cp's OEIS Frontend

A001653 Numbers k such that 2*k^2 - 1 is a square.

%I A001653 M3955 N1630 #463 Aug 26 2025 04:42:15
%S A001653 1,5,29,169,985,5741,33461,195025,1136689,6625109,38613965,225058681,
%T A001653 1311738121,7645370045,44560482149,259717522849,1513744654945,
%U A001653 8822750406821,51422757785981,299713796309065,1746860020068409,10181446324101389,59341817924539925
%N A001653 Numbers k such that 2*k^2 - 1 is a square.
%C A001653 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z values.
%C A001653 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 A001653 (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 A001653 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 A001653 Numbers n such that 2*n^2 = ceiling(sqrt(2)*n*floor(sqrt(2)*n)). - _Benoit Cloitre_, May 10 2003
%C A001653 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 A001653 If x is in the sequence then so is x*(8*x^2-3). - _James R. Buddenhagen_, Jan 13 2005
%C A001653 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 A001653 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 A001653 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 A001653 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 A001653 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 A001653 Apparently Ljunggren shows that 169 is the last square term.
%C A001653 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 A001653 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 A001653 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 A001653 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 A001653 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 A001653 Apart from initial 1: subsequence of A198389, see also A198385. - _Reinhard Zumkeller_, Oct 25 2011
%C A001653 (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 A001653 Area of the Fibonacci snowflake of order n. - _José Luis Ramírez Ramírez_, Dec 13 2012
%C A001653 Area of the 3-generalized Fibonacci snowflake of order n, n >= 3. - _José Luis Ramírez Ramírez_, Dec 13 2012
%C A001653 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 A001653 Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 4 = 0. - _Colin Barker_, Feb 04 2014
%C A001653 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 A001653 Positive values of x (or y) satisfying x^2 - 34xy + y^2 + 144 = 0. - _Colin Barker_, Mar 04 2014
%C A001653 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 A001653 Positive integers z such that z^2 is a centered square number (A001844). - _Colin Barker_, Feb 12 2015
%C A001653 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 A001653 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 A001653 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 A001653 From _Wolfdieter Lang_, Jun 13 2018: (Start)
%C A001653 (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 A001653 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 A001653 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 A001653 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 A001653 A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
%D A001653 John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 188.
%D A001653 W. Ljunggren, "Zur Theorie der Gleichung x^2+1=Dy^4", Avh. Norske Vid. Akad. Oslo I. 5, 27pp.
%D A001653 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A001653 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D A001653 P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - _N. J. A. Sloane_, Mar 08 2022
%D A001653 David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 91.
%H A001653 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 A001653 I. Adler, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/7-2/adler.pdf">Three Diophantine equations - Part II</a>, Fib. Quart., 7 (1969), pp. 181-193.
%H A001653 César Aguilera, <a href="https://doi.org/10.31219/osf.io/7etk8">Notes on Perfect Numbers</a>, OSF Preprints, 2023, p 21.
%H A001653 R. C. Alperin, <a href="https://www.fq.math.ca/Papers/57-4/alperin07132019.pdf">A family of nonlinear recurrences and their linear solutions</a>, Fib. Q., 57:4 (2019), 318-321.
%H A001653 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 A001653 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 A001653 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 A001653 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 A001653 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 A001653 Daniel C. Fielder, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/6-3/fielder.pdf">Special integer sequences controlled by three parameters</a>, Fibonacci Quarterly 6, 1968, 64-70.
%H A001653 Daniel C. Fielder, <a href="https://web.archive.org/web/2024*/https://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 A001653 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 A001653 T. W. Forget and T. A. Larkin, <a href="https://web.archive.org/web/2024*/https://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 A001653 G. Frobenius, <a href="https://edoc.bbaw.de/frontdoor/index/index/docId/5095">Uber die Markoff'schen Zahlen</a>, Sitzungsber. Konig. Preuss. Akad. Wiss. (1913) p. 348-387, Parag. 9
%H A001653 L. J. Gerstein, <a href="http://www.jstor.org/stable/30044157">Pythagorean triples and inner products</a>, Math. Mag., 78 (2005), 205-213.
%H A001653 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 A001653 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 A001653 R. J. Hetherington, <a href="/A000129/a000129.pdf">Letter to N. J. A. Sloane, Oct 26 1974</a>
%H A001653 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 A001653 INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=403">Encyclopedia of Combinatorial Structures 403</a>
%H A001653 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A001653 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%H A001653 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 A001653 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 A001653 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 A001653 A. Martin, <a href="/A001652/a001652.pdf">Table of prime rational right-angled triangles</a> (annotated scans of a few pages).
%H A001653 Sam Northshield, <a href="https://web.archive.org/web/2024*/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 A001653 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 A001653 James M. Parks, <a href="https://arxiv.org/abs/2107.06891">Computing Pythagorean Triples</a>, arXiv:2107.06891 [math.GM], 2021.
%H A001653 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 A001653 Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H A001653 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 A001653 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 A001653 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 A001653 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 A001653 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 A001653 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 A001653 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/NSWNumber.html">NSW Number</a>
%H A001653 Wikipedia, <a href="http://en.wikipedia.org/wiki/Tree_of_primitive_Pythagorean_triples">Tree of primitive Pythagorean triples</a>.
%H A001653 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 A001653 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 A001653 <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>
%H A001653 <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H A001653 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-1).
%F A001653 G.f.: x*(1-x)/(1-6*x+x^2).
%F A001653 a(n) = 6*a(n-1) - a(n-2) with a(1)=1, a(2)=5.
%F A001653 4*a(n) = A077445(n).
%F A001653 Can be extended backwards by a(-n+1) = a(n).
%F A001653 a(n) = sqrt((A002315(n)^2 + 1)/2). [Inserted by _N. J. A. Sloane_, May 08 2000]
%F A001653 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 A001653 a(n) = A000129(2n+1). - _Ira M. Gessel_, Sep 27 2002
%F A001653 a(n) ~ (1/4)*sqrt(2)*(sqrt(2) + 1)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
%F A001653 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 A001653 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 A001653 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 A001653 From _Charlie Marion_, Jul 16 2003: (Start)
%F A001653 For n >= k >= 0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g., 169^2 = 5741*5 - 144.
%F A001653 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 A001653 Sum_{k=0..n} ((-1)^(n-k)*a(k)) = A079291(n+1); e.g., -1 + 5 - 29 + 169 = 144.
%F A001653 A001652(n) + A046090(n) - a(n) = A001542(n); e.g., 119 + 120 - 169 = 70.
%F A001653 (End)
%F A001653 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 A001653 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 A001653 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 A001653 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 A001653 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 A001653 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 A001653 a(n)*a(n+3) = 24 + a(n+1)*a(n+2). - _Ralf Stephan_, May 29 2004
%F A001653 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 A001653 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 A001653 a(n) = Sum_{k=0..n} binomial(n+k, 2*k)4^k. - _Paul Barry_, Aug 30 2004 [offset 0]
%F A001653 a(n) = Sum_{k=0..n} binomial(2*n+1, 2*k+1)*2^k. - _Paul Barry_, Sep 30 2004 [offset 0]
%F A001653 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 A001653 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 A001653 a(n) = Pell(2*n+1) = Pell(n)^2 + Pell(n+1)^2. - _Paul Barry_, Jul 18 2005 [offset 0]
%F A001653 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 A001653 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 A001653 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 A001653 If k is in the sequence, then the next term is floor(k*(3+2*sqrt(2))). - _Lekraj Beedassy_, Jul 19 2005
%F A001653 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 A001653 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 A001653 a(n) = round(sqrt(A002315(n)^2/2)). - _Lekraj Beedassy_, Jul 15 2006
%F A001653 a(n) = A079291(n) + A079291(n+1). - _Lekraj Beedassy_, Aug 14 2006
%F A001653 a(n+1) = 3*a(n) + sqrt(8*a(n)^2-4), a(1)=1. - _Richard Choulet_, Sep 18 2007
%F A001653 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 A001653 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 A001653 [a(n), A001109(n)] = [1,4; 1,5]^n * [1,0]. - _Gary W. Adamson_, Mar 21 2008
%F A001653 From _Charlie Marion_, Apr 10 2009: (Start)
%F A001653 In general, for n >= k, a(n+k) = 2*A001541(k)*a(n)-a(n-k);
%F A001653 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 A001653 (End)
%F A001653 G.f.: sqrt(x)*tan(4*arctan(sqrt(x)))/4. - _Johannes W. Meijer_, Aug 01 2010
%F A001653 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 A001653 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 A001653 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 A001653 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 A001653 a(k+j+1) = (A001541(k)*A001541(j) + A002315(k)*A002315(j))/2. - _Charlie Marion_, Jun 25 2012
%F A001653 a(n)^2 = 2*A182435(n)*(A182435(n)-1)+1. - _Bruno Berselli_, Oct 23 2012
%F A001653 a(n) = A143608(n-1)*A143608(n) + 1 = A182190(n-1)+1. - _Charlie Marion_, Dec 11 2012
%F A001653 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 A001653 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 A001653 a(n)^2 = A001652(n-1)^2 + (A001652(n-1)+1)^2. - _Hermann Stamm-Wilbrandt_, Aug 31 2014
%F A001653 Sum_{n >= 2} 1/( a(n) - 1/a(n) ) = 1/4. - _Peter Bala_, Mar 25 2015
%F A001653 a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 2^k * 2^floor(k/2). - _David Pasino_, Jul 09 2016
%F A001653 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 A001653 a(n+2) = (a(n+1)^2 + 4)/a(n). - _Vladimir M. Zarubin_, Sep 06 2016
%F A001653 a(n) = 2*A053141(n)+1. - _R. J. Mathar_, Aug 16 2019
%F A001653 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 A001653 a(n) = round(((2+sqrt(2))*(3+2*sqrt(2))^(n-1))/4). - _Paul Weisenhorn_, May 23 2020
%F A001653 a(n+1) = Sum_{k >= n} binomial(2*k,2*n)*(1/2)^(k+1). Cf. A102591. - _Peter Bala_, Nov 29 2021
%F A001653 a(n+1) = 3*a(n) + A077444(n). - _César Aguilera_, Jul 13 2023
%e A001653 From _Muniru A Asiru_, Mar 19 2018: (Start)
%e A001653 For k=1, 2*1^2 - 1 = 2 - 1 = 1 = 1^2.
%e A001653 For k=5, 2*5^2 - 1 = 50 - 1 = 49 = 7^2.
%e A001653 For k=29, 2*29^2 - 1 = 1682 - 1 = 1681 = 41^2.
%e A001653 ... (End)
%e A001653 G.f. = x + 5*x^2 + 29*x^3 + 169*x^4 + 985*x^5 + 5741*x^6 + ... - _Michael Somos_, Jun 26 2022
%p A001653 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 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 A001653 LinearRecurrence[{6,-1}, {1,5}, 40] (* _Harvey P. Dale_, Jul 12 2011 *)
%t A001653 a[ n_] := -(-1)^n ChebyshevU[2 n - 2, I]; (* _Michael Somos_, Jul 22 2018 *)
%t A001653 Numerator[{1} ~Join~
%t A001653 Table[FromContinuedFraction[Flatten[Table[{1, 4}, n]]], {n, 1, 40}]]; (* _Greg Dresden_, Sep 10 2019 *)
%o A001653 (PARI) {a(n) = subst(poltchebi(n-1) + poltchebi(n), x, 3)/4}; /* _Michael Somos_, Nov 02 2002 */
%o A001653 (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 A001653 (PARI) {a(n) = -(-1)^n * polchebyshev(2*n-2, 2, I)}; /* _Michael Somos_, Jun 26 2022 */
%o A001653 (Haskell)
%o A001653 a001653 n = a001653_list !! n
%o A001653 a001653_list = 1 : 5 : zipWith (-) (map (* 6) $ tail a001653_list) a001653_list
%o A001653 -- _Reinhard Zumkeller_, May 07 2013
%o A001653 (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 A001653 (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 A001653 Other two sides are A001652, A046090.
%Y A001653 Cf. A001519, A001109, A005054, A122074, A056220, A056869 (subset of primes).
%Y A001653 Cf. A000217, A000290, A002315, A002559, A005319, A069894.
%Y A001653 Row 6 of array A094954.
%Y A001653 Row 1 of array A188647.
%Y A001653 Cf. similar sequences listed in A238379.
%K A001653 nonn,easy,nice,changed
%O A001653 1,2
%A A001653 _N. J. A. Sloane_
%E A001653 Additional comments from _Wolfdieter Lang_, Feb 10 2000
%E A001653 Better description from _Harvey P. Dale_, Jan 15 2002
%E A001653 Edited by _N. J. A. Sloane_, Nov 02 2002