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%I M4217 N1760 #587 Mar 01 2024 05:21:00
%S 0,1,6,35,204,1189,6930,40391,235416,1372105,7997214,46611179,
%T 271669860,1583407981,9228778026,53789260175,313506783024,
%U 1827251437969,10650001844790,62072759630771,361786555939836,2108646576008245,12290092900109634,71631910824649559
%N a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.
%C 8*a(n)^2 + 1 = 8*A001110(n) + 1 = A055792(n+1) is a perfect square. - _Gregory V. Richardson_, Oct 05 2002
%C 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
%C (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
%C 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]
%C n such that 8*n^2 = floor(sqrt(8)*n*ceiling(sqrt(8)*n)). - _Benoit Cloitre_, May 10 2003
%C 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
%C 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
%C Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference). - _Emeric Deutsch_, Jun 19 2005
%C 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
%C 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
%C 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
%C One half the bisection of the Pell numbers (A000129). - _Franklin T. Adams-Watters_, Jan 08 2006
%C Pell trapezoids: for n > 0, a(n) = (A000129(n-1)+A000129(n+1))*A000129(n)/2; see also A084158. - _Charlie Marion_, Apr 01 2006
%C 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
%C 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
%C 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
%C 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
%C 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
%C 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
%C 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<c then a+b = c-d and ((d+b)^2, d^2-b^2) is a solution too. If (a,b), (c,d) and (e,f) are three consecutive solutions of the Diophantine equation 0+1+2+...+x = y^2 with a < c < e then (8*d^2, d*(f-b)) is a solution too. - _Mohamed Bouhamida_, Aug 29 2009
%C 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
%C a(n)/A002315(n) converges to cos^2(Pi/8) (see A201488). - _Gary Detlefs_, Nov 25 2009
%C Binomial transform of A086347. - _Johannes W. Meijer_, Aug 01 2010
%C 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
%C In general, if b(0)=1, b(1)=k and for n > 1, b(n) = 6*b(n-1) - b(n-2), then
%C for n > 0, b(n) = a(n)*k-a(n-1); e.g.,
%C for k=2, when b(n) = A038725(n), 2 = 1*2 - 0, 11 = 6*2 - 1, 64 = 35*2 - 6, 373 = 204*2 - 35;
%C for k=3, when b(n) = A001541(n), 3 = 1*3 - 0, 17 = 6*3 - 1; 99 = 35*3 - 6; 577 = 204*3 - 35;
%C for k=4, when b(n) = A038723(n), 4 = 1*4 - 0, 23 = 6*4 - 1; 134 = 35*4 - 6; 781 = 204*4 - 35;
%C for k=5, when b(n) = A001653(n), 5 = 1*5 - 0, 29 = 6*5 - 1; 169 = 35*5 - 6; 985 = 204*5 - 35.
%C See also A002315, A054488, A038761, A054489, A054490.
%C - _Charlie Marion_, Dec 08 2010
%C 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
%C 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
%C 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
%C From _Richard R. Forberg_, Aug 30 2013: (Start)
%C 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:
%C a(2n)^2 = A001333(2n)^2 * (A001333(2n)^2 - 1)/2;
%C a(2n+1)^2 = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2. (End)
%C 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
%C 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
%C 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
%C 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
%C Behera & Panda call these the balancing numbers, and A001541 are the balancers. - _Michel Marcus_, Nov 07 2017
%C 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
%C 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
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%H G. K. Panda and Ravi Kumar Davala, <a href="https://www.fq.math.ca/Papers1/53-3/PandaDavala04232015.pdf">Perfect Balancing Numbers</a>, Fibonacci Quart. 53 (2015), no. 3, 261-264.
%H Ashish Kumar Pandey and B. K. Sharma, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL26/Pandey/pandey14.html">On Inequalities Related to a Generalized Euler Totient Function and Lucas Sequences</a>, J. Int. Seq. (2023) Vol. 26, Art. 23.8.6.
%H Poo-Sung Park, <a href="https://www.jstor.org/stable/30044886">Ramanujan's Continued Fraction for a Puzzle</a>, College Mathematics Journal, 2005, 363-365.
%H Michael Penn, <a href="https://www.youtube.com/watch?v=jMfZ9jRsHSI">Balancing Numbers</a>, Youtube video, 2020.
%H Robert Phillips, <a href="https://web.archive.org/web/20100713033314/http://www.usca.edu/math/~mathdept/bobp/pdf/polgonal.pdf">Polynomials of the form 1+4ke+4ke^2</a>, 2008.
%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H B. Polster and M. Ross, <a href="http://arxiv.org/abs/1503.04658">Marching in squares</a>, arXiv preprint arXiv:1503.04658 [math.HO], 2015.
%H Helmut Prodinger, <a href="https://arxiv.org/abs/2008.03916">How to sum powers of balancing numbers efficiently</a>, arXiv:2008.03916 [math.NT], 2020.
%H Rajesh Ram, <a href="https://web.archive.org/web/20131021224033/http://users.tellurian.net/hsejar/maths/triangle/">Triangle Numbers that are Perfect Squares</a>
%H K. J. Ramsey, <a href="http://groups.yahoo.com/group/Triangular_and_Fibonacci_Numbers/message/23">Relation of Mersenne Primes To Square Triangular Numbers</a> [edited by K. J. Ramsey, May 14 2011]
%H Kenneth Ramsay and Andras Erszegi, <a href="/A001109/a001109.txt">Relation of Square Triangular Numbers To Mersenne Primes</a>, digest of 4 messages in Triangular_and_Fibonacci_Numbers Yahoo Group, May 15 - Jun 28, 2006.
%H Kenneth Ramsey, <a href="http://groups.yahoo.com/group/Triangular_and_Fibonacci_Numbers/message/62">Generalized Proof re Square Triangular Numbers</a>
%H Kenneth Ramsey, <a href="/A001108/a001108.txt">Generalized Proof re Square Triangular Numbers</a>, digest of 2 messages in Triangular_and_Fibonacci_Numbers Yahoo group, May 27, 2005 - Oct 10, 2011.
%H Salah E. Rihane, Bernadette Faye, Florian Luca, and Alain Togbe, <a href="https://arxiv.org/abs/1811.03015">An exponential Diophantine equation related to the difference between powers of two consecutive Balancing numbers</a>, arXiv:1811.03015 [math.NT], 2018.
%H A. Sandhya, <a href="https://www.angelfire.com/ak/ashoksandhya/maths2.html">Puzzle 4: A problem Srinivasa Ramanujan, the famous 20th century Indian Mathematician Solved</a>
%H Sci.math Newsgroup, <a href="https://web.archive.org/web/20130721085940/http://www.math.niu.edu/~rusin/known-math/98/sq_tri">Square numbers which are triangular</a>
%H Sci.math Newsgroup, <a href="/A000217/a000217_1.txt">Square numbers which are triangular</a> [Cached copy]
%H R. A. Sulanke, <a href="https://web.archive.org/web/20180416202341/http://math.boisestate.edu/~sulanke/PAPERS/cutpasteII.pdf">Moments, Narayana numbers and the cut and paste for lattice paths</a>.
%H R. A. Sulanke, <a href="https://doi.org/10.37236/1385">Bijective recurrences concerning Schroeder paths</a>, Electron. J. Combin. 5 (1998), Research Paper 47, 11 pp.
%H Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, <a href="https://hal.archives-ouvertes.fr/hal-02918958/document#page=18">Integer sequences and ellipse chains inside a hyperbola</a>, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
%H Ahmet Tekcan, Merve Tayat, and Meltem E. Ozbek, <a href="https://dx.doi.org/10.1155/2014/897834">The diophantine equation 8x^2-y^2+8x(1+t)+(2t+1)^2=0 and t-balancing numbers</a>, ISRN Combinatorics, Volume 2014, Article ID 897834, 5 pages.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BinomialCoefficient.html">Binomial coefficient</a>.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/SquareTriangularNumber.html">Square Triangular Number</a>.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/TriangularNumber.html">Triangular Number</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Triangular_square_number">Triangular square number</a>
%H Rick Young, <a href="https://web.archive.org/web/20120818220024/http://www.cob.ohio-state.edu/~young_53/Quote.ram.html">Relevant quotation from biography of Ramanujan</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-1).
%F G.f.: x / (1 - 6*x + x^2). - _Simon Plouffe_ in his 1992 dissertation.
%F 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).
%F a(n) = sqrt(A001110(n)).
%F a(n) = A001542(n)/2.
%F a(n) = sqrt((A001541(n)^2-1)/8) (cf. Richardson comment).
%F a(n) = 3*a(n-1) + sqrt(8*a(n-1)^2+1). - _R. J. Mathar_, Oct 09 2000
%F a(n) = A000129(n)*A001333(n) = A000129(n)*(A000129(n)+A000129(n-1)) = ceiling(A001108(n)/sqrt(2)). - _Henry Bottomley_, Apr 19 2000
%F a(n) ~ (1/8)*sqrt(2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
%F Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - _Gregory V. Richardson_, Oct 05 2002
%F 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
%F a(2*n) = a(n)*A003499(n). 4*a(n) = A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003
%F a(n) = floor((3+2*sqrt(2))^n/(4*sqrt(2))). - _Lekraj Beedassy_, Apr 23 2003
%F a(-n) = -a(n). - _Michael Somos_, Apr 07 2003
%F For n >= 1, a(n) = Sum_{k=0..n-1} A001653(k). - _Charlie Marion_, Jul 01 2003
%F For n > 0, 4*a(2*n) = A001653(n)^2 - A001653(n-1)^2. - _Charlie Marion_, Jul 16 2003
%F For n > 0, a(n) = Sum_{k = 0..n-1}((2*k+1)*A001652(n-1-k)) + A000217(n). - _Charlie Marion_, Jul 18 2003
%F a(2*n+1) = a(n+1)^2 - a(n)^2. - _Charlie Marion_, Jan 12 2004
%F 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
%F 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
%F From _Paul Barry_, Feb 06 2004: (Start)
%F a(n) = A000129(2*n)/2;
%F a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))*sqrt(2)/8;
%F a(n) = Sum_{i=0..n} Sum_{j=0..n} A000129(i+j)*n!/(i!*j!*(n-i-j)!)/2. (End)
%F E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - _Paul Barry_, Apr 21 2004
%F A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - _Creighton Dement_, Sep 16 2004
%F a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1)*2^(k-1). - _Paul Barry_, Oct 01 2004
%F 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
%F a(n+1) = Sum_{k=0..n} A001850(k)*A001850(n-k), self convolution of central Delannoy numbers. - _Benoit Cloitre_, Sep 28 2005
%F 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
%F a(n) = 5*(a(n-1) + a(n-2)) - a(n-3). - _Mohamed Bouhamida_, Sep 20 2006
%F 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
%F 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
%F 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
%F [A001653(n), a(n)] = [1,4; 1,5]^n * [1,0]. - _Gary W. Adamson_, Mar 21 2008
%F a(n) = Sum_{k=0..n-1} 4^k*binomial(n+k,2*k+1). - _Paul Barry_, Apr 20 2009
%F a(n+1)^2 - 6*a(n+1)*a(n) + a(n)^2 = 1. - _Charlie Marion_, Dec 14 2010
%F 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
%F 16*a(n)^2 + 1 = A056771(n). - _James R. Buddenhagen_, Dec 09 2011
%F A010054(A000290(a(n))) = 1. - _Reinhard Zumkeller_, Dec 17 2011
%F In general, a(n+k)^2 - A003499(k)*a(n+k)*a(n) + a(n)^2 = a(k)^2. - _Charlie Marion_, Jan 11 2012
%F a(n+1) = Sum_{k=0..n} A101950(n,k)*5^k. - _Philippe Deléham_, Feb 10 2012
%F 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
%F 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
%F 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
%F From _Peter Bala_, Dec 23 2012: (Start)
%F Product_{n >= 1} (1 + 1/a(n)) = 1 + sqrt(2).
%F Product_{n >= 2} (1 - 1/a(n)) = (1/3)*(1 + sqrt(2)). (End)
%F 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
%F 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
%F 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
%F 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
%F Dirichlet g.f.: (PolyLog(s,3+2*sqrt(2)) - PolyLog(s,3-2*sqrt(2)))/(4*sqrt(2)). - _Ilya Gutkovskiy_, Jun 27 2016
%F 4*a(n)^2 - 1 = A278310(n) for n > 0. - _Bruno Berselli_, Nov 24 2016
%F From _Klaus Purath_, Jan 18 2020: (Start)
%F a(n) = (a(n-3) + a(n+3))/198.
%F a(n) = Sum_{i=1..n} A001653(i), n>=1.
%F a(n) = sinh( 2 * n * arccsch(1) ) / ( 2 * sqrt(2) ). - _Federico Provvedi_, Feb 01 2021
%F (End)
%F a(n) = A002965(2*n)*A002965(2*n+1). - _Jon E. Schoenfield_, Jan 08 2022
%F a(n) = A002965(4*n)/2. - _Gerry Martens_, Jul 14 2023
%F a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n+k, 2*k+1)*8^k. - _Peter Bala_, Jul 17 2023
%e G.f. = x + 6*x^2 + 35*x^3 + 204*x^4 + 1189*x^5 + 6930*x^6 + 40391*x^7 + ...
%e 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
%p 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_
%p with (combinat):seq(fibonacci(2*n,2)/2, n=0..20); # _Zerinvary Lajos_, Apr 20 2008
%t Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{-1,6},#]}]&, {0,1}, 30]][[1]] (* _Harvey P. Dale_, Mar 23 2011 *)
%t CoefficientList[Series[x/(1-6x+x^2),{x,0,30}],x] (* _Harvey P. Dale_, Mar 23 2011 *)
%t LinearRecurrence[{6, -1}, {0, 1}, 50] (* _Vladimir Joseph Stephan Orlovsky_, Feb 12 2012 *)
%t a[ n_]:= ChebyshevU[n-1, 3]; (* _Michael Somos_, Sep 02 2012 *)
%t Table[Fibonacci[2n, 2]/2, {n, 0, 20}] (* _Vladimir Reshetnikov_, Sep 16 2016 *)
%t TrigExpand@Table[Sinh[2 n ArcCsch[1]]/(2 Sqrt[2]), {n, 0, 10}] (* _Federico Provvedi_, Feb 01 2021 *)
%o (PARI) {a(n) = imag((3 + quadgen(32))^n)}; /* _Michael Somos_, Apr 07 2003 */
%o (PARI) {a(n) = subst( poltchebi( abs(n+1)) - 3 * poltchebi( abs(n)), x, 3) / 8}; /* _Michael Somos_, Apr 07 2003 */
%o (PARI) {a(n) = polchebyshev( n-1, 2, 3)}; /* _Michael Somos_, Sep 02 2012 */
%o (PARI) is(n)=ispolygonal(n^2,3) \\ _Charles R Greathouse IV_, Nov 03 2016
%o (Sage) [lucas_number1(n,6,1) for n in range(27)] # _Zerinvary Lajos_, Jun 25 2008
%o (Sage) [chebyshev_U(n-1,3) for n in (0..20)] # _G. C. Greubel_, Dec 23 2019
%o (Haskell)
%o a001109 n = a001109_list !! n :: Integer
%o a001109_list = 0 : 1 : zipWith (-)
%o (map (* 6) $ tail a001109_list) a001109_list
%o -- _Reinhard Zumkeller_, Dec 17 2011
%o (Magma) [n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Jul 25 2015
%o (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
%Y Cf. A001108, A001542, A001653, A001850, A002315, A002965, A278310.
%Y 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).
%Y Cf. A323182.
%K nonn,easy,nice
%O 0,3
%A _N. J. A. Sloane_
%E Additional comments from _Wolfdieter Lang_, Feb 10 2000
%E Duplication of a formula removed by _Wolfdieter Lang_, Feb 10 2015
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