*Q(0)*x*(1-x)/(1-3*x), where Q(k) = 1 + 1/( 1 - x*(8*k-9)/( x*(8*k-1) - 3/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 10 2013

A103999 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2M x 2N Klein bottle.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 16, 34, 1, 1, 54, 196, 198, 1, 1, 196, 1666, 2704, 1154, 1, 1, 726, 16384, 64152, 37636, 6726, 1, 1, 2704, 171394, 1844164, 2549186, 524176, 39202, 1, 1, 10086, 1844164, 57523158, 220581904, 101757654, 7300804, 228486, 1

Offset: 0

Ralf Stephan, Feb 26 2005

			Array begins:
  1,   1,     1,        1,           1,             1,                1, ...
  1,   6,    34,      198,        1154,          6726,            39202, ...
  1,  16,   196,     2704,       37636,        524176,          7300804, ...
  1,  54,  1666,    64152,     2549186,     101757654,       4064620168, ...
  1, 196, 16384,  1844164,   220581904,   26743369156,    3252222705664, ...
  1, 726,171394, 57523158, 21050622914, 7902001927776, 2988827208115522, ...

Cliff, Danny and Zoe Stoll, About Klein bottles
W. T. Lu and F. Y. Wu, Dimer statistics on the Moebius strip and the Klein bottle, arXiv:cond-mat/9906154 [cond-mat.stat-mech], 1999.

Rows include A003499, A067902+2. Columns include A003500+2.
Main diagonal gives A340557.
Cf. A099390, A103997.

T[m_, n_] := Product[4 Sin[(4k-1) Pi/(4n)]^2 + 4 Cos[j Pi/(2m+1)]^2, {j, 1, m}, {k, 1, n}] // Round;
Table[T[m-n, n], {m, 0, 9}, {n, 0, m}] // Flatten (* Jean-François Alcover, Aug 20 2018 *)

default(realprecision, 120);
{T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*sin((4*a-1)*Pi/(4*n))^2+4*sin((2*b-1)*Pi/(2*k))^2)))} \\ Seiichi Manyama, Jan 11 2021

T(M, N) = Product_{m=1..M} Product_{n=1..N} ( 4sin(Pi*(4n-1)/(4N))^2 + 4sin(Pi*(2m-1)/(2M))^2 ).

A237599 Positive integers k such that x^2 - 6xy + y^2 + k = 0 has integer solutions.

Original entry on oeis.org

4, 7, 8, 16, 23, 28, 31, 32, 36, 47, 56, 63, 64, 68, 71, 72, 79, 92, 100, 103, 112, 119, 124, 127, 128, 136, 144, 151, 164, 167, 175, 184, 188, 191, 196, 199, 200, 207, 223, 224, 239, 248, 252, 256, 263, 271, 272, 279, 284, 287, 288, 292, 311, 316, 324, 328

Offset: 1

Colin Barker, Feb 10 2014

nonn

Nonnegative numbers of the form 8x^2 - y^2. - Jon E. Schoenfield, Jun 03 2022

			4 is in the sequence because x^2 - 6xy + y^2 + 4 = 0 has integer solutions, for example (x, y) = (1, 5).

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

Cf. A001653 (k = 4), A006452 (k = 7), A001541 (k = 8), A075870 (k = 16), A156066 (k = 23), A217975 (k = 28), A003499 (k = 32), A075841 (k = 36), A077443 (k = 56).
Cf. A031363, A084917, A237351, A237606, A237609, A237610.
For primes see A007522 and A141175.
For a list of sequences giving numbers and/or primes represented by binary quadratic forms, see the "Binary Quadratic Forms and OEIS" link.

A067902 a(n) = 14*a(n-1) - a(n-2); a(0) = 2, a(1) = 14.

Original entry on oeis.org

2, 14, 194, 2702, 37634, 524174, 7300802, 101687054, 1416317954, 19726764302, 274758382274, 3826890587534, 53301709843202, 742397047217294, 10340256951198914, 144021200269567502, 2005956546822746114, 27939370455248878094, 389145229826661547202, 5420093847118012782734

Offset: 0

Lekraj Beedassy, May 13 2003

Solves for x in x^2 - 3*y^2 = 4. [Complete nonnegative solutions are in A003500 and A052530. - Wolfdieter Lang, Sep 05 2021]
For n>0, a(n)+2 is the number of dimer tilings of a 4 X 2n Klein bottle (cf. A103999).
This is the Lucas sequence V(14,1). In addition to the comment above: If x = a(n) then y(n) = (a(n+1) - a(n-1))/24, n >= 1. - Klaus Purath, Aug 17 2021

			G.f. = 2 + 14*x + 194*x^2 + 2702*x^3 + 37634*x^4 + 524174*x^5 + ...

G. C. Greubel, Table of n, a(n) for n = 0..850
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (14,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A067900, A302332.
Row 2 * 2 of array A188644.
Cf. A001075, A001353, A003500, A052530.

m:=7;; a:=[2,14];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

[Floor((2+Sqrt(3))^(2*n)+(1+Sqrt(3))^(-n)): n in [0..19]]; // Vincenzo Librandi, Mar 31 2011

a := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(14) fi: 14*a(n-1)-a(n-2): end: for n from 0 to 30 do printf(`%d,`,a(n)) od:
seq( simplify(2*ChebyshevT(n, 7)), n=0..20); # G. C. Greubel, Dec 23 2019

a[0]=2; a[1]=14; a[n_]:= 14a[n-1] -a[n-2]; Table[a[n], {n,0,20}] (* Robert G. Wilson v, Jan 30 2004 *)
2*ChebyshevT[Range[21] -1, 7] (* G. C. Greubel, Dec 23 2019 *)

vector( 21, n, 2*polchebyshev(n-1, 1, 7) ) \\ G. C. Greubel, Dec 23 2019

[lucas_number2(n,14,1) for n in range(0,20)] # Zerinvary Lajos, Jun 26 2008

[2*chebyshev_T(n,7) for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: 2*(1-7*x)/(1-14*x+x^2). - N. J. A. Sloane, Nov 22 2006
a(n) = p^n + q^n, where p = 7 + 4*sqrt(3) and q = 7 - 4*sqrt(3). - Tanya Khovanova, Feb 06 2007
a(n) = 2*A011943(n+1). - R. J. Mathar, Sep 27 2014
From Peter Bala, Oct 16 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 7 - 4*sqrt(3). This sequence gives the partial denominators in the simple continued fraction expansion of 1 + F(alpha) = 2.07140228197873197080... = 2 + 1/(14 + 1/(194 + 1/(2702 + ...))). Cf. A005248.
12*Sum_{n >= 1} 1/(a(n) - 16/a(n)) = 1.
16*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 12/a(n)) = 1.
Series acceleration formula for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/12 - 16*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 16)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(7-4*sqrt(3)))^2 - 1 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499.
(End)
From Klaus Purath, Aug 17 2021: (Start)
a(n) = (a(n-1)*a(n-2) + 2688)/a(n-3), n >= 3.
a(n) = (a(n-1)^2 + 192)/a(n-2), n >= 2.
a(2*n) = A302332(n-1) + A302332(n), n >= 1.
a(2*n+1) = 14*A302332(n). (End)
a(n) = A003500(2*n) = S(2*n,4) - S(2*n-2, 4) = 2*T(2*n,2), for n >= 0, with Chebyshev S and T. S(n, 4) = A001353(n+1) and T(n, 2) = A001075(n). - Wolfdieter Lang, Sep 06 2021

A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

Original entry on oeis.org

2, 10, 98, 970, 9602, 95050, 940898, 9313930, 92198402, 912670090, 9034502498, 89432354890, 885289046402, 8763458109130, 86749292044898, 858729462339850, 8500545331353602, 84146723851196170, 832966693180608098, 8245520207954884810

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 11 2003

a(n+1)/a(n) converges to (5+sqrt(24)) = 9.8989794... a(0)/a(1)=2/10; a(1)/a(2)=10/98; a(2)/a(3)=98/970; a(3)/a(4)=970/9602; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.10102051... = 1/(5+sqrt(24)) = (5-sqrt(24)).
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 96 = 0. - Colin Barker, Feb 25 2014
A triangle whose sides are a(n) - 1, a(n) and a(n) + 1 is nearly Fleenor-Heronian since its area is the product of an integer and the square root of 2. See A003500. - Charlie Marion, Dec 18 2020

			a(4) = 9602 = 10*a(3) - a(2) = 10*970 - 98 = (5+sqrt(24))^4 + (5-sqrt(24))^4.

T. D. Noe, Table of n, a(n) for n = 0..200
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A005248, A006242, A036336, A086927, A135927, A174503.

I:=[2,10]; [n le 2 select I[n] else 10*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Nov 07 2018

a[0] = 2; a[1] = 10; a[n_] := 10a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{10,-1}, {2,10}, 30] (* G. C. Greubel, Nov 07 2018 *)

polsym(x^2 - 10*x + 1,20) \\ Charles R Greathouse IV, Jun 11 2011

{a(n) = 2 * real( (5 + 2 * quadgen(24))^n )}; /* Michael Somos, Feb 25 2014 */

[lucas_number2(n,10,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.
G.f.: (2-10*x)/(1-10*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 5 - sqrt(24). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.09989 80642 72052 68138 ... = 2 + 1/(10 + 1/(98 + 1/(970 + ...))).
Also F(-alpha) = 0.89989 78538 78393 34715 ... has the continued fraction representation 1 - 1/(10 - 1/(98 - 1/(970 - ...))) and the simple continued fraction expansion 1/(1 + 1/((10-2) + 1/(1 + 1/((98-2) + 1/(1 + 1/((970-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((10^2-4) + 1/(1 + 1/((98^2-4) + 1/(1 + 1/((970^2-4) + 1/(1 + ...))))))). Cf. A174503 and A005248. (End)
a(-n) = a(n). - Michael Somos, Feb 25 2014
From Peter Bala, Oct 16 2019: (Start)
8*Sum_{n >= 1} 1/(a(n) - 12/a(n)) = 1.
12*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 8/a(n)) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/8 - 12*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 12))  and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/12 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 8)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(5-sqrt(24)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(sqrt(24)-5))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499. (End)
E.g.f.: 2*exp(5*x)*cosh(2*sqrt(6)*x). - Stefano Spezia, Oct 18 2019
From Peter Bala, Mar 29 2022: (Start)
a(n) = 2*T(n,5), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(2^n) = A135927(n+1) and a(3^n) = A006242(n+1), both for n >= 0. (End)

More terms from Colin Barker, Feb 25 2014

A097733 Pell equation solutions (7b(n))^2 - 2(5a(n))^2 = -1 with b(n)=A097732(n), n >= 0. Note that D=50=25^2 is not squarefree.

Original entry on oeis.org

1, 197, 39005, 7722793, 1529074009, 302748930989, 59942759261813, 11868363584907985, 2349876047052519217, 465263588952813896981, 92119840736610099083021, 18239263202259846804541177

Offset: 0

Wolfdieter Lang, Aug 31 2004

			(x,y) = (7,1), (1393,197), (275807,39005), ... give the positive integer solutions to x^2 - 50*y^2 =-1.

Indranil Ghosh, Table of n, a(n) for n = 0..434
Tanya Khovanova, Recursive Sequences
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
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 (198,-1).

Cf. A097731 for S(n, 198).
Row 7 of array A188647.
Cf. A000129, A100047.

a:=[1,197];; for n in [3..20] do a[n]:=198*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

I:=[1,197]; [n le 2 select I[n] else 198*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Aug 01 2019

LinearRecurrence[{198, -1},{1, 197},20] (* Ray Chandler, Aug 11 2015 *)

my(x='x+O('x^20)); Vec((1-x)/(1-198*x+x^2)) \\ G. C. Greubel, Aug 01 2019

((1-x)/(1-198*x+x^2)).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

a(n) = S(n, 2*99) - S(n-1, 2*99) = T(2*n+1, 5*sqrt(2))/(5*sqrt(2)), with Chebyshev polynomials of the 2nd and first kind. See A049310 for the triangle of S(n, x)= U(n, x/2) coefficients. S(-1, x) := 0 =: U(-1, x); and A053120 for the T-triangle.
a(n) = ((-1)^n)*S(2*n, 14*i) with the imaginary unit i and Chebyshev polynomials S(n, x) with coefficients shown in A049310.
G.f.: (1-x)/(1-198*x+x^2).
a(n) = 198*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=197. - Philippe Deléham, Nov 18 2008
a(n) = k^n + k^(-n) - a(n-1) = A003499(3n) - a(n-1), where k = (sqrt(2)+1)^6 = 99 + 70*sqrt(2) and a(0)=1. - Charles L. Hohn, Apr 05 2011
From Peter Bala, Mar 23 2015: (Start)
a(n) = ( Pell(6*n + 6 - 2*k) - Pell(6*n + 2*k) )/( Pell(6 - 2*k) - Pell(2*k) ), for k an arbitrary integer.
a(n) = ( Pell(6*n + 6 - 2*k - 1) + Pell(6*n + 2*k + 1) )/( Pell(6 - 2*k - 1) + Pell(2*k + 1) ), for k an arbitrary integer.
The aerated sequence (b(n))n>=1 = [1, 0, 197, 0, 39005, 0, 7722793, 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 = -200, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. (End)
Sum_{n >= 1} 1/( a(n) - 1/a(n) ) = 1/196. - Peter Bala, Mar 26 2015

A065113 Sum of the squares of the a(n)-th and the (a(n)+1)st triangular numbers (A000217) is a perfect square.

Original entry on oeis.org

6, 40, 238, 1392, 8118, 47320, 275806, 1607520, 9369318, 54608392, 318281038, 1855077840, 10812186006, 63018038200, 367296043198, 2140758220992, 12477253282758, 72722761475560, 423859315570606, 2470433131948080, 14398739476117878, 83922003724759192

Offset: 1

Robert G. Wilson v, Nov 12 2001

The sequence of square roots of the sum of the squares of the n-th and the (n+1)st triangular numbers is A046176.

			T6 = 21 and T7 = 28, 21^2 + 28^2 = 441 + 784 = 1225 = 35^2.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
M. Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT] (2008)
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001652, A002315, A003499 (first differences), A065651.

CoefficientList[ Series[2*(x - 3)/(-1 + 7x - 7x^2 + x^3), {x, 0, 24} ], x]
LinearRecurrence[{7,-7,1},{6,40,238},41] (* Harvey P. Dale, Dec 27 2011 *)

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

Vec(2*x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^40)) \\ Colin Barker, Mar 05 2016

a(n) = 2*A001652(n) = -1 + A002315(n).
a(n) - a(n-1) = A003499(n).
From Michael Somos, Apr 07 2003: (Start)
G.f.: 2*x*(3-x)/((1-6*x+x^2)*(1-x)).
a(n) = 6*a(n-1) - a(n-2) + 4.
a(-1-n) = -a(n) - 2. (End)
a(1)=6, a(2)=40, a(3)=238, a(n) = 7*a(n-1)-7*a(n-2)+a(n-3). - Harvey P. Dale, Dec 27 2011
a(n)^2 + (a(n)+2)^2 = A075870(n+1)^2 = A165518(n+1). - Joerg Arndt, Feb 15 2012
a(n) = (-2-(3-2*sqrt(2))^n*(-1+sqrt(2))+(1+sqrt(2))*(3+2*sqrt(2))^n)/2. - Colin Barker, Mar 05 2016
From Klaus Purath, Sep 05 2021: (Start)
(a(n+1) - a(n) - a(n-1) + a(n-2))/8 = A005319(n), for n >= 3.
((a(n) - a(n-1))^2)/2 - 2 = A005319(n)^2 = 2*A132592(n), for n>= 2.
a(n) = A265278(2*n+1).
a(n) = A293004(2*n+1).
a(n) = A213667(2*n).
a(n) = Sum_{k=1..n} A003499(k). (End)

A081554 a(n) = sqrt(2)( (3+2sqrt(2))^n - (3-2*sqrt(2))^n ).

Original entry on oeis.org

0, 8, 48, 280, 1632, 9512, 55440, 323128, 1883328, 10976840, 63977712, 372889432, 2173358880, 12667263848, 73830224208, 430314081400, 2508054264192, 14618011503752, 85200014758320, 496582077046168, 2894292447518688

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003

Numbers m such that ceiling( sqrt(m*m/2) )^2 = 4 + m*m/2. - Ctibor O. Zizka, Nov 09 2009

Numbers m such that 2*m^2+16 is a square. - Bruno Berselli, Dec 17 2014

G. C. Greubel, Table of n, a(n) for n = 0..1000
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Tanya Khovanova, Recursive Sequences
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

m:=50; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(8*x/(1-6*x+x^2))); // G. C. Greubel, Aug 16 2018

a = 3 + 2Sqrt[2]; b = 3 - 2Sqrt[2]; Table[Simplify[Sqrt[2](a^n - b^n)], {n, 0, 25}]
CoefficientList[Series[8x/(1-6x+x^2),{x,0,40}],x]  (* Harvey P. Dale, Mar 11 2011 *)
Table[4 Fibonacci[2 n, 2], {n, 0, 50}] (* G. C. Greubel, Aug 16 2018 *)

x='x+O('x^50); concat([0], Vec(8*x/(1-6*x+x^2))) \\ G. C. Greubel, Aug 16 2018

a(n)^2 = 2*A003499(n)^2 - 8.
a(n) = 8*A001109(n).
G.f.: 8*x/(1-6*x+x^2). - Philippe Deléham, Nov 17 2008
a(0)=0, a(1)=8, a(n) = 6*a(n-1) - a(n-2) for n>1. - Philippe Deléham, Sep 19 2009
a(n) = 4*Pell(2*n) = 4*A000129(2*n). - G. C. Greubel, Aug 16 2018

A106329 Numbers k such that k^2 = 8*j^2 + 9.

Original entry on oeis.org

3, 9, 51, 297, 1731, 10089, 58803, 342729, 1997571, 11642697, 67858611, 395508969, 2305195203, 13435662249, 78308778291, 456417007497, 2660193266691, 15504742592649, 90368262289203, 526704831142569, 3069860724566211, 17892459516254697, 104284896372961971

Offset: 1

Pierre CAMI, Apr 29 2005

The ratio a(n)/(2*j(n)) tends to sqrt(2) as n increases.
After 3, first differences of A301383. - Bruno Berselli, Mar 22 2018
For n > 0, a(n+1) is the n-th almost Lucas-balancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

Colin Barker, Table of n, a(n) for n = 1..1000
Tanya Khovanova, Recursive Sequences
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A003499, A103328, A301383.

CoefficientList[Series[3 x (1 - 3 x)/(1 - 6 x + x^2), {x, 0, 23}], x] (* Michael De Vlieger, Nov 02 2020 *)

Vec((3-9*x)/(1-6*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Dec 28 2011

a(1)=3, a(2)=9 then a(n) = 6*a(n-1)-a(n-2).
G.f.: 3*x*(1 - 3*x)/(1 - 6*x + x^2). - Philippe Deléham, Nov 17 2008
a(n) = (3/2)*A003499(n-1).
a(n) = 3*((3-2*sqrt(2))^(n-1) + (3+2*sqrt(2))^(n-1))/2. - Colin Barker, Oct 13 2015
E.g.f.: 3*exp(3*x)*(3*cosh(2*sqrt(2)*x) - 2*sqrt(2)*sinh(2*sqrt(2)*x)) - 9. - Stefano Spezia, Nov 25 2022

A081555 a(n) = 6*a(n-1) - a(n-2) - 4, a(0)=3, a(1)=7.

Original entry on oeis.org

3, 7, 35, 199, 1155, 6727, 39203, 228487, 1331715, 7761799, 45239075, 263672647, 1536796803, 8957108167, 52205852195, 304278004999, 1773462177795, 10336495061767, 60245508192803, 351136554095047, 2046573816377475, 11928306344169799, 69523264248641315

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Mar 24 2003

2*(a(2*n+1) + 1) is a perfect square.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A003499, A051927.

a:=[3,7];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]-4; od; a; # G. C. Greubel, Aug 13 2019

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (3-14*x+7*x^2)/((1-x)*(1-6*x+x^2)) )); // G. C. Greubel, Aug 13 2019

seq(coeff(series((3-14*x+7*x^2)/((1-x)*(1-6*x+x^2)), x, n+1), x, n), n = 0 ..30); # G. C. Greubel, Aug 13 2019

a[n_]:= a[n] = 6*a[n-1] -a[n-2] -4; a[0] = 3; a[1] = 7; Table[a[n], {n, 0, 25}]
LinearRecurrence[{7,-7,1}, {3,7,35}, 30] (* G. C. Greubel, Aug 13 2019 *)

PARI
```
a(n)=1+2*real((3+quadgen(32))^n)
```
PARI
```
a(n)=1+2*subst(poltchebi(abs(n)),x,3)
```

a(n)=if(n<0,a(-n),1+polsym(1-6*x+x^2,n)[n+1])

def A081555_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((3-14*x+7*x^2)/((1-x)*(1-6*x+x^2))).list()
A081555_list(30) # G. C. Greubel, Aug 13 2019

a(n) = A051927(2n).
a(n) = A003499(n) + 1.
a(2n) + 1 = A003499(n)^2.
a(n) = (3 + 2*sqrt(2))^n + (3 - 2*sqrt(2))^n + 1.
G.f.: (3-14*x+7*x^2)/((1-x)*(1-6*x+x^2)).

cp's OEIS Frontend

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A067902 a(n) = 14*a(n-1) - a(n-2); a(0) = 2, a(1) = 14.

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A097733 Pell equation solutions (7*b(n))^2 - 2*(5*a(n))^2 = -1 with b(n)=A097732(n), n >= 0. Note that D=50=2*5^2 is not squarefree.

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A065113 Sum of the squares of the a(n)-th and the (a(n)+1)st triangular numbers (A000217) is a perfect square.

A097733 Pell equation solutions (7b(n))^2 - 2(5a(n))^2 = -1 with b(n)=A097732(n), n >= 0. Note that D=50=25^2 is not squarefree.

A081554 a(n) = sqrt(2)( (3+2sqrt(2))^n - (3-2*sqrt(2))^n ).