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 A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1. (Formerly M4217 N1760) 193
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS 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 a 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 REFERENCES 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). P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022 LINKS 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. 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. 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). FORMULA 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->infinity} 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 EXAMPLE 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 MAPLE 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 MATHEMATICA 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 *) PROG (PARI) {a(n) = imag((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */ (PARI) {a(n) = subst( poltchebi( abs(n+1)) - 3 * poltchebi( abs(n)), x, 3) / 8}; /* Michael Somos, Apr 07 2003 */ (PARI) {a(n) = polchebyshev( n-1, 2, 3)}; /* Michael Somos, Sep 02 2012 */ (PARI) is(n)=ispolygonal(n^2, 3) \\ Charles R Greathouse IV, Nov 03 2016 (Sage) [lucas_number1(n, 6, 1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008 (Sage) [chebyshev_U(n-1, 3) for n in (0..20)] # G. C. Greubel, Dec 23 2019 (Haskell) a001109 n = a001109_list !! n :: Integer a001109_list = 0 : 1 : zipWith (-) (map (* 6) \$ tail a001109_list) a001109_list -- Reinhard Zumkeller, Dec 17 2011 (Magma) [n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015 (GAP) 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 CROSSREFS Cf. 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. Sequence in context: A161727 A121838 A242629 * A352972 A180033 A354134 Adjacent sequences: A001106 A001107 A001108 * A001110 A001111 A001112 KEYWORD nonn,easy,nice,changed AUTHOR N. J. A. Sloane EXTENSIONS Additional comments from Wolfdieter Lang, Feb 10 2000 Duplication of a formula removed by Wolfdieter Lang, Feb 10 2015 STATUS approved

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Last modified September 17 15:47 EDT 2024. Contains 375987 sequences. (Running on oeis4.)