A174501 -id:A174501 - cp's OEIS frontend

A003499 a(n) = 6*a(n-1) - a(n-2), with a(0) = 2, a(1) = 6.

2, 6, 34, 198, 1154, 6726, 39202, 228486, 1331714, 7761798, 45239074, 263672646, 1536796802, 8957108166, 52205852194, 304278004998, 1773462177794, 10336495061766, 60245508192802, 351136554095046, 2046573816377474, 11928306344169798, 69523264248641314

Offset: 0

N. J. A. Sloane

Two times Chebyshev polynomials of the first kind evaluated at 3.
Also 2(a(2*n)-2) and a(2*n+1)-2 are perfect squares. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
Chebyshev polynomials of the first kind evaluated at 3, then multiplied by 2. - Michael Somos, Apr 07 2003
Also gives solutions > 2 to the equation x^2 - 3 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004
Output of Lu and Wu's formula for the number of perfect matchings of an m X n Klein bottle where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009
It appears that for prime P = 8*n +- 3, that a((P-1)/2) == -6 (mod P) and for all composites C = 8*n +- 3, there is at least one i < (C-1)/2 such that a(i) == -6 (mod P). Only a few of the primes P of the form 8*n +-3, e.g., 29, had such an i less than (P-1)/2. As for primes P = 8*n +- 1, it seems that the sum of the two adjacent terms, a((P-1)/2) and a((P+1)/2), is congruent to 8 (mod P). - Kenneth J Ramsey, Feb 14 2012 and Mar 05 2012
For n >= 1, a(n) is also the curvature of circles (rounded to the nearest integer) successively inscribed toward angle 90 degree of tangent lines, starting with a unit circle. The expansion factor is 5.828427... or 1/(3 - 2*sqrt(2)), which is also 3 + 2*sqrt(2) or A156035. See illustration in links. - Kival Ngaokrajang, Sep 04 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 32 = 0. - Colin Barker, Feb 08 2014

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 198.
Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002; pp. 480-481.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, 2001, Wiley, pp. 77-79.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E. Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.
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.
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence K_3.
S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Refik Keskin and Olcay Karaatli, Some New Properties of Balancing Numbers and Square Triangular Numbers, Journal of Integer Sequences, Vol. 15 (2012), #12.1.4.
Tanya Khovanova, Recursive Sequences
W. Lu and F. Y. Wu, Close-packed dimers on nonorientable surfaces, arXiv:cond-mat/0110035 [cond-mat.stat-mech], 2001-2002; Physics Letters A, 293(2002), 235-246. [From Sarah-Marie Belcastro, Jul 04 2009]
Kival Ngaokrajang, Illustration of initial terms
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
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
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 recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).

A081555(n) = 1 + a(n).
Bisection of A002203.
First row of array A103999.
Row 1 * 2 of array A188645. A174501.

a:=[2,6];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 16 2020

I:=[2,6]; [n le 2 select I[n] else 6*Self(n-1) -Self(n-2): n in [1..25]]; // G. C. Greubel, Jan 16 2020

R:=PowerSeriesRing(Integers(), 25); Coefficients(R!( (2-6*x)/(1 - 6*x + x^2) )); // Marius A. Burtea, Jan 16 2020

A003499:=-2*(-1+3*z)/(1-6*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

a[0]=2; a[1]=6; a[n_]:= 6a[n-1] -a[n-2]; Table[a[n], {n,0,25}] (* Robert G. Wilson v, Jan 30 2004 *)
Table[Tr[MatrixPower[{{6, -1}, {1, 0}}, n]], {n, 25}] (* Artur Jasinski, Apr 22 2008 *)
LinearRecurrence[{6, -1}, {2, 6}, 25] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2012 *)
CoefficientList[Series[(2-6x)/(1-6x+x^2), {x,0,25}], x] (* Vincenzo Librandi, Jun 07 2013 *)
(* From Eric W. Weisstein, Apr 17 2018 *)
Table[(3-2Sqrt[2])^n + (3+2Sqrt[2])^n, {n,0,25}]//Expand
Table[(1+Sqrt[2])^(2n) + (1-Sqrt[2])^(2n), {n,0,25}]//FullSimplify
Join[{2}, Table[Fibonacci[4n, 2]/Fibonacci[2n, 2], {n, 25}]]
2*ChebyshevT[Range[0, 25], 3] (* End *)

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

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

[lucas_number2(n,6,1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

G.f.: (2-6*x)/(1 - 6*x + x^2).
a(n) = (3+2*sqrt(2))^n + (3-2*sqrt(2))^n = 2*A001541(n).
For all sequence elements n, 2*n^2 - 8 is a perfect square. Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 06 2002
a(2*n)+2 is a perfect square, 2(a(2*n+1)+2) is a perfect square. a(n), a(n-1) and A077445(n), n > 0, satisfy the Diophantine equation x^2 + y^2 - 3*z^2 = -8. - Mario Catalani (mario.catalani(AT)unito.it), Mar 24 2003
a(n+1) is the trace of n-th power of matrix {{6, -1}, {1, 0}}. - Artur Jasinski, Apr 22 2008
a(n) = Product_{r=1..n} (4*sin^2((4*r-1)*Pi/(4*n)) + 4). [Lu/Wu] - Sarah-Marie Belcastro, Jul 04 2009
a(n) = (1 + sqrt(2))^(2*n) + (1 + sqrt(2))^(-2*n). - Gerson Washiski Barbosa, Sep 19 2010
For n > 0, a(n) = A001653(n) + A001653(n+1). - Charlie Marion, Dec 27 2011
For n > 0, a(n) = b(4*n)/b(2*n) where b(n) is the Pell sequence, A000129. - Kenneth J Ramsey, Feb 14 2012
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 3 - 2*sqrt(2). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.16585 37786 96882 80543 ... = 2 + 1/(6 + 1/(34 + 1/(198 + ...))). Cf. A174501.
Also F(-alpha) = 0.83251219269380007634 ... has the continued fraction representation 1 - 1/(6 - 1/(34 - 1/(198 - ...))) and the simple continued fraction expansion 1/(1 + 1/((6-2) + 1/(1 + 1/((34-2) + 1/(1 + 1/((198-2) + 1/(1 + ...))))))). Cf. A174501 and A003500.
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((6^2-4) + 1/(1 + 1/((34^2-4) + 1/(1 + 1/((198^2-4) + 1/(1 + ...))))))).
(End)
G.f.: G(0), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013
Inverse binomial transform of A228568 [Bhadouria]. - R. J. Mathar, Nov 10 2013
From Peter Bala, Oct 16 2019: (Start)
4*Sum_{n >= 1} 1/(a(n) - 8/a(n)) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 4/a(n)) = 1.
Series acceleration formulas for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/4 - 8*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 8)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 4*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 4)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(3-2*sqrt(2)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(2*sqrt(2)-3))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A067902.
(End)
E.g.f.: 2*exp(3*x)*cosh(2*sqrt(2)*x). - Stefano Spezia, Oct 18 2019
a(2*n)+2 = a(n)^2. - Greg Dresden and Shraya Pal, Jun 29 2021

A174500 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003500(n)) ), where A003500(n) = (2+sqrt(3))^n + (2-sqrt(3))^n.

Original entry on oeis.org

1, 2, 1, 12, 1, 50, 1, 192, 1, 722, 1, 2700, 1, 10082, 1, 37632, 1, 140450, 1, 524172, 1, 1956242, 1, 7300800, 1, 27246962, 1, 101687052, 1, 379501250, 1, 1416317952, 1, 5285770562, 1, 19726764300, 1, 73621286642, 1, 274758382272, 1

Offset: 1

Paul D. Hanna, Mar 20 2010

			Let L = Sum_{n>=1} 1/(n*A003500(n)) or, more explicitly,
L = 1/4 + 1/(2*14) + 1/(3*52) + 1/(4*194) + 1/(5*724) + 1/(6*2702) +...
so that L = 0.2937696594138291094177057532058145970820225289928...
then exp(L) = 1.3414748719687236691269115428250035920032300984596...
equals the continued fraction expansion given by this sequence:
exp(L) = [1;2,1,12,1,50,1,192,1,722,1,2700,1,10082,1,...]; i.e.,
exp(L) = 1 + 1/(2 + 1/(1 + 1/(12 + 1/(1 + 1/(50 + 1/(1 +...)))))).
Compare these partial quotients to A003500(n), n=1,2,3,...:
[4,14,52,194,724,2702,10084,37634,140452,524174,1956244,...].

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,5,0,-5,0,1).

Cf. A003500, A174501.

a[n_?OddQ] = 1; a[n_?EvenQ] := a[n] = 4*a[n-2] - a[n-4] + 4;  a[2] = 2; a[4] = 12; Table[a[n], {n, 1, 41}] (* Jean-François Alcover, May 15 2014, after the first conjecture *)

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((2+sqrt(3))^m+(2-sqrt(3))^m))));contfrac(exp(L))[n]}

a(2n-1) = 1, a(2n) = A003500(n) - 2, for n>=1 [conjecture].
From Peter Bala, Jan 04 2013: (Start)
The above conjectures are correct. The real number exp( Sum_{n>=1} 1/(n*A003500(n)) ) is equal to the infinite product F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) evaluated at x = 2 - sqrt(3). Ramanujan has given a continued fraction expansion for F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(N - sqrt(N^2 - 4))), N an integer greater than 3. The present case is when N = 4. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F((2 - sqrt(3))^k), k = 1, 2, 3, ...: if [1; c(1), 1, c(2), 1, c(3), 1, ...] denotes the present sequence then the simple continued fraction expansion of F((2 - sqrt(3))^k) is given by [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. (End)
a(n) = 5*a(n-2)-5*a(n-4)+a(n-6). G.f.: -x*(x^4+2*x^3-4*x^2+2*x+1) / ((x-1)*(x+1)*(x^4-4*x^2+1)). [Colin Barker, Jan 20 2013]

A174502 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086903(n)) ), where A086903(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.

Original entry on oeis.org

1, 6, 1, 60, 1, 486, 1, 3840, 1, 30246, 1, 238140, 1, 1874886, 1, 14760960, 1, 116212806, 1, 914941500, 1, 7203319206, 1, 56711612160, 1, 446489578086, 1, 3515205012540, 1, 27675150522246, 1, 217885999165440, 1, 1715412842801286, 1

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A086903(n)) or, more explicitly,
L = 1/8 + 1/(2*62) + 1/(3*488) + 1/(4*3842) + 1/(5*30248) +...
so that L = 0.1338200441271648228100625767561479963630539052200...
then exp(L) = 1.1431870779045667085973926071888878662387686835715...
equals the continued fraction given by this sequence:
exp(L) = [1;6,1,60,1,486,1,3840,1,30246,1,238140,1,...]; i.e.,
exp(L) = 1 + 1/(6 + 1/(1 + 1/(60 + 1/(1 + 1/(486 + 1/(1 +...)))))).
Compare these partial quotients to A086903(n), n=1,2,3,...:
[8,62,488,3842,30248,238142,1874888,14760962,116212808,914941502,...].

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,9,0,-9,0,1).

Cf. A086903, A174500, A174501, A174503, A174508.

LinearRecurrence[{0,9,0,-9,0,1},{1,6,1,60,1,486},50] (* Harvey P. Dale, Jun 09 2013 *)

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((4+sqrt(15))^m+(4-sqrt(15))^m))));contfrac(exp(L))[n]}

a(2n-2) = 1, a(2n-1) = A086903(n) - 2, for n>=1 [conjecture].
The above conjectures are correct. See the Bala link for details. - Peter Bala, Jan 08 2013
a(n) = 9*a(n-2)-9*a(n-4)+a(n-6). G.f.: -(x^4+6*x^3-8*x^2+6*x+1) / ((x-1)*(x+1)*(x^4-8*x^2+1)). [Colin Barker, Jan 20 2013]

A174507 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A085447(n)) ), where A085447(n) = (3+sqrt(10))^n + (3-sqrt(10))^n.

Original entry on oeis.org

1, 5, 37, 1, 233, 1441, 1, 8885, 54757, 1, 337433, 2079361, 1, 12813605, 78960997, 1, 486579593, 2998438561, 1, 18477210965, 113861704357, 1, 701647437113, 4323746327041, 1, 26644125399365, 164188498723237, 1, 1011775117738793

Offset: 0

Paul D. Hanna, Mar 21 2010

			Let L = Sum_{n>=1} 1/(n*A085447(n)) or, more explicitly,
L = 1/6 + 1/(2*38) + 1/(3*234) + 1/(4*1442) + 1/(5*8886) +...
so that L = 0.1814484777922995750614847484088330644558009487798...
then exp(L) = 1.1989527624251050123398509513177598419795554140316...
equals the continued fraction given by this sequence:
exp(L) = [1;5,37,1,233,1441,1,8885,54757,1,337433,...]; i.e.,
exp(L) = 1 + 1/(5 + 1/(37 + 1/(1 + 1/(233 + 1/(1441 + 1/(1 +...)))))).
Compare these partial quotients to A085447(n), n=1,2,3,...:
[6,38,234,1442,8886,54758,337434,2079362,12813606,78960998,...].

Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1.
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 2.
Index entries for linear recurrences with constant coefficients, signature (0,0,39,0,0,-39,0,0,1).

Cf. A085447, A174501, A174506, A174508.

LinearRecurrence[{0,0,39,0,0,-39,0,0,1},{1,5,37,1,233,1441,1,8885,54757},30] (* Harvey P. Dale, Aug 07 2016 *)

{a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((3+sqrt(10))^m+(3-sqrt(10))^m))));contfrac(exp(L))[n]}

a(3n-3) = 1, a(3n-2) = A085447(2n-1) - 1, a(3n-1) = A085447(2n) - 1, for n>=1 [conjecture].
a(n) = 39*a(n-3)-39*a(n-6)+a(n-9). G.f.: -(x^2-x+1) * (x^6 -6*x^5 -6*x^4 -2*x^3 +42*x^2 +6*x +1) / ((x-1)*(x^2+x+1)*(x^6-38*x^3+1)). [Colin Barker, Jan 20 2013]
From Peter Bala, Jan 25 2013: (Start)
The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A085447(n)) ) is equal to the infinite product F(x) := product {n >= 0} (1 + x^(4*n+3))/(1 - x^(4*n+1)) evaluated at x = sqrt(10) - 3. Ramanujan has given a continued fraction expansion for the product F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(sqrt(N^2 + 4) - N)), N a positive integer. The present case is when N = 6. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(10) - 3}^(2*k+1)), k = 1, 2, 3, ...: if [1; c(1), c(2), 1, c(3), c(4), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({sqrt(10) - 3}^(2*k+1)) is given by [1; c(2*k+1), c(2*(2*k+1)), 1, c(3*(2*k+1)), c(4*(2*k+1)), 1, ...].
(End)

cp's OEIS Frontend

A003499 a(n) = 6*a(n-1) - a(n-2), with a(0) = 2, a(1) = 6.

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A174500 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003500(n)) ), where A003500(n) = (2+sqrt(3))^n + (2-sqrt(3))^n.

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A174502 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086903(n)) ), where A086903(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.

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A174507 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A085447(n)) ), where A085447(n) = (3+sqrt(10))^n + (3-sqrt(10))^n.

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