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%I M3955 N1630 #444 Mar 03 2024 08:03:44
%S 1,5,29,169,985,5741,33461,195025,1136689,6625109,38613965,225058681,
%T 1311738121,7645370045,44560482149,259717522849,1513744654945,
%U 8822750406821,51422757785981,299713796309065,1746860020068409,10181446324101389,59341817924539925
%N Numbers k such that 2*k^2 - 1 is a square.
%C Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z values.
%C The defining equation is X^2 + (X+1)^2 = Z^2, which when doubled gives 2Z^2 = (2X+1)^2 + 1. So the sequence gives Z's such that 2Z^2 = odd square + 1 (A069894).
%C (x,y) = (a(n), a(n+1)) are the solutions with x < y of x/(yz) + y/(xz) + z/(xy)=3 with z=2. - _Floor van Lamoen_, Nov 29 2001
%C Consequently the sum n^2*(2n^2 - 1) of the first n odd cubes (A002593) is also a square. - _Lekraj Beedassy_, Jun 05 2002
%C Numbers n such that 2*n^2 = ceiling(sqrt(2)*n*floor(sqrt(2)*n)). - _Benoit Cloitre_, May 10 2003
%C Also, number of domino tilings in S_5 X P_2n. - _Ralf Stephan_, Mar 30 2004. Here S_5 is the star graph on 5 vertices with the edges {1,2}, {1,3}, {1,4}, {1,5}.
%C If x is in the sequence then so is x*(8*x^2-3). - _James R. Buddenhagen_, Jan 13 2005
%C In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - _Paul Barry_, Mar 13 2005
%C a(n) = L(n,6), where L is defined as in A108299; see also A002315 for L(n,-6). - _Reinhard Zumkeller_, Jun 01 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 largest T-circle that intersects C(n-1). C(n) has radius a(n) and the coordinates of its points of intersection with C(n-1) are A001108(n) and A055997(n). Cf. A001109. - _Charlie Marion_, Sep 14 2005
%C Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5} which do not end in 0. - _Tanya Khovanova_, Jan 10 2007
%C The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators = A002315 and denominators = {a(n)}. - _Clark Kimberling_, Aug 26 2008
%C Apparently Ljunggren shows that 169 is the last square term.
%C If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p < r then s-r = p+q+1. - _Mohamed Bouhamida_, Aug 29 2009
%C If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = Y^2 with p < r then r = 3p+2q+1 and s = 4p+3q+2. - _Mohamed Bouhamida_, Sep 02 2009
%C Equals INVERT transform of A005054: (1, 4, 20, 100, 500, 2500, ...) and INVERTi transform of A122074: (1, 6, 40, 268, 1796, ...). - _Gary W. Adamson_, Jul 22 2010
%C a(n) is the number of compositions of n when there are 5 types of 1 and 4 types of other natural numbers. - _Milan Janjic_, Aug 13 2010
%C The remainder after division of a(n) by a(k) appears to belong to a periodic sequence: 1, 5, ..., a(k-1), 0, a(k)-a(k-1), ..., a(k)-1, a(k)-1, ..., a(k)-a(k-1), 0, a(k-1), ..., 5, 1. See Bouhamida's Sep 01 2009 comment. - _Charlie Marion_, May 02 2011
%C Apart from initial 1: subsequence of A198389, see also A198385. - _Reinhard Zumkeller_, Oct 25 2011
%C (a(n+1), 2*b(n+1)) and (a(n+2), 2*b(n+1)), n >= 0, with b(n):= A001109(n), give the (u(2*n), v(2*n)) and (u(2*n+1), v(2*n+1)) sequences, respectively, for Pythagorean triples (x,y,z), where x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, which are generated from (u(0)=1, v(0)=2) by the substitution rule (u,v) -> (2*v+u,v) if u < v and (u,v) -> (u,2*u+v) if u > v. This leads to primitive triples because gcd(u,v) = 1 is respected. This corresponds to (primitive) Pythagorean triangles with |x-y|=1 (the catheti differ by one length unit). This (u,v) sequence starts with (1,2), (5,2), (5,12), (29,12), (29,70) ... - _Wolfdieter Lang_, Mar 06 2012
%C Area of the Fibonacci snowflake of order n. - _José Luis Ramírez Ramírez_, Dec 13 2012
%C Area of the 3-generalized Fibonacci snowflake of order n, n >= 3. - _José Luis Ramírez Ramírez_, Dec 13 2012
%C For the o.g.f. given by _Johannes W. Meijer_, Aug 01 2010, in the formula section see a comment under A077445. - _Wolfdieter Lang_, Jan 18 2013
%C Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 4 = 0. - _Colin Barker_, Feb 04 2014
%C Length of period of the continued fraction expansion of a(n)*sqrt(2) is 1, the corresponding repeating value is A077444(n). - _Ralf Stephan_, Feb 20 2014
%C Positive values of x (or y) satisfying x^2 - 34xy + y^2 + 144 = 0. - _Colin Barker_, Mar 04 2014
%C The value of the hypotenuse in each triple of the Tree of primitive Pythagorean triples (cf. Wikipedia link) starting with root (3,4,5) and recursively selecting the central branch at each triple node of the tree. - _Stuart E Anderson_, Feb 05 2015
%C Positive integers z such that z^2 is a centered square number (A001844). - _Colin Barker_, Feb 12 2015
%C The aerated sequence (b(n)) n >= 1 = [1, 0, 5, 0, 29, 0, 169, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. - _Peter Bala_, Mar 25 2015
%C A002315(n-1)/a(n) is the closest rational approximation of sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001541 and A001542 give a complete set of closest rational approximations of sqrt(2) with restricted numerator or denominator. A002315(n-1)/a(n) < sqrt(2). - _A.H.M. Smeets_, May 28 2017
%C Equivalently, numbers x such that (x-1)*x/2 + x*(x+1)/2 = y^2 + (y+1)^2. y-values are listed in A001652. Example: for x=29 and y=20, 28*29/2 + 29*30/2 = 20^2 + 21^2. - _Bruno Berselli_, Mar 19 2018
%C From _Wolfdieter Lang_, Jun 13 2018: (Start)
%C (a(n), a(n+1)), with a(0):= 1, give all proper positive solutions m1 = m1(n) and m2 = m2(n), with m1 < m2 and n >= 0, of the Markoff triple (m, m1, m2) (see A002559) for m = 2, i.e., m1^2 - 6*m1*m2 + m2^2 = -4. Hence the unique Markoff triple with largest value m = 2 is (1, 1, 2) (for general m from A002559 this is the famous uniqueness conjecture).
%C For X = m2 - m1 and Y = m2 this becomes the reduced indefinite quadratic form representation X^2 + 4*X*Y - 4*Y^2 = -4, with discriminant 32, and the only proper fundamental solution (X(0), Y(0)) = (0, 1). For all nonnegative proper (X(n), Y(n)) solutions see (A005319(n) = a(n+1) - a(n), a(n+1)), for n >= 0. (End)
%C Each Pell(2*k+1) = a(k+1) number with k >= 3 appears as largest number of an ordered Markoff (Markov) triple [x, y, m] with smallest value x = 2 as [2, Pell(2*k-1), Pell(2*k+1)]. This known result follows also from all positive proper solutions of the Pell equation q^2 - 2*m^2 = -1 which are q = q(k) = A002315(k) and m = m(k) = Pell(2*k+1), for k >= 0. y = y(k) = m(k) - 2*q(k) = Pell(2*k-1), with Pell(-1) = 1. The k = 0 and 1 cases do not satisfy x=2 <= y(k) <= m(k). The numbers 1 and 5 appear also as largest Markoff triple members because they are also Fibonacci numbers, and for these triples x=1. - _Wolfdieter Lang_, Jul 11 2018
%C All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0 < a < b < c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=a(n+1), z=A002315(n) with 0 < n. - _Michael Somos_, Jun 26 2022
%D R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
%D A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
%D W. Ljunggren, "Zur Theorie der Gleichung x^2+1=Dy^4", Avh. Norske Vid. Akad. Oslo I. 5, 27pp.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - _N. J. A. Sloane_, Mar 08 2022
%D David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 91.
%H T. D. Noe and Eric Chen, <a href="/A001653/b001653.txt">Table of n, a(n) for n = 1..1000 (terms 1..201 from T. D. Noe)</a>
%H I. Adler, <a href="http://www.fq.math.ca/Scanned/7-2/adler.pdf">Three Diophantine equations - Part II</a>, Fib. Quart., 7 (1969), pp. 181-193.
%H César Aguilera, <a href="https://doi.org/10.31219/osf.io/7etk8">Notes on Perfect Numbers</a>, OSF Preprints, 2023, p 21.
%H S. Barbero, U. Cerruti, and N. Murru, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Barbero2/barbero7.html">A Generalization of the Binomial Interpolated Operator and its Action on Linear Recurrent Sequences</a>, J. Int. Seq. 13 (2010) # 10.9.7, proposition 16.
%H A. Blondin-Massé, S. Brlek, S. Labbé, and M. Mendès France, <a href="http://www.labmath.uqam.ca/~annales/volumes/35-2/141.pdf">Fibonacci snowflakes</a>, Special Issue dedicated to Paulo Ribenboim, Annales des Sciences Mathématiques du Québec 35, No 2 (2011).
%H A. J. C. Cunningham, <a href="https://archive.org/details/binomialfactoris01cunn/page/n46/mode/1up">Binomial Factorisations</a>, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
%H J.-P. Ehrmann et al., <a href="http://forumgeom.fau.edu/POLYA/ProblemCenter/POLYA003.html">POLYA003: Integers of the form a/(bc) + b/(ca) + c/(ab)</a>.
%H S. Falcon, <a href="http://dx.doi.org/10.4236/am.2014.515216">Relationships between Some k-Fibonacci Sequences</a>, Applied Mathematics, 2014, 5, 2226-2234.
%H Daniel C. Fielder, <a href="http://www.fq.math.ca/Scanned/6-3/fielder.pdf">Special integer sequences controlled by three parameters</a>, Fibonacci Quarterly 6, 1968, 64-70.
%H Daniel C. Fielder, <a href="http://www.fq.math.ca/Scanned/6-3/errata.pdf">Errata:Special integer sequences controlled by three parameters</a>, Fibonacci Quarterly 6, 1968, 64-70.
%H Alex Fink, Richard K. Guy, and Mark Krusemeyer, <a href="https://doi.org/10.11575/cdm.v3i2.61940">Partitions with parts occurring at most thrice</a>, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
%H T. W. Forget and T. A. Larkin, <a href="http://www.fq.math.ca/Scanned/6-3/forget.pdf">Pythagorean triads of the form X, X+1, Z described by recurrence sequences</a>, Fib. Quart., 6 (No. 3, 1968), 94-104.
%H L. J. Gerstein, <a href="http://www.jstor.org/stable/30044157">Pythagorean triples and inner products</a>, Math. Mag., 78 (2005), 205-213.
%H Glass, Darren B. <a href="https://doi.org/10.1016/j.ejc.2016.09.010">Critical groups of graphs with dihedral actions. II</a>. Eur. J. Comb. 61, 25-46 (2017).
%H M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, <a href="http://www.jstor.org/stable/2968551">Problem 47</a>, Amer. Math. Monthly, 4 (1897), 25-28.
%H R. J. Hetherington, <a href="/A000129/a000129.pdf">Letter to N. J. A. Sloane, Oct 26 1974</a>
%H H. J. Hindin, <a href="/A006062/a006062.pdf">Stars, hexes, triangular numbers and Pythagorean triples</a>, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=403">Encyclopedia of Combinatorial Structures 403</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%H Giuseppe Lancia and Paolo Serafini, <a href="https://pdfs.semanticscholar.org/91a3/5543a09ee02c713a2a204057cdef88b00d2c.pdf">Polyhedra</a>. Chapter 2 of Compact Extended Linear Programming Models (2018). EURO Advanced Tutorials on Operational Research. Springer, Cham., 11.
%H Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2019volume19/FG201902index.html">Integer Sequences and Circle Chains Inside a Hyperbola</a>, Forum Geometricorum (2019) Vol. 19, 11-16.
%H A. Martin, <a href="https://web.archive.org/web/20171111031012/http://www.mathunion.org/ICM/ICM1912.2/Main/icm1912.2.0040.0058.ocr.pdf">Table of prime rational right-angled triangles</a>, The Mathematical Magazine, 2 (1910), 297-324.
%H A. Martin, <a href="/A001652/a001652.pdf">Table of prime rational right-angled triangles</a> (annotated scans of a few pages).
%H Sam Northshield, <a href="https://www.fq.math.ca/Papers1/58-5/northshield.pdf">Topographs; Conway and Otherwise</a>, Fibonacci Quart. 58 (2020), no. 5, 172-189. See p. 16.
%H J.-C. Novelli and J.-Y. Thibon, <a href="http://arxiv.org/abs/1403.5962">Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions</a>, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
%H James M. Parks, <a href="https://arxiv.org/abs/2107.06891">Computing Pythagorean Triples</a>, arXiv:2107.06891 [math.GM], 2021.
%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 José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, <a href="http://arxiv.org/abs/1212.1368">A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake</a>, arXiv preprint arXiv:1212.1368 [cs.DM], 2012-2014.
%H Dan Romik, <a href="https://doi.org/10.1090/S0002-9947-08-04467-X">The dynamics of Pythagorean Triples</a>, Trans. Amer. Math. Soc. 360 (2008), 6045-6064.
%H Michael Z. Spivey and Laura L. Steil, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL9/Spivey/spivey7.html">The k-Binomial Transforms and the Hankel Transform</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
%H P. E. Trier, <a href="https://archim.org.uk/eureka/archive/Eureka-4.pdf">"Almost Isosceles" Right-Angled Triangles</a>, Eureka, No. 4, May 1940, pp. 9 - 11.
%H Michel Waldschmidt, <a href="http://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/ContinuedFractionsOujda2015.pdf">Continued fractions</a>, Ecole de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/NSWNumber.html">NSW Number</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Tree_of_primitive_Pythagorean_triples">Tree of primitive Pythagorean triples</a>.
%H H. C. Williams and R. K. Guy, <a href="http://dx.doi.org/10.1142/S1793042111004587">Some fourth-order linear divisibility sequences</a>, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
%H H. C. Williams and R. K. Guy, <a href="http://www.emis.de/journals/INTEGERS/papers/a17self/a17self.Abstract.html ">Some Monoapparitic Fourth Order Linear Divisibility Sequences</a>, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</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-x)/(1-6*x+x^2).
%F a(n) = 6*a(n-1) - a(n-2) with a(1)=1, a(2)=5.
%F 4*a(n) = A077445(n).
%F Can be extended backwards by a(-n+1) = a(n).
%F a(n) = sqrt((A002315(n)^2 + 1)/2). [Inserted by _N. J. A. Sloane_, May 08 2000]
%F a(n+1) = S(n, 6)-S(n-1, 6), n>=0, with S(n, 6) = A001109(n+1), S(-2, 6) := -1. S(n, x)=U(n, x/2) are Chebyshev's polynomials of the second kind. Cf. triangle A049310. a(n+1) = T(2*n+1, sqrt(2))/sqrt(2), n>=0, with T(n, x) Chebyshev's polynomials of the first kind. [Offset corrected by _Wolfdieter Lang_, Mar 06 2012]
%F a(n) = A000129(2n+1). - _Ira M. Gessel_, Sep 27 2002
%F a(n) ~ (1/4)*sqrt(2)*(sqrt(2) + 1)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
%F a(n) = (((3 + 2*sqrt(2))^(n+1) - (3 - 2*sqrt(2))^(n+1)) - ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n)) / (4*sqrt(2)). Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - _Gregory V. Richardson_, Oct 12 2002
%F Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 4) = a(n). - _Benoit Cloitre_, Nov 10 2002
%F For n and j >= 1, Sum_{k=0..j} a(k)*a(n) - Sum_{k=0..j-1} a(k)*a(n-1) = A001109(j+1)*a(n) - A001109(j)*a(n-1) = a(n+j); e.g., (1+5+29)*5 - (1+5)*1=169. - _Charlie Marion_, Jul 07 2003
%F From _Charlie Marion_, Jul 16 2003: (Start)
%F For n >= k >= 0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g., 169^2 = 5741*5 - 144.
%F For n > 0, a(n) ^2 - a(n-1)^2 = 4*Sum_{k=0..2*n-1} a(k) = 4*A001109(2n); e.g., 985^2 - 169^2 = 4*(1 + 5 + 29 + ... + 195025) = 4*235416.
%F Sum_{k=0..n} ((-1)^(n-k)*a(k)) = A079291(n+1); e.g., -1 + 5 - 29 + 169 = 144.
%F A001652(n) + A046090(n) - a(n) = A001542(n); e.g., 119 + 120 - 169 = 70.
%F (End)
%F Sum_{k=0...n} ((2k+1)*a(n-k)) = A001333(n+1)^2 - (1 + (-1)^(n+1))/2; e.g., 1*169 + 3*29 + 5*5 + 7*1 = 288 = 17^2 - 1; 1*29 + 3*5 + 5*1 = 49 = 7^2. - _Charlie Marion_, Jul 18 2003
%F Sum_{k=0...n} a(k)*a(n) = Sum_{k=0..n} a(2k) and Sum_{k=0..n} a(k)*a(n+1) = Sum_{k=0..n} a(2k+1); e.g., (1+5+29)*29 = 1+29+985 and (1+5+29)*169 = 5+169+5741. - _Charlie Marion_, Sep 22 2003
%F For n >= 3, a_{n} = 7(a_{n-1} - a_{n-2}) + a_{n-3}, with a_1 = 1, a_2 = 5 and a_3 = 29. a(n) = ((-1+2^(1/2))/2^(3/2))*(3 - 2^(3/2))^n + ((1+2^(1/2))/2^(3/2))*(3 + 2^(3/2))^n. - _Antonio Alberto Olivares_, Oct 13 2003
%F Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. - _Charlie Marion_, Jul 01 2003
%F Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for n > 0, a(n)*b(n)*c(n)/(a(n)+b(n)+c(n)) = Sum_{k=0..n} c(2*k+1); e.g., 20*21*29/(20+21+29) = 5+169 = 174; a(n)*b(n)*c(n)/(a(n-1)+b(n-1)+c(n-1)) = Sum_{k=0..n} c(2*k); e.g., 119*120*169/(20+21+29) = 1+29+985+33461 = 34476. - _Charlie Marion_, Dec 01 2003
%F Also solutions x > 0 of the equation floor(x*r*floor(x/r))==floor(x/r*floor(x*r)) where r=1+sqrt(2). - _Benoit Cloitre_, Feb 15 2004
%F a(n)*a(n+3) = 24 + a(n+1)*a(n+2). - _Ralf Stephan_, May 29 2004
%F For n >= k, a(n)*a(n+2*k+1) - a(n+k)*a(n+k+1) = a(k)^2-1; e.g., 29*195025-985*5741 = 840 = 29^2-1; 1*169-5*29 = 24 = 5^2-1; a(n)*a(n+2*k)-a(n+k)^2 = A001542(k)^2; e.g., 169*195025-5741^2 = 144 = 12^2; 1*29-5^2 = 4 = 2^2. - _Charlie Marion_ Jun 02 2004
%F For all k, a(n) is a factor of a((2n+1)*k+n). a((2*n+1)*k+n) = a(n)*(Sum_{j=0..k-1} (-1)^j*(a((2*n+1)*(k-j)) + a((2*n+1)*(k-j)-1))+(-1)^k); e.g., 195025 = 5*(33461+5741-169-29+1); 7645370045 = 169*(6625109+1136689-1).- _Charlie Marion_, Jun 04 2004
%F a(n) = Sum_{k=0..n} binomial(n+k, 2*k)4^k. - _Paul Barry_, Aug 30 2004 [offset 0]
%F a(n) = Sum_{k=0..n} binomial(2*n+1, 2*k+1)*2^k. - _Paul Barry_, Sep 30 2004 [offset 0]
%F For n < k, a(n)*A001541(k) = A011900(n+k)+A053141(k-n-1); e.g., 5*99 = 495 = 493+2. For n >= k, a(n)*A001541(k) = A011900(n+k)+A053141(n-k); e.g., 29*3 = 87 = 85+2. - _Charlie Marion_, Oct 18 2004
%F a(n) = (-1)^n*U(2*n, i*sqrt(4)/2) = (-1)^n*U(2*n, i), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - _Paul Barry_, Mar 13 2005 [offset 0]
%F a(n) = Pell(2*n+1) = Pell(n)^2 + Pell(n+1)^2. - _Paul Barry_, Jul 18 2005 [offset 0]
%F a(n)*a(n+k) = A000129(k)^2 + A000129(2n+k+1)^2; e.g., 29*5741 = 12^2+169^2. - _Charlie Marion_, Aug 02 2005
%F Let a(n)*a(n+k) = x. Then 2*x^2-A001541(k)*x+A001109(k)^2 = A001109(2*n+k+1)^2; e.g., let x=29*985; then 2x^2-17x+6^2 = 40391^2; cf. A076218. - _Charlie Marion_, Aug 02 2005
%F With a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = (a^((2n+1)/2)+b^((2n+1)/2))/(2*sqrt(2)). a(n) = A001109(n+1)-A001109(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
%F If k is in the sequence, then the next term is floor(k*(3+2*sqrt(2))). - _Lekraj Beedassy_, Jul 19 2005
%F a(n) = Jacobi_P(n,-1/2,1/2,3)/Jacobi_P(n,-1/2,1/2,1). - _Paul Barry_, Feb 03 2006 [offset 0]
%F a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(n,j)*C(n-j,k)*Pell(n-j+1), where Pell = A000129. - _Paul Barry_, May 19 2006 ]offset 0]
%F a(n) = round(sqrt(A002315(n)^2/2)). - _Lekraj Beedassy_, Jul 15 2006
%F a(n) = A079291(n) + A079291(n+1). - _Lekraj Beedassy_, Aug 14 2006
%F a(n+1) = 3*a(n) + sqrt(8*a(n)^2-4), a(1)=1. - _Richard Choulet_, Sep 18 2007
%F 6*a(n)*a(n+1) = a(n)^2+a(n+1)^2+4; e.g., 6*5*29 = 29^2+5^2+4; 6*169*985 = 169^2+985^2+4. - _Charlie Marion_, Oct 07 2007
%F 2*A001541(k)*a(n)*a(n+k) = a(n)^2+a(n+k)^2+A001542(k)^2; e.g., 2*3*5*29 = 5^2+29^2+2^2; 2*99*29*5741 = 2*99*29*5741=29^2+5741^2+70^2. - _Charlie Marion_, Oct 12 2007
%F [a(n), A001109(n)] = [1,4; 1,5]^n * [1,0]. - _Gary W. Adamson_, Mar 21 2008
%F From _Charlie Marion_, Apr 10 2009: (Start)
%F In general, for n >= k, a(n+k) = 2*A001541(k)*a(n)-a(n-k);
%F e.g., a(n+0) = 2*1*a(n)-a(n); a(n+1) = 6*a(n)-a(n-1); a(6+0) = 33461 = 2*33461-33461; a(5+1) = 33461 = 6*5741-985; a(4+2) = 33461 = 34*985-29; a(3+3) = 33461 = 198*169-1.
%F (End)
%F G.f.: sqrt(x)*tan(4*arctan(sqrt(x)))/4. - _Johannes W. Meijer_, Aug 01 2010
%F Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = a(n-1)*k-((k-1)/(k^n)). - _Charles L. Hohn_, Mar 06 2011
%F Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(n) - a(n-1)). - _Charles L. Hohn_, Apr 04 2011
%F Let T(n) be the n-th triangular number; then, for n > 0, T(a(n)) + A001109(n-1) = A046090(n)^2. See also A046090. - _Charlie Marion_, Apr 25 2011
%F For k > 0, a(n+2*k-1) - a(n) = 4*A001109(n+k-1)*A002315(k-1); a(n+2*k) - a(n) = 4*A001109(k)*A002315(n+k-1). - _Charlie Marion_, Jan 06 2012
%F a(k+j+1) = (A001541(k)*A001541(j) + A002315(k)*A002315(j))/2. - _Charlie Marion_, Jun 25 2012
%F a(n)^2 = 2*A182435(n)*(A182435(n)-1)+1. - _Bruno Berselli_, Oct 23 2012
%F a(n) = A143608(n-1)*A143608(n) + 1 = A182190(n-1)+1. - _Charlie Marion_, Dec 11 2012
%F G.f.: G(0)*(1-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 a(n+1) = 4*A001652(n) + 3*a(n) + 2 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - _Hermann Stamm-Wilbrandt_, Jul 27 2014
%F a(n)^2 = A001652(n-1)^2 + (A001652(n-1)+1)^2. - _Hermann Stamm-Wilbrandt_, Aug 31 2014
%F Sum_{n >= 2} 1/( a(n) - 1/a(n) ) = 1/4. - _Peter Bala_, Mar 25 2015
%F a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 2^k * 2^floor(k/2). - _David Pasino_, Jul 09 2016
%F E.g.f.: (sqrt(2)*sinh(2*sqrt(2)*x) + 2*cosh(2*sqrt(2)*x))*exp(3*x)/2. - _Ilya Gutkovskiy_, Jul 09 2016
%F a(n+2) = (a(n+1)^2 + 4)/a(n). - _Vladimir M. Zarubin_, Sep 06 2016
%F a(n) = 2*A053141(n)+1. - _R. J. Mathar_, Aug 16 2019
%F For n>1, a(n) is the numerator of the continued fraction [1,4,1,4,...,1,4] with (n-1) repetitions of 1,4. For the denominators see A005319. - _Greg Dresden_, Sep 10 2019
%F a(n) = round(((2+sqrt(2))*(3+2*sqrt(2))^(n-1))/4). - _Paul Weisenhorn_, May 23 2020
%F a(n+1) = Sum_{k >= n} binomial(2*k,2*n)*(1/2)^(k+1). Cf. A102591. - _Peter Bala_, Nov 29 2021
%F a(n+1) = 3*a(n) + A077444(n). - _César Aguilera_, Jul 13 2023
%e From _Muniru A Asiru_, Mar 19 2018: (Start)
%e For k=1, 2*1^2 - 1 = 2 - 1 = 1 = 1^2.
%e For k=5, 2*5^2 - 1 = 50 - 1 = 49 = 7^2.
%e For k=29, 2*29^2 - 1 = 1682 - 1 = 1681 = 41^2.
%e ... (End)
%e G.f. = x + 5*x^2 + 29*x^3 + 169*x^4 + 985*x^5 + 5741*x^6 + ... - _Michael Somos_, Jun 26 2022
%p a[0]:=1: a[1]:=5: 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
%p A001653:=-(-1+5*z)/(z**2-6*z+1); # Conjectured (correctly) by _Simon Plouffe_ in his 1992 dissertation; gives sequence except for one of the leading 1's
%t LinearRecurrence[{6,-1}, {1,5}, 40] (* _Harvey P. Dale_, Jul 12 2011 *)
%t a[ n_] := -(-1)^n ChebyshevU[2 n - 2, I]; (* _Michael Somos_, Jul 22 2018 *)
%t Numerator[{1} ~Join~
%t Table[FromContinuedFraction[Flatten[Table[{1, 4}, n]]], {n, 1, 40}]]; (* _Greg Dresden_, Sep 10 2019 *)
%o (PARI) {a(n) = subst(poltchebi(n-1) + poltchebi(n), x, 3)/4}; /* _Michael Somos_, Nov 02 2002 */
%o (PARI) a(n)=([5,2;2,1]^(n-1))[1,1] \\ Lambert Klasen (lambert.klasen(AT)gmx.de), corrected by _Eric Chen_, Jun 14 2018
%o (PARI) {a(n) = -(-1)^n * polchebyshev(2*n-2, 2, I)}; /* _Michael Somos_, Jun 26 2022 */
%o (Haskell)
%o a001653 n = a001653_list !! n
%o a001653_list = 1 : 5 : zipWith (-) (map (* 6) $ tail a001653_list) a001653_list
%o -- _Reinhard Zumkeller_, May 07 2013
%o (Magma) I:=[1,5]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Feb 22 2014
%o (GAP) a:=[1,5];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # _Muniru A Asiru_, Mar 19 2018
%Y Other two sides are A001652, A046090.
%Y Cf. A001519, A001109, A005054, A122074, A056220, A056869 (subset of primes).
%Y Cf. A000217, A000290, A002315, A002559, A005319, A069894.
%Y Row 6 of array A094954.
%Y Row 1 of array A188647.
%Y Cf. similar sequences listed in A238379.
%K nonn,easy,nice
%O 1,2
%A _N. J. A. Sloane_
%E Additional comments from _Wolfdieter Lang_, Feb 10 2000
%E Better description from _Harvey P. Dale_, Jan 15 2002
%E Edited by _N. J. A. Sloane_, Nov 02 2002
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