A038761 -id:A038761 - cp's OEIS frontend

A129345 a(2n) = A001542(n+1), a(2n+1) = A038761(n+1); a Pellian-related sequence.

2, 9, 12, 53, 70, 309, 408, 1801, 2378, 10497, 13860, 61181, 80782, 356589, 470832, 2078353, 2744210, 12113529, 15994428, 70602821, 93222358, 411503397, 543339720, 2398417561, 3166815962, 13979001969, 18457556052, 81475594253, 107578520350, 474874563549

Offset: 0

Creighton Dement, Apr 10 2007

Summation of -a(n) and A129346 returns twice Pell numbers A000129 (apart from signs; starting from 2nd term of A000129).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A129346, A000129, A001542, A038761.

CoefficientList[Series[(2 + 9 x - x^3)/((x^2 + 2 x - 1) (x^2 - 2 x - 1)), {x, 0, 29}], x] (* Michael De Vlieger, May 26 2016 *)
LinearRecurrence[{0,6,0,-1},{2,9,12,53},30] (* Harvey P. Dale, Apr 29 2025 *)

Vec((2+9*x-x^3)/((x^2+2*x-1)*(x^2-2*x-1)) + O(x^40)) \\ Colin Barker, May 26 2016

G.f.: (2+9*x-x^3)/((x^2+2*x-1)*(x^2-2*x-1)).
From Colin Barker, May 26 2016: (Start)
a(n) = ((-1-sqrt(2))^(1+n)-(-1+sqrt(2))^(1+n)+(1-sqrt(2))^n*(-4+3*sqrt(2))+(1+sqrt(2))^n*(4+3*sqrt(2)))/(2*sqrt(2)).
a(n) = 6*a(n-2)-a(n-4) for n>3.
(End)

A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245, 12290092900109634, 71631910824649559, 417501372047787720

Offset: 0

N. J. A. Sloane

8*a(n)^2 + 1 = 8*A001110(n) + 1 = A055792(n+1) is a perfect square. - Gregory V. Richardson, Oct 05 2002
For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006
(a(n), b(n)) where b(n) = A082291(n) are the integer solutions of the equation 2*binomial(b,a) = binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003
This sequence gives the values of y in solutions of the Diophantine equation x^2 - 8y^2 = 1. It also gives the values of the product xy where (x,y) satisfies x^2 - 2y^2 = +-1, i.e., a(n) = A001333(n)*A000129(n). a(n) also gives the inradius r of primitive Pythagorean triangles having legs whose lengths are consecutive integers, with corresponding semiperimeter s = a(n+1) = {A001652(n) + A046090(n) + A001653(n)}/2 and area rs = A029549(n) = 6*A029546(n). - Lekraj Beedassy, Apr 23 2003 [edited by Jon E. Schoenfield, May 04 2014]
n such that 8*n^2 = floor(sqrt(8)*n*ceiling(sqrt(8)*n)). - Benoit Cloitre, May 10 2003
For n > 0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy, Sep 09 2003
a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series). - Creighton Dement, Dec 16 2004
Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch, Jun 19 2005
Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have altogether six D steps on the line y=x (shown between parentheses). - Emeric Deutsch, Jul 07 2005
Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n > 0, define C(n) to be the smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion, Sep 14 2005
Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance, t(20)=2*t(14)=210, so 6 is in the sequence. - Floor van Lamoen, Oct 13 2005
One half the bisection of the Pell numbers (A000129). - Franklin T. Adams-Watters, Jan 08 2006
Pell trapezoids: for n > 0, a(n) = (A000129(n-1)+A000129(n+1))*A000129(n)/2; see also A084158. - Charlie Marion, Apr 01 2006
Tested for 2 < p < 27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J Ramsey, May 16 2006
If 2^p - 1 is prime then M(p) divides a(2^(p-1)-1). - Kenneth J Ramsey, Jun 08 2006; comment corrected by Robert Israel, Mar 18 2007
If 8*n+5 and 8*n+7 are twin primes then their product divides a(4*n+3). - Kenneth J Ramsey, Jun 08 2006
If p is an odd prime, then if p == 1 or 7 (mod 8), then a((p-1)/2) == 0 (mod p) and a((p+1)/2) == 1 (mod p); if p == 3 or 5 (mod 8), then a((p-1)/2) == 1 (mod p) and a((p+1)/2) == 0 (mod p). Kenneth J Ramsey's comment about twin primes follows from this. - Robert Israel, Mar 18 2007
a(n)*(a(n+b) - a(b-2)) = (a(n+1)+1)*(a(n+b-1) - a(b-1)). This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ramsey, Oct 17 2007
For n < 0, let a(n) = -a(-n). Then (a(n+j) + a(k+j)) * (a(n+b+k+j) - a(b-j-2)) = (a(n+j+1) + a(k+j+1)) * (a(n+b+k+j-1) - a(b-j-1)). - Charlie Marion, Mar 04 2011
Sequence gives y values of the Diophantine equation: 0+1+2+...+x = y^2. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with aMohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with p < r then r = 3*p+4*q+1 and s = 2*p+3*q+1. - Mohamed Bouhamida, Sep 02 2009
a(n)/A002315(n) converges to cos^2(Pi/8) (see A201488). - Gary Detlefs, Nov 25 2009
Binomial transform of A086347. - Johannes W. Meijer, Aug 01 2010
If x=a(n), y=A055997(n+1) and z = x^2+y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
In general, if b(0)=1, b(1)=k and for n > 1, b(n) = 6*b(n-1) - b(n-2), then
  for n > 0, b(n) = a(n)*k-a(n-1); e.g.,
  for k=2, when b(n) = A038725(n), 2 = 1*2 - 0, 11 = 6*2 - 1,  64 = 35*2 - 6, 373 = 204*2 - 35;
  for k=3, when b(n) = A001541(n), 3 = 1*3 - 0, 17 = 6*3 - 1;  99 = 35*3 - 6; 577 = 204*3 - 35;
  for k=4, when b(n) = A038723(n), 4 = 1*4 - 0, 23 = 6*4 - 1; 134 = 35*4 - 6; 781 = 204*4 - 35;
  for k=5, when b(n) = A001653(n), 5 = 1*5 - 0, 29 = 6*5 - 1; 169 = 35*5 - 6; 985 = 204*5 - 35.
  See also A002315, A054488, A038761, A054489, A054490.
  - Charlie Marion, Dec 08 2010
See a Wolfdieter Lang comment on A001653 on a sequence of (u,v) values for Pythagorean triples (x,y,z) with x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, generated from (u(0)=1,v(0)=2), the triple (3,4,5), by a substitution rule given there. The present a(n) appears there as b(n). The corresponding generated triangles have catheti differing by one length unit. - Wolfdieter Lang, Mar 06 2012
a(n)*a(n+2k) + a(k)^2 and a(n)*a(n+2k+1) + a(k)*a(k+1) are triangular numbers. Generalizes description of sequence. - Charlie Marion, Dec 03 2012
a(n)*a(n+2k) + a(k)^2 is the triangular square A001110(n+k). a(n)*a(n+2k+1) + a(k)*a(k+1) is the triangular oblong A029549(n+k). - Charlie Marion, Dec 05 2012
From Richard R. Forberg, Aug 30 2013: (Start)
The squares of a(n) are the result of applying triangular arithmetic to the squares, using A001333 as the "guide" on what integers to square, as follows:
a(2n)^2 = A001333(2n)^2 * (A001333(2n)^2 - 1)/2;
a(2n+1)^2 = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2. (End)
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,5}. - Milan Janjic, Jan 25 2015
Panda and Rout call these "balancing numbers" and note that the period of the sequence modulo a prime p is the same as that modulo p^2 when p = 13, 31, 1546463. But these are precisely the p in A238736 such that p^2 divides A000129(p - (2/p)), where (2/p) is a Jacobi symbol. In light of the above observation by Franklin T. Adams-Watters that the present sequence is one half the bisection of the Pell numbers, i.e., a(n) = A000129(2*n)/2, it follows immediately that modulo a fixed prime p, or any power thereof, the period of a(n) is half that of A000129(n). - John Blythe Dobson, Mar 06 2015
The triangular number = square number identity Tri((T(n, 3) - 1)/2) = S(n-1, 6)^2 with Tri, T, and S given in A000217, A053120 and A049310, is the special case k = 1 of the k-family of identities Tri((T(n, 2*k+1) - 1)/2) = Tri(k)*S(n-1, 2*(2*k+1))^2, k >= 0, n >= 0, with S(-1, x) = 0. For k=2 see A108741(n) for S(n-1, 10)^2. This identity boils down to the identities S(n-1, 2*x)^2 = (T(2*n, x) - 1)/(2*(x^2-1)) and  2*T(n, x)^2 - 1 = T(2*n, x) with x = 2*k+1. - Wolfdieter Lang, Feb 01 2016
a(2)=6 is perfect. For n=2*k, k > 0, k not equal to 1, a(n) is a multiple of a(2) and since every multiple (beyond 1) of a perfect number is abundant, then a(n) is abundant. sigma(a(4)) = 504 > 408 = 2*a(4). For n=2*k+1, k > 0, a(n) mod 10 = A000012(n), so a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. sigma(a(5)) = 1260 < 2378 = 2*a(5). - Muniru A Asiru, Apr 14 2016
Behera & Panda call these the balancing numbers, and A001541 are the balancers. - Michel Marcus, Nov 07 2017
In general, a second-order linear recurrence with constant coefficients having a signature of (c,d) will be duplicated by a third-order recurrence having a signature of (x,c^2-c*x+d,-d*x+c*d). The formulas of Olivares and Bouhamida in the formula section which have signatures of (7,-7,1) and (5,5,-1), respectively, are specific instances of this general rule for x = 7 and x = 5. - Gary Detlefs, Jan 29 2021
Note that 6 is the largest triangular number in the sequence, because it is proved that 8 and 9 are the largest perfect powers which are consecutive (Catalan's conjecture). 0 and 1 are also in the sequence because they are also perfect powers and 0*1/2 = 0^2 and 8*9/2 = (2*3)^2. - Metin Sariyar, Jul 15 2021

			G.f. = x + 6*x^2 + 35*x^3 + 204*x^4 + 1189*x^5 + 6930*x^6 + 40391*x^7 + ...
6 is in the sequence since 6^2 = 36 is a triangular number: 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8. - _Michael B. Porter_, Jul 02 2016

Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, pp. 193, 197.
D. M. Burton, The History of Mathematics, McGraw Hill, (1991), p. 213.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 10.
P. Franklin, E. F. Beckenbach, H. S. M Coxeter, N. H. McCoy, K. Menger, and J. L. Synge, Rings And Ideals, No 8, The Carus Mathematical Monographs, The Mathematical Association of America, (1967), pp. 144-146.
A. Patra, G. K. Panda, and T. Khemaratchatakumthorn. "Exact divisibility by powers of the balancing and Lucas-balancing numbers." Fibonacci Quart., 59:1 (2021), 57-64; see B(n).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

Indranil Ghosh, Table of n, a(n) for n = 0..1304 (terms 0..200 from T. D. Noe)
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Irving Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E. Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See v_n in Lemma 7.9 p. 21.
Jean-Paul Allouche, Zeta-regularization of arithmetic sequences, EPJ Web of Conferences (2020) Vol. 244, 01008.
Dario Alpern for Diophantine equation a^4+b^3=c^2.
Kasper Andersen, Lisa 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.
Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, Joel Iiams, and Julia Peterson, Powers of Two as Sums of Two Balancing Numbers, Combinatorics, Graph Theory and Computing (SEICCGTC 2021) Springer Proc. Math. Stat., Vol 448, pp. 383-392. See p. 384.
Raymond A. Beauregard and Vladimir A. Dobrushkin, Powers of a Class of Generating Functions, Mathematics Magazine, Volume 89, Number 5, December 2016, pp. 359-363.
A. Behera and G. K. Panda, On the Square Roots of Triangular Numbers, Fib. Quart., 37 (1999), pp. 98-105.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Elwyn Berlekamp and Joe P. Buhler, Puzzle Column, Emissary, MSRI Newsletter, Fall 2005. Problem 1, (6 MB).
Kisan Bhoi and Prasanta Kumar Ray, On the Diophantine equation Bn1+Bn2=2^a1+2^a2+2^a3, arXiv:2212.06372 [math.NT], 2022.
Daniel Birmajer, Juan B. Gil, and Michael D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, example 12.
Alexander Bogomolny, There exist triangular numbers that are also squares
John C. Butcher, On Ramanujan, continued Fractions and an interesting number
Paula Catarino, Helena Campos, and Paulo Vasco, On some identities for balancing and cobalancing numbers, Annales Mathematicae et Informaticae, 45 (2015) pp. 11-24.
E. K. Çetinalp, N. Yilmaz, and Ö. Deveci, The balancing-like sequences in groups, Acta Univ. Apulensis Math. (2023) No. 73, 139-153. See p. 144.
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 301, 302, P_{13}).
Mahadi Ddamulira, Repdigits as sums of three balancing numbers, Mathematica Slovaca, (2019) hal-02405969.
Tomislav Doslic, Planar polycyclic graphs and their Tutte polynomials, Journal of Mathematical Chemistry, Volume 51, Issue 6, 2013, pp. 1599-1607.
D. B. Eperson, Triangular numbers, Math. Gaz., 47 (1963), 236-237.
Leonhard Euler, De solutione problematum diophanteorum per numeros integros, Par. 19.
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Bernadette Faye, Florian Luca, and Pieter Moree, On the discriminator of Lucas sequences, arXiv:1708.03563 [math.NT], 2017.
Morgan Fiebig, aBa Mbirika, and Jürgen Spilker, Period patterns, entry points, and orders in the Lucas sequences: theory and applications, arXiv:2408.14632 [math.NT], 2024. See p. 5.
Rigoberto Flórez, Robinson A. Higuita, and Antara Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
Aviezri S. Fraenkel, On the recurrence f(m+1)= b(m)*f(m)-f(m-1) and applications, Discrete Mathematics 224 (2000), pp. 273-279.
Robert Frontczak, A Note on Hybrid Convolutions Involving Balancing and Lucas-Balancing Numbers, Applied Mathematical Sciences, Vol. 12, 2018, No. 25, 1201-1208.
Robert Frontczak, Sums of Balancing and Lucas-Balancing Numbers with Binomial Coefficients, International Journal of Mathematical Analysis (2018) Vol. 12, No. 12, 585-594.
Robert Frontczak, Powers of Balancing Polynomials and Some Consequences for Fibonacci Sums, International Journal of Mathematical Analysis (2019) Vol. 13, No. 3, 109-115.
Robert Frontczak and Taras Goy, Additional close links between balancing and Lucas-balancing polynomials, arXiv:2007.14048 [math.NT], 2020.
Robert Frontczak and Taras Goy, More Fibonacci-Bernoulli relations with and without balancing polynomials, arXiv:2007.14618 [math.NT], 2020.
Robert Frontczak and Taras Goy, Lucas-Euler relations using balancing and Lucas-balancing polynomials, arXiv:2009.09409 [math.NT], 2020.
Robert Frontczak and Kalika Prasad, Balancing polynomials, Fibonacci numbers and some new series for $\pi$, Mediterranean Journal of Mathematics (2023) Vol. 20, Article number: 207.
Bill Gosper, The Triangular Squares, 2014.
H. Harborth, Fermat-like binomial equations, Applications of Fibonacci numbers, Proc. 2nd Int. Conf., San Jose/Ca., August 1986, 1-5 (1988).
Brian Hayes, Calculemus!, American Scientist, 96 (Sep-Oct 2008), 362-366.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Michael A. Jones, Proof Without Words: The Square of a Balancing Number Is a Triangular Number, The College Mathematics Journal, Vol. 43, No. 3 (May 2012), p. 212.
Vedat Kabasakal and Fatih Yılmaz, On balancing numbers and application to coding theory, Ch. 16, Recent Developments in Mathematics, Pegem Akademi (Turkiye, 2024), 200-209. See p. 201.
Refik Keskin and Olcay Karaatli, Some New Properties of Balancing Numbers and Square Triangular Numbers, Journal of Integer Sequences, Vol. 15 (2012), Article #12.1.4.
Omar Khadir, Kalman Liptai, and Laszlo Szalay, On the Shifted Product of Binary Recurrences, J. Int. Seq. 13 (2010), 10.6.1.
Tanya Khovanova, Recursive Sequences
Phil Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
Kalman Liptai, Fibonacci Balancing Numbers, Fib. Quart. 42 (4) (2004) 330-340.
Madras College, St Andrews, Square Triangular Numbers
aBa Mbirika, Janeè Schrader, and Jürgen Spilker, Pell and associated Pell braid sequences as GCDs of sums of k consecutive Pell, balancing, and related numbers, arXiv:2301.05758 [math.NT], 2023. See also J. Int. Seq. (2023) Vol. 26, Art. 23.6.4.
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
G. K. Panda, Sequence balancing and cobalancing numbers, Fib. Q., Vol. 45, No. 3 (2007), 265-271. See p. 266.
G. K. Panda and S. S. Rout, Periodicity of Balancing Numbers, Acta Mathematica Hungarica 143 (2014), 274-286.
G. K. Panda and Ravi Kumar Davala, Perfect Balancing Numbers, Fibonacci Quart. 53 (2015), no. 3, 261-264.
Ashish Kumar Pandey and B. K. Sharma, On Inequalities Related to a Generalized Euler Totient Function and Lucas Sequences, J. Int. Seq. (2023) Vol. 26, Art. 23.8.6.
Poo-Sung Park, Ramanujan's Continued Fraction for a Puzzle, College Mathematics Journal, 2005, 363-365.
Michael Penn, Balancing Numbers, Youtube video, 2020.
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
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
B. Polster and M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658 [math.HO], 2015.
Kalika Prasad, Munesh Kumari, Ritanjali Mohanty, and Hrishikesh Mahato, Spinor Algebra of k-Balancing and k-Lucas-Balancing Numbers, Journal of Algebra and Its Applications (2024).
Kalika Prasad, Munesh Kumari, and Jagmohan Tanti, Octonions and hyperbolic octonions with the k-balancing and k-Lucas balancing numbers, The Journal of Analysis, (2024), Vol. 32, No. 3, 1281-1296.
Kalika Prasad, Munesh Kumari, and Jagmohan Tanti, Generalized k-balancing and k-Lucas balancing numbers and associated polynomials, Kyungpook Mathematical Journal (2023), Vol. 63, No. 4, 539-550.
Kalika Prasad, Hrishikesh Mahato, and Munesh Kumari, Some properties of r-circulant matrices with k-balancing and k-Lucas balancing numbers, Boletín de la Sociedad Matemática Mexicana (2023), Vol. 29, No. 2, Article no. 44.
Helmut Prodinger, How to sum powers of balancing numbers efficiently, arXiv:2008.03916 [math.NT], 2020.
Rajesh Ram, Triangle Numbers that are Perfect Squares
K. J. Ramsey, Relation of Mersenne Primes To Square Triangular Numbers [edited by K. J. Ramsey, May 14 2011]
Kenneth Ramsay and Andras Erszegi, Relation of Square Triangular Numbers To Mersenne Primes, digest of 4 messages in Triangular_and_Fibonacci_Numbers Yahoo Group, May 15 - Jun 28, 2006.
Kenneth Ramsey, Generalized Proof re Square Triangular Numbers
Kenneth Ramsey, Generalized Proof re Square Triangular Numbers, digest of 2 messages in Triangular_and_Fibonacci_Numbers Yahoo group, May 27, 2005 - Oct 10, 2011.
Salah E. Rihane, Bernadette Faye, Florian Luca, and Alain Togbe, An exponential Diophantine equation related to the difference between powers of two consecutive Balancing numbers, arXiv:1811.03015 [math.NT], 2018.
A. Sandhya, Puzzle 4: A problem Srinivasa Ramanujan, the famous 20th century Indian Mathematician Solved
Sci.math Newsgroup, Square numbers which are triangular
Sci.math Newsgroup, Square numbers which are triangular [Cached copy]
R. A. Sulanke, Moments, Narayana numbers and the cut and paste for lattice paths.
R. A. Sulanke, Bijective recurrences concerning Schroeder paths, Electron. J. Combin. 5 (1998), Research Paper 47, 11 pp.
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, Merve Tayat, and Meltem E. Ozbek, The diophantine equation 8x^2-y^2+8x(1+t)+(2t+1)^2=0 and t-balancing numbers, ISRN Combinatorics, Volume 2014, Article ID 897834, 5 pages.
Bibhu Prasad Tripathy and Bijan Kumar Patel, On balancing and Lucas-balancing numbers expressible as product of two k-Fibonacci numbers, arXiv:2508.12238 [math.NT], 2025.
Eric Weisstein's World of Mathematics, Binomial coefficient.
Eric Weisstein's World of Mathematics, Square Triangular Number.
Eric Weisstein's World of Mathematics, Triangular Number.
Wikipedia, Triangular square number
Rick Young, Relevant quotation from biography of Ramanujan
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000217, A000290, A001108, A001542, A001653, A001850, A002315, A002965, A278310.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), this sequence (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).
Cf. A323182.

a:=[0,1];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Dec 18 2018

a001109 n = a001109_list !! n :: Integer
a001109_list = 0 : 1 : zipWith (-)
   (map (* 6) $ tail a001109_list) a001109_list
-- Reinhard Zumkeller, Dec 17 2011

[n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015

a[0]:=1: a[1]:=6: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n],n=0..26); # Emeric Deutsch
with (combinat):seq(fibonacci(2*n,2)/2, n=0..20); # Zerinvary Lajos, Apr 20 2008

Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{-1,6},#]}]&, {0,1}, 30]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
CoefficientList[Series[x/(1-6x+x^2),{x,0,30}],x]  (* Harvey P. Dale, Mar 23 2011 *)
LinearRecurrence[{6, -1}, {0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
a[ n_]:= ChebyshevU[n-1, 3]; (* Michael Somos, Sep 02 2012 *)
Table[Fibonacci[2n, 2]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
TrigExpand@Table[Sinh[2 n ArcCsch[1]]/(2 Sqrt[2]), {n, 0, 10}] (* Federico Provvedi, Feb 01 2021 *)

{a(n) = imag((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */

{a(n) = subst( poltchebi( abs(n+1)) - 3 * poltchebi( abs(n)), x, 3) / 8}; /* Michael Somos, Apr 07 2003 */

{a(n) = polchebyshev( n-1, 2, 3)}; /* Michael Somos, Sep 02 2012 */

is(n)=ispolygonal(n^2,3) \\ Charles R Greathouse IV, Nov 03 2016

[lucas_number1(n,6,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,3) for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: x / (1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n-1, 6) = U(n-1, 3) with U(n, x) Chebyshev's polynomials of the second kind. S(-1, x) := 0. Cf. triangle A049310 for S(n, x).
a(n) = sqrt(A001110(n)).
a(n) = A001542(n)/2.
a(n) = sqrt((A001541(n)^2-1)/8) (cf. Richardson comment).
a(n) = 3*a(n-1) + sqrt(8*a(n-1)^2+1). - R. J. Mathar, Oct 09 2000
a(n) = A000129(n)*A001333(n) = A000129(n)*(A000129(n)+A000129(n-1)) = ceiling(A001108(n)/sqrt(2)). - Henry Bottomley, Apr 19 2000
a(n) ~ (1/8)*sqrt(2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 05 2002
a(n) = ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) / (4*sqrt(2)). - Gregory V. Richardson, Oct 13 2002. Corrected for offset 0, and rewritten. - Wolfdieter Lang, Feb 10 2015
a(2*n) = a(n)*A003499(n). 4*a(n) = A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003
a(n) = floor((3+2*sqrt(2))^n/(4*sqrt(2))). - Lekraj Beedassy, Apr 23 2003
a(-n) = -a(n). - Michael Somos, Apr 07 2003
For n >= 1, a(n) = Sum_{k=0..n-1} A001653(k). - Charlie Marion, Jul 01 2003
For n > 0, 4*a(2*n) = A001653(n)^2 - A001653(n-1)^2. - Charlie Marion, Jul 16 2003
For n > 0, a(n) = Sum_{k = 0..n-1}((2*k+1)*A001652(n-1-k)) + A000217(n). - Charlie Marion, Jul 18 2003
a(2*n+1) = a(n+1)^2 - a(n)^2. - Charlie Marion, Jan 12 2004
a(k)*a(2*n+k) = a(n+k)^2 - a(n)^2; e.g., 204*7997214 = 40391^2 - 35^2. - Charlie Marion, Jan 15 2004
For j < n+1, a(k+j)*a(2*n+k-j) - Sum_{i = 0..j-1} a(2*n-(2*i+1)) = a(n+k)^2 - a(n)^2. - Charlie Marion, Jan 18 2004
From Paul Barry, Feb 06 2004: (Start)
a(n) = A000129(2*n)/2;
a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))*sqrt(2)/8;
a(n) = Sum_{i=0..n} Sum_{j=0..n} A000129(i+j)*n!/(i!*j!*(n-i-j)!)/2. (End)
E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - Paul Barry, Apr 21 2004
A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - Creighton Dement, Sep 16 2004
a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1)*2^(k-1). - Paul Barry, Oct 01 2004
a(n) = A001653(n+1) - A038723(n); (a(n)) = chuseq[J]( 'ii' + 'jj' + .5'kk' + 'ij' - 'ji' + 2.5e ), apart from initial term. - Creighton Dement, Nov 19 2004, modified by Davide Colazingari, Jun 24 2016
a(n+1) = Sum_{k=0..n} A001850(k)*A001850(n-k), self convolution of central Delannoy numbers. - Benoit Cloitre, Sep 28 2005
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3), a(1) = 0, a(2) = 1, a(3) = 6, n > 3. Also a(n) = ( (1 + sqrt(2) )^(2*n) - (1 - sqrt(2) )^(2*n) ) / (4*sqrt(2)). - Antonio Alberto Olivares, Oct 23 2003
a(n) = 5*(a(n-1) + a(n-2)) - a(n-3). - Mohamed Bouhamida, Sep 20 2006
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),3), see second formula. - Marcos Carreira, Dec 27 2006
The perfect median m(n) can be expressed in terms of the Pell numbers P() = A000129() by m(n) = P(n + 2) * (P(n + 2) + P(n + 1)) for n >= 0. - Winston A. Richards (ugu(AT)psu.edu), Jun 11 2007
For k = 0..n, a(2*n-k) - a(k) = 2*a(n-k)*A001541(n). Also, a(2*n+1-k) - a(k) = A002315(n-k)*A001653(n). - Charlie Marion, Jul 18 2007
[A001653(n), a(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = Sum_{k=0..n-1} 4^k*binomial(n+k,2*k+1). - Paul Barry, Apr 20 2009
a(n+1)^2 - 6*a(n+1)*a(n) + a(n)^2 = 1. - Charlie Marion, Dec 14 2010
a(n) = A002315(m)*A011900(n-m-1) + A001653(m)*A001652(n-m-1) - a(m) = A002315(m)*A053141(n-m-1) + A001653(m)*A046090(n-m-1) + a(m) with m < n; otherwise a(n) = A002315(m)*A053141(m-n) - A001653(m)*A011900(m-n) + a(m) = A002315(m)*A053141(m-n) - A001653(m)*A046090(m-n) - a(m) = (A002315(n) - A001653(n))/2. - Kenneth J Ramsey, Oct 12 2011
16*a(n)^2 + 1 = A056771(n). - James R. Buddenhagen, Dec 09 2011
A010054(A000290(a(n))) = 1. - Reinhard Zumkeller, Dec 17 2011
In general, a(n+k)^2 - A003499(k)*a(n+k)*a(n) + a(n)^2 = a(k)^2. - Charlie Marion, Jan 11 2012
a(n+1) = Sum_{k=0..n} A101950(n,k)*5^k. - Philippe Deléham, Feb 10 2012
PSUM transform of a(n+1) is A053142. PSUMSIGN transform of a(n+1) is A084158. BINOMIAL transform of a(n+1) is A164591. BINOMIAL transform of A086347 is a(n+1). BINOMIAL transform of A057087(n-1). - Michael Somos, May 11 2012
a(n+k) = A001541(k)*a(n) + sqrt(A132592(k)*a(n)^2 + a(k)^2). Generalizes formula dated Oct 09 2000. - Charlie Marion, Nov 27 2012
a(n) + a(n+2*k) = A003499(k)*a(n+k); a(n) + a(n+2*k+1) = A001653(k+1)*A002315(n+k). - Charlie Marion, Nov 29 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = 1 + sqrt(2).
Product_{n >= 2} (1 - 1/a(n)) = (1/3)*(1 + sqrt(2)). (End)
G.f.: G(0)*x/(2-6*x), 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
G.f.: H(0)*x/2, where H(k) = 1 + 1/( 1 - x*(6-x)/(x*(6-x) + 1/H(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 18 2014
a(n) = (a(n-1)^2 - a(n-3)^2)/a(n-2) + a(n-4) for n > 3. - Patrick J. McNab, Jul 24 2015
a(n-k)*a(n+k) + a(k)^2 = a(n)^2, a(n+k) + a(n-k) = A003499(k)*a(n), for n >= k >= 0. - Alexander Samokrutov, Sep 30 2015
Dirichlet g.f.: (PolyLog(s,3+2*sqrt(2)) - PolyLog(s,3-2*sqrt(2)))/(4*sqrt(2)). - Ilya Gutkovskiy, Jun 27 2016
4*a(n)^2 - 1 = A278310(n) for n > 0. - Bruno Berselli, Nov 24 2016
From Klaus Purath, Jan 18 2020: (Start)
a(n) = (a(n-3) + a(n+3))/198.
a(n) = Sum_{i=1..n} A001653(i), n>=1.
a(n) = sinh( 2 * n * arccsch(1) ) / ( 2 * sqrt(2) ). - Federico Provvedi, Feb 01 2021
(End)
a(n) = A002965(2*n)*A002965(2*n+1). - Jon E. Schoenfield, Jan 08 2022
a(n) = A002965(4*n)/2. - Gerry Martens, Jul 14 2023
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n+k, 2*k+1)*8^k. - Peter Bala, Jul 17 2023

Additional comments from Wolfdieter Lang, Feb 10 2000

Duplication of a formula removed by Wolfdieter Lang, Feb 10 2015

A156035 Decimal expansion of 3 + 2*sqrt(2).

Original entry on oeis.org

5, 8, 2, 8, 4, 2, 7, 1, 2, 4, 7, 4, 6, 1, 9, 0, 0, 9, 7, 6, 0, 3, 3, 7, 7, 4, 4, 8, 4, 1, 9, 3, 9, 6, 1, 5, 7, 1, 3, 9, 3, 4, 3, 7, 5, 0, 7, 5, 3, 8, 9, 6, 1, 4, 6, 3, 5, 3, 3, 5, 9, 4, 7, 5, 9, 8, 1, 4, 6, 4, 9, 5, 6, 9, 2, 4, 2, 1, 4, 0, 7, 7, 7, 0, 0, 7, 7, 5, 0, 6, 8, 6, 5, 5, 2, 8, 3, 1, 4, 5, 4, 7, 0, 0, 2

Offset: 1

Klaus Brockhaus, Feb 02 2009

Limit_{n -> oo} b(n+1)/b(n) = 3+2*sqrt(2) for b = A155464, A155465, A155466.
Limit_{n -> oo} b(n)/b(n-1) = 3+2*sqrt(2) for b = A001652, A001653, A002315, A156156, A156157, A156158. - Klaus Brockhaus, Sep 23 2009
From Richard R. Forberg, Aug 14 2013: (Start)
Ratios b(n+1)/b(n) for all sequences of the form b(n) = 6*b(n-1) - b(n-2), for any initial values of b(0) and b(1), converge to this ratio.
Ratios b(n+1)/b(n) for all sequences of the form b(n) = 5*b(n-1) + 5*b(n-2) + b(n-3), for all b(0), b(1) and b(2) also converge to 3 + 2*sqrt(2). For example see A084158 (Pell Triangles).
Ratios of alternating values, b(n+2)/b(n), for all sequences of the form b(n) = 2*b(n-1) + b(n-2), also converge to 3 + 2*sqrt(2). These include A000129 (Pell Numbers). Also see A014176. (End)
Let ABCD be a square inscribed in a circle. When P is the midpoint of the arc AB, then the ratio (PC*PD)/(PA*PB) is equal to 3+2*sqrt(2). See the Mathematical Reflections link. - Michel Marcus, Jan 10 2017
Limit of ratios of successive terms of A001652 when n-> infinity. - Harvey P. Dale, Jun 16 2017; improved by Bernard Schott, Feb 28 2022
A quadratic integer with minimal polynomial x^2 - 6x + 1. - Charles R Greathouse IV, Jul 11 2020
Ratio between radii of the large circumscribed circle R and the small internal circle r drawn on the Sangaku tablet at Isaniwa Jinjya shrine in Ehime Prefecture (pictures in links). - Bernard Schott, Feb 25 2022

			3 + 2*sqrt(2) = 5.828427124746190097603377448...

Diogo Queiros-Condé and Michel Feidt, Fractal and Trans-scale Nature of Entropy, Iste Press and Elsevier, 2018, page 45.

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Mathematical Reflections, Solution to Problem J286, Issue 1, 2014, p. 5.
Bernard Schott, Sangaku at Isaniwa Jinya, The six circles.
Terakoya Suzu, Sangaku (mathematics tablet) II, Sangaku at Isaniwa Jinya shrine.
Wikipedia, Sangaku.
Bernard Ycart, Les Sangakus, Sangaku du Temple Isaniwa Jinya (in French).
Index entries for algebraic numbers, degree

Cf. A002193 (sqrt(2)), A090488, A010466, A014176.
Cf. A155464, A155465, A155466.
Cf. A104178 (decimal expansion of log_10(3+2*sqrt(2))).
Cf. A000129, A001109, A001541, A001542, A001652, A001653, A002315, A005319, A075870, A038723, A038725, A038761, A054488, A054489, A075848, A077413, A084158, A106328, A156156, A156157, A156158.
Cf. A242412 (sangaku).

SetDefaultRealField(RealField(100));  3 + 2*Sqrt(2); // G. C. Greubel, Aug 21 2018

RealDigits[N[3+2*Sqrt[2],200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2011 *)

3+sqrt(8) \\ Charles R Greathouse IV, Jun 10 2011

Equals 1 + A090488 = 3 + A010466. - R. J. Mathar, Feb 19 2009
Equals exp(arccosh(3)), since arccosh(x) = log(x+sqrt(x^2-1)). - Stanislav Sykora, Nov 01 2013
Equals (1+sqrt(2))^2, that is, A014176^2. - Michel Marcus, May 08 2016
The periodic continued fraction is [5; [1, 4]]. - Stefano Spezia, Mar 17 2024

A048654 a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 9, 22, 53, 128, 309, 746, 1801, 4348, 10497, 25342, 61181, 147704, 356589, 860882, 2078353, 5017588, 12113529, 29244646, 70602821, 170450288, 411503397, 993457082, 2398417561, 5790292204

Offset: 0

Barry E. Williams

Generalized Pellian with second term equal to 4.

The generalized Pellian with second term equal to s has the terms a(n) = A000129(n)*s + A000129(n-1). The generating function is -(1+s*x-2*x)/(-1+2*x+x^2). - R. J. Mathar, Nov 22 2007

T. D. Noe, Table of n, a(n) for n = 0..300
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252.
A. F. Horadam, Basic Properties of a Certain Generalized Sequence of Numbers, Fibonacci Quarterly, Vol. 3, No. 3, 1965, pp. 161-176.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2,1).

Cf. A000129, A001333, A048655, A038761, A084214, A100525.

a048654 n = a048654_list !! n
a048654_list =
   1 : 4 : zipWith (+) a048654_list (map (* 2) $ tail a048654_list)
-- Reinhard Zumkeller, Aug 01 2011

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!((1+2*x)/(1-2*x-x^2))); // G. C. Greubel, Jul 26 2018

LinearRecurrence[{2,1},{1,4},30] (* Harvey P. Dale, Jul 27 2011 *)

a[0]:1$
a[1]:4$
a[n]:=2*a[n-1]+a[n-2]$
A048654(n):=a[n]$
makelist(A048654(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n)=(([0, 1; 1,2]^n)*[1,4]~)[1] \\ Charles R Greathouse IV, May 18 2015

[lucas_number1(n+1,2,-1) +2*lucas_number1(n,2,-1) for n in (0..40)] # G. C. Greubel, Aug 09 2022

a(n) = ((3+sqrt(2))*(1+sqrt(2))^n - (3-sqrt(2))*(1-sqrt(2))^n)/2*sqrt(2).
a(n) = 2*A000129(n+2) - 3*A000129(n+1). - Creighton Dement, Oct 27 2004
G.f.: (1+2*x)/(1-2*x-x^2). - Philippe Deléham, Nov 03 2008
a(n) = binomial transform of 1, 3, 2, 6, 4, 12, ... . - Al Hakanson (hawkuu(AT)gmail.com), Aug 08 2009
E.g.f.: exp(x)*cosh(sqrt(2)*x) + 3*exp(x)*sinh(sqrt(2)*x)/sqrt(2). - Vaclav Kotesovec, Feb 16 2015
a(n) is the denominator of the continued fraction [4, 2, ..., 2, 4] with n-1 2's in the middle. For the numerators, see A221174. - Greg Dresden and Tongjia Rao, Sep 02 2021
a(n) = A001333(n) + A000129(n). - G. C. Greubel, Aug 09 2022

A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 23, 134, 781, 4552, 26531, 154634, 901273, 5253004, 30616751, 178447502, 1040068261, 6061962064, 35331704123, 205928262674, 1200237871921, 6995498968852, 40772755941191, 237641036678294, 1385073464128573, 8072799748093144, 47051725024430291, 274237550398488602

Offset: 0

Barry E. Williams, May 02 2000

This sequence gives one half of all positive solutions y = y1 = a(n) of the first class of the Pell equation x^2 - 2*y^2 = -7. For the corresponding x=x1 terms see A054490(n). Therefore it also gives one fourth of all positive solutions x = x1 of the first class of the Pell equation x^2 - 2*y^2 = 14, with the y=y1 terms given by A054490. - Wolfdieter Lang, Feb 26 2015

			n = 2: A054490(2)^2 - 2*(2*a(2))^2 =
       65^2 - 2*(2*23)^2 = -7,
      (4*a(2))^2 - 2*A054490(2)^2 =
      (4*23)^2 - 2*65^2 = 14. - _Wolfdieter Lang_, Feb 26 2015
a(2) = (A253811(1) + A101386(1))/2 = (19 + 27)/2 = 23. - _Wolfdieter Lang_, Mar 19 2015

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

Stefano Spezia, Table of n, a(n) for n = 0..1300
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
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, A001653, A054490.

A038725(n) = a(-n).

a[0]:=1: a[1]:=4: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{6,-1},{1,4},30] (* Harvey P. Dale, Aug 06 2020 *)

{a(n) = real((3 + 2*quadgen(8))^n * (1 + quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */

{a(n) = polchebyshev(n, 1, 3) + polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */

a(n) = ((4+sqrt(2))/8)*(3+2*sqrt(2))^(n-1) + ((4-sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008
a(n) = A001653(n+1) - A001109(n). - Antonio Alberto Olivares, Mar 29 2008
Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v. - Michael Somos, Sep 28 2008
G.f.: (1 - 2*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). - Michael Somos, Sep 28 2008
a(n) = Sum_{k = 0..n} A238731(n,k)*3^k. - Philippe Deléham, Mar 05 2014
a(n) = S(n,6) - 2*S(n-1, 6), n >= 0, with the Chebyshev polynomials S(n, x) (A049310) with S(-1, x) = 0 evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell comment above. - Wolfdieter Lang, Feb 26 2015
a(0) = -(A038761(0) - A038762(0))/2, a(n) = (A253811(n-1) + A101386(n-1))/2, n >= 1. See the Mar 19 2015 comment on A054490. - Wolfdieter Lang, Mar 19 2015
E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, Apr 30 2020

More terms from James Sellers, May 03 2000

A054490 Expansion of (1+5x)/(1-6x+x^2).

Original entry on oeis.org

1, 11, 65, 379, 2209, 12875, 75041, 437371, 2549185, 14857739, 86597249, 504725755, 2941757281, 17145817931, 99933150305, 582453083899, 3394785353089, 19786259034635, 115322768854721, 672150354093691, 3917579355707425, 22833325780150859

Offset: 0

Barry E. Williams, May 04 2000

A Pellian-related second-order recursive sequence.
Third binomial transform of 1,8,8,64,64,512. - Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009
Binomial transform of A164607. - R. J. Mathar, Oct 26 2011
Pisano period lengths: 1, 1, 4, 2, 6, 4, 3, 2, 12, 6, 12, 4, 14, 3, 12, 2, 8, 12, 20, 6, ... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Feb 26 2015: (Start)
This sequence gives all positive solutions x = x1 = a(n) of the first class of the (generalized) Pell equation x^2 - 2*y^2 = -7. For the corresponding y1 terms see 2*A038723(n). All positive solutions of the second class are given by (x2(n), y2(n)) = (A255236(n), A038725(n+1)), n >= 0. See (A254938(1), 2*A255232(1)) for the fundamental solution (1, 2) of the first class. See the Nagell reference, Theorem 111, p. 210, Theorem 110, p. 208, Theorem 108a, pp. 206-207.
This sequence also gives all positive solutions y = y1 of the first class of the Pell equation x^2 - 2*y^2 = 14. The corresponding solutions x1 are given in 4*A038723. This follows from the preceding comment. (End)
From Wolfdieter Lang, Mar 19 2015: (Start)
a(0) = -(2*A038761(0) - A038762(0)), a(n) = 2*A253811(n-1) + A101386(n-1), for n >= 1.
This follows from the general trivial fact that if X^2 - D*Y^2 = N (X, Y positive integers, D > 1, not a square, and N a non-vanishing integer) then x:= D*Y +/- X and y:= Y +/- X (correlated signs) satisfy x^2 - D*y^2 = -(D-1)*N. with integers x and y. Here D = 2 and N = 7. (End)

			n = 2: sqrt(8*23^2-7) = 65.
2*19 + 27  = 65. - _Wolfdieter Lang_, Mar 19 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N. Y., 1964, pp. 122-125, 194-196.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

G. C. Greubel, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See y(n) in Lemma 7.9 p. 21.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000129, A001109, A038723, A038725, A054488, A054489, A101386, A253811, A255236.

a:=[1,11];; for n in [3..30] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 20 2020

I:=[1,11]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

a[0]:=1: a[1]:=11: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006

CoefficientList[Series[(1+5x)/(1-6x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 20 2015 *)
LinearRecurrence[{6, -1}, {1, 11}, 30] (* G. C. Greubel, Jul 26 2018 *)

my(x='x+O('x^30)); Vec((1+5*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jul 26 2018

[lucas_number1(2*n+1,2,-1) + 3*lucas_number1(2*n,2,-1) for n in (0..30)] # G. C. Greubel, Jan 20 2020

a(n) = 6*a(n-1) - a(n-2) for n>1, a(0)=1, a(1)=11.
a(n) = sqrt(8*A038723(n)^2 - 7).
a(n) = (11*((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) = 11*S(n, 6) + 5*S(n-1, 6), n >= 0, with Chebyshev's polynomials S(n, x) (A049310) evaluated at x=6: S(n, 6) = A001109(n-1). See the g.f. and the Pell equation comments above. - Wolfdieter Lang, Feb 26 2015
a(n) = 2*A253811(n-1) + A101386(n-1), for n >= 1. See the Mar 19 2015 comment above. - Wolfdieter Lang, Mar 19 2015
From G. C. Greubel, Jan 20 2020: (Start)
a(n) = Pell(2*n+1) + 3*Pell(2*n).
a(n) = ChebyshevU(n,3) + 5*ChebyshevU(n-1,3).
E.g.f.: exp(3*x)*( cosh(2*sqrt(2)*x) + 2*sqrt(2)*sinh(2*sqrt(2)*x) ). (End)

More terms from James Sellers, May 05 2000

More terms from Vincenzo Librandi, Mar 20 2015

A038725 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 11, 64, 373, 2174, 12671, 73852, 430441, 2508794, 14622323, 85225144, 496728541, 2895146102, 16874148071, 98349742324, 573224305873, 3340996092914, 19472752251611, 113495517416752, 661500352248901, 3855506596076654, 22471539224211023, 130973728749189484

Offset: 0

Barry E. Williams, May 02 2000

From Wolfdieter Lang, Feb 26 2015: (Start)
The sequence {2*a(n+1)}_{n >= 0}, gives all positive solutions y = y2(n) = 2*a(n+1) of the second class of the Pell equation x^2 - 2*y^2  = -7. For the corresponding terms x = x2(n) see A255236(n).
See A255236 for comments on the first class solutions and the relation to the Pell equation x^2 - 2*y^2 = 14. (End)

			n = 2: a(3) = sqrt((181^2 + 7)/2)/2 = 64.
a(3) = (53 + 75)/2 = 64. - _Wolfdieter Lang_, Mar 19 2015

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

Stefano Spezia, Table of n, a(n) for n = 0..1300
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See x'(n) in Lemma 7.9 p. 21.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001653 and A001541. Cf. A001109.

A038723(n) = a(-n).

a[0]:=1: a[1]:=2: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006

Union[Flatten[NestList[{#[[2]],#[[3]],6#[[3]]-#[[2]]}&,{1,2,11},25]]]  (* Harvey P. Dale, Mar 04 2011 *)
LinearRecurrence[{6,-1},{1,2},30] (* Harvey P. Dale, Jun 12 2017 *)

{a(n) = real((3 + 2*quadgen(8))^n * (1 - quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */

{a(n) = polchebyshev(n, 1, 3) - polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3); a(n) = ((4-sqrt(2))/8)*(3+2*sqrt(2))^(n-1)+((4+sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008
From Michael Somos, Sep 28 2008: (Start)
Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v.
G.f.: (1 - 4*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). (End)
From Wolfdieter Lang, Feb 26 2015: (Start)
a(n) = S(n, 6) - 4*S(n-1, 6), n>=0, with the Chebyshev polynomials S(n, x) (A049310), with S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n-1). See the g.f. and the Pell equation comment above.
a(n) = 6*a(n-1) - a(n-2), n >= 1, a(-1) = 4, a(0) = 1. (See the name.) (End)
From Wolfdieter Lang, Mar 19 2015: (Start)
a(n+1) = sqrt((A255236(n)^2 + 7)/2)/2, n >= 0.
a(n+1) = (A038761(n) + A038762(n))/2, n >= 0. See the Mar 19 2015 comment on A054490. (End)
E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) - sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, May 01 2020

A101386 Expansion of g.f.: (5 - 3x)/(1 - 6x + x^2).

Original entry on oeis.org

5, 27, 157, 915, 5333, 31083, 181165, 1055907, 6154277, 35869755, 209064253, 1218515763, 7102030325, 41393666187, 241259966797, 1406166134595, 8195736840773, 47768254910043, 278413792619485, 1622714500806867, 9457873212221717, 55124524772523435, 321289275422918893

Offset: 0

Creighton Dement, Jan 23 2005

A floretion-generated sequence relating to NSW numbers and numbers n such that (n^2 - 8)/2 is a square. It is also possible to label this sequence as the "tesfor-transform of the zero-sequence" under the floretion given in the program code, below. This is because the sequence "vesseq" would normally have been A046184 (indices of octagonal numbers which are also a square) using the floretion given. This floretion, however, was purposely "altered" in such a way that the sequence "vesseq" would turn into A000004. As (a(n)) would not have occurred under "natural" circumstances, one could speak of it as the transform of A000004.
Floretion Algebra Multiplication Program FAMP code: - tesforseq[ + 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e], Note: vesforseq = A000004, lesforseq = A002315, jesforseq = A077445
From Wolfdieter Lang, Feb 05 2015: (Start)
All positive solutions x = a(n) of the (generalized) Pell equation x^2 - 2*y^2 = +7 based on the  fundamental solution (x2,y2) = (5,3) of the second class of (proper) solutions. The corresponding y solutions are given by y(n) = A253811(n).
All other positive solutions come from the first class of (proper) solutions based on the fundamental solution (x1,y1) = (3,1). These are given in A038762 and A038761.
All solutions of this Pell equation are found in A077443(n+1) and A077442(n), for n >= 0. See, e.g., the Nagell reference on how to find all solutions.
(End)

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Colin Barker, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
Tanya Khovanova, Recursive Sequences
Morris Newman, Daniel Shanks, and H. C. Williams, Simple groups of square order and an interesting sequence of primes, Acta Arith., 38 (1980/1981) 129-140. MR82b:20022.
Eric Weisstein's World of Mathematics, NSW Number.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000004, A000129, A001109, A002315, A038761, A038762, A046184, A054490, A164737.

Cf. A077442, A077443(n+1), A077445, A253811.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!((5 - 3*x)/(1-6*x+x^2))); // G. C. Greubel, Jul 26 2018

A101386:= (n) -> simplify(5*ChebyshevU(n, 3) - 3*ChebyshevU(n-1, 3)); seq( A101386(n), n = 0..30); # G. C. Greubel, Mar 17 2020

CoefficientList[ Series[(5-3x)/(1-6x+x^2), {x,0,30}], x] (* Robert G. Wilson v, Jan 29 2005 *)
LinearRecurrence[{6,-1},{5,27},30] (* Harvey P. Dale, Apr 23 2016 *)

Vec((5-3*x)/(1-6*x+x^2) + O(x^30)) \\ Colin Barker, Feb 05 2015

[5*chebyshev_U(n,3) -3*chebyshev_U(n-1,3) for n in (0..30)] # G. C. Greubel, Mar 17 2020

a(n) = A002315(n) + A077445(n+1). Note: the offset of A077445 is 1.
a(n+1) - a(n) = 2*A054490(n+1).
a(n) = 6*a(n-1) - a(n-2), a(0)=5, a(1)=27. - Philippe Deléham, Nov 17 2008
From Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009: (Start)
a(n) = ((5+sqrt(18))*(3 + sqrt(8))^n + (5-sqrt(18))*(3 - sqrt(8))^n)/2.
Third binomial transform of A164737. (End)
a(n) = rational part of z(n), with z(n) = (5+3*sqrt(2))*(3+2*sqrt(2))^n, n >= 0, the general positive solutions of the second class of proper solutions. See the preceding formula. - Wolfdieter Lang, Feb 05 2015
a(n) = 5*A001109(n+1) - 3*A001109(n). - G. C. Greubel, Mar 17 2020
a(n) = Pell(2*n+2) + 3*Pell(2*n+1), where Pell(n) = A000129(n). - G. C. Greubel, Apr 17 2020
E.g.f.: exp(3*x)*(5*cosh(2*sqrt(2)*x) + 3*sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Mar 16 2024

More terms from Robert G. Wilson v, Jan 29 2005

A038762 a(n) = 6*a(n-1) - a(n-2) for n >= 2, with a(0)=3, a(1)=13.

Original entry on oeis.org

3, 13, 75, 437, 2547, 14845, 86523, 504293, 2939235, 17131117, 99847467, 581953685, 3391874643, 19769294173, 115223890395, 671574048197, 3914220398787, 22813748344525, 132968269668363, 774995869665653, 4517006948325555, 26327045820287677, 153445267973400507

Offset: 0

Barry E. Williams, May 03 2000

This gives part of the (increasingly sorted) positive solutions x to the Pell equation x^2 - 2*y^2 = +7. For the y solutions see A038761. The other part of solutions is found in A101386 and A253811. - Wolfdieter Lang, Feb 05 2015

			a(3)^2 - 2*A038761(3)^2 = 437^2 - 2*309^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
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, A038725, A038761, A077443, A101386, A253811.

I:=[3, 13]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

LinearRecurrence[{6,-1},{3,13},40] (* Vincenzo Librandi, Nov 16 2011 *)

x='x+O('x^30); Vec((3-5*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jul 26 2018

a(n) = sqrt(2*(A038761(n))^2+7).
a(n) = (13*((3+2*sqrt(2))^n -(3-2*sqrt(2))^n)-3*((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = A077443(2n) = A038725(n)+A038725(n+1).
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3); a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^(n-1)+(1/2)*(3-sqrt(2))*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Apr 20 2008
G.f.: (3-5*x)/(1-6*x+x^2). - Philippe Deléham, Nov 03 2008, corrected by R. J. Mathar, Nov 06 2011
a(n) = -5*A001109(n) +3*A001109(n+1). - R. J. Mathar, Nov 06 2011
a(n) = rational part of z(n) = (3 + sqrt(2))*(3 + 2*sqrt(2))^n, n >= 0. z(n) gives only one part of the positive solutions to the Pell equation x^2 - 2*y^2 = 7. See the Nagell reference on how to find z(n), and a comment above. - Wolfdieter Lang, Feb 05 2015
E.g.f.: exp(3*x)*(3*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Mar 16 2024

More terms from James Sellers, May 04 2000

Unspecific Pell comment replaced by Wolfdieter Lang, Feb 05 2015

A077442 2*a(n)^2 + 7 is a square.

Original entry on oeis.org

1, 3, 9, 19, 53, 111, 309, 647, 1801, 3771, 10497, 21979, 61181, 128103, 356589, 746639, 2078353, 4351731, 12113529, 25363747, 70602821, 147830751, 411503397, 861620759, 2398417561, 5021893803, 13979001969, 29269742059, 81475594253

Offset: 0

Gregory V. Richardson, Nov 06 2002

Lim. n -> Inf. a(n)/a(n-2) = 3 + 2*Sqrt(2) = R1*R2. Lim. k -> Inf. a(2*k-1)/a(2*k) = (9 + 4*Sqrt(2))/7 = R1 (ratio #1). Lim. k -> Inf. a(2*k)/a(2*k-1) = (11 + 6*Sqrt(2))/7 = R2 (ratio #2).

a(n) gives for n >= 0 all positive y-values solving the (generalized) Pell equation x^2 - 2*y^2 = 7. A077443(n+1) gives the corresponding x-values. See, e.g., the Nagell reference on how to find all solutions. - Wolfdieter Lang, Feb 05 2015

			a(4)^2 - 2*a(3)^2 = 27^2 - 2*19^2  = +7. - _Wolfdieter Lang_, Feb 05 2015

L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
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.
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.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Colin Barker, Table of n, a(n) for n = 0..1000
J. J. O'Connor and E. F. Robertson, History of Pell's Equation
J. P. Robertson, Solving the Generalized Pell Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A077443, A038762, A038761, A101386, A253811.

CoefficientList[Series[(1+3 x+3 x^2+x^3)/ (1-6 x^2+x^4),{x,0,50}],x]  (* Harvey P. Dale, Mar 12 2011 *)
LinearRecurrence[{0, 6, 0, -1},{1,3,9,19},50] (* Sture Sjöstedt, Oct 08 2012 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[1;3;9;19])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

Vec((x+1)^3/(x^2+2*x-1)/(x^2-2*x-1) + O(x^50)) \\ Colin Barker, Mar 27 2016

For n>0, a(2n) = A046090(n) + A001653(n) + A001652(n-1); a(2n+1) = A001652(n+1) - A001652(n-1) - A001653(n-1); e.g. 53=21+29+3; 111=119-3-5. - Charlie Marion, Aug 14 2003
The same recurrences hold for the odd and even indices respectively : a(n+2) = 6*a(n+1) - a(n), a(n+1) = 3*a(n) + 2*(2*a(n)^2+7)^0.5. - Richard Choulet, Oct 11 2007
G.f.: (x+1)^3/(x^2+2*x-1)/(x^2-2*x-1). a(n)= [ -A077985(n)-3*A077985(n-1)+3*A000129(n+1)+A000129(n)]/2. - R. J. Mathar, Nov 16 2007
a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=3, a(3)=9, a(4)=19. - Sture Sjöstedt, Oct 08 2012
a(n) = ((-(-1 - sqrt(2))^n*(-2+sqrt(2)) - (-1+sqrt(2))^n*(2+sqrt(2)) + (1-sqrt(2))^n*(-4+3*sqrt(2)) + (1+sqrt(2))^n*(4+3*sqrt(2))))/(4*sqrt(2)). - Colin Barker, Mar 27 2016

Edited: n in Name replaced by a(n). Pell comments moved to comment section. - Wolfdieter Lang, Feb 05 2015

cp's OEIS Frontend

A129345 a(2n) = A001542(n+1), a(2n+1) = A038761(n+1); a Pellian-related sequence.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

PARI

PARI

PARI

Sage

Sage

Formula

Extensions

A156035 Decimal expansion of 3 + 2*sqrt(2).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A048654 a(n) = 2*a(n-1) + a(n-2); a(0)=1, a(1)=4.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

Maxima

PARI

SageMath

Formula

A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

Extensions

A054490 Expansion of (1+5x)/(1-6x+x^2).

A101386 Expansion of g.f.: (5 - 3x)/(1 - 6x + x^2).