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%I A002531 M1340 N0513 #144 Nov 04 2023 20:11:33
%S A002531 1,1,2,5,7,19,26,71,97,265,362,989,1351,3691,5042,13775,18817,51409,
%T A002531 70226,191861,262087,716035,978122,2672279,3650401,9973081,13623482,
%U A002531 37220045,50843527,138907099,189750626,518408351,708158977,1934726305
%N A002531 a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1); a(0) = a(1) = 1.
%C A002531 Numerators of continued fraction convergents to sqrt(3), for n >= 1.
%C A002531 For the denominators see A002530.
%C A002531 Consider the mapping f(a/b) = (a + 3*b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the convergents 1/1, 2/1, 5/3, 7/4, 19/11, ... converging to sqrt(3). Sequence contains the numerators. - _Amarnath Murthy_, Mar 22 2003
%C A002531 In the Murthy comment if we take a = 0, b = 1 then the denominator of the reduced fraction is a(n+1). A083336(n)/a(n+1) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 26 2003
%C A002531 If signs are disregarded, all terms of A002316 appear to be elements of this sequence. - _Creighton Dement_, Jun 11 2007
%C A002531 2^(-floor(n/2))*(1 + sqrt(3))^n = a(n) + A002530(n)*sqrt(3); integers in the real quadratic number field Q(sqrt(3)). - _Wolfdieter Lang_, Feb 10 2018
%C A002531 Let T(n) = A000034(n), U(n) = A002530(n), V(n) = a(n), x(n) = U(n)/V(n). Then T(n*m) * U(n+m) = U(n)*V(m) + U(m)*V(n), T(n*m) * V(n+m) = 3*U(n)*U(m) + V(m)*V(n), x(n+m) = (x(n) + x(m))/(1 + 3*x(n)*x(m)). - _Michael Somos_, Nov 29 2022
%D A002531 I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 181.
%D A002531 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A002531 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D A002531 A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.
%H A002531 Harry J. Smith, Table of n, a(n) for n = 0..2000
%H A002531 MacTutor, D'Arcy Thompson on Greek irrationals
%H A002531 Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H A002531 Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
%H A002531 Albert Tarn, Approximations to certain square roots and the series of numbers connected therewith [Annotated scanned copy]
%H A002531 D'Arcy Thompson, Excess and Defect: Or the Little More and the Little Less, Mind, New Series, Vol. 38, No. 149 (Jan., 1929), pp. 43-55 (13 pages). See page 48.
%H A002531 Hein van Winkel, Q-quadrangles inscribed in a circle, 2014. See Table 1. [Reference from Antreas Hatzipolakis, Jul 14 2014]
%H A002531 Index entries for "core" sequences
%H A002531 Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).
%H A002531 Index entries for sequences related to Chebyshev polynomials.
%F A002531 G.f.: (1 + x - 2*x^2 + x^3)/(1 - 4*x^2 + x^4).
%F A002531 a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1), n > 0.
%F A002531 a(2*n) = (1/2)*((2 + sqrt(3))^n+(2 - sqrt(3))^n); a(2*n) = A003500(n)/2; a(2*n+1) = round(1/(1 + sqrt(3))*(2 + sqrt(3))^n). - _Benoit Cloitre_, Dec 15 2002
%F A002531 a(n) = ((1 + sqrt(3))^n + (1 - sqrt(3))^n)/(2*2^floor(n/2)). - _Bruno Berselli_, Nov 10 2011
%F A002531 a(n) = A080040(n)/(2*2^floor(n/2)). - _Ralf Stephan_, Sep 08 2013
%F A002531 a(2*n) = (-1)^n*T(2*n,u) and a(2*n+1) = (-1)^n*1/u*T(2*n+1,u), where u = sqrt(-1/2) and T(n,x) denotes the Chebyshev polynomial of the first kind. - _Peter Bala_, May 01 2012
%F A002531 a(n) = (-sqrt(2)*i)^n*T(n, sqrt(2)*i/2)*2^(-floor(n/2)) = A026150(n)*2^(-floor(n/2)), n >= 0, with i = sqrt(-1) and the Chebyshev T polynomials (A053120). - _Wolfdieter Lang_, Feb 10 2018
%F A002531 From _Franck Maminirina Ramaharo_, Nov 14 2018: (Start)
%F A002531 a(n) = ((1 - sqrt(2))*(-1)^n + 1 + sqrt(2))*(((sqrt(2) - sqrt(6))/2)^n + ((sqrt(6) + sqrt(2))/2)^n)/4.
%F A002531 E.g.f.: cosh(sqrt(3/2)*x)*(sqrt(2)*sinh(x/sqrt(2)) + cosh(x/sqrt(2))). (End)
%F A002531 a(n) = (-1)^n*a(-n) for all n in Z. - _Michael Somos_, Mar 22 2022
%F A002531 a(n) = 4*a(n-2) - a(n-4). - _Boštjan Gec_, Sep 21 2023
%e A002531 1 + 1/(1 + 1/(2 + 1/(1 + 1/2))) = 19/11 so a(5) = 19.
%e A002531 Convergents are 1, 2, 5/3, 7/4, 19/11, 26/15, 71/41, 97/56, 265/153, 362/209, 989/571, 1351/780, 3691/2131, ... = A002531/A002530.
%e A002531 G.f. = 1 + x + 2*x^2 + 5*x^3 + 7*x^4 + 19*x^5 + 26*x^6 + 71*x^7 + ... - _Michael Somos_, Mar 22 2022
%p A002531 A002531 := proc(n) option remember; if n=0 then 0 elif n=1 then 1 elif n=2 then 1 elif type(n,odd) then A002531(n-1)+A002531(n-2) else 2*A002531(n-1)+A002531(n-2) fi; end; [ seq(A002531(n), n=0..50) ];
%p A002531 with(numtheory): tp := cfrac (tan(Pi/3),100): seq(nthnumer(tp,i), i=-1..32 ); # _Zerinvary Lajos_, Feb 07 2007
%p A002531 A002531:=(1+z-2*z**2+z**3)/(1-4*z**2+z**4); # _Simon Plouffe_; see his 1992 dissertation
%t A002531 Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[3], n]]], {n, 1, 40}], 1, 1] (* _Stefan Steinerberger_, Apr 01 2006 *)
%t A002531 Join[{1},Numerator[Convergents[Sqrt[3],40]]] (* _Harvey P. Dale_, Jan 23 2012 *)
%t A002531 CoefficientList[Series[(1 + x - 2 x^2 + x^3)/(1 - 4 x^2 + x^4), {x, 0, 30}], x] (* _Vincenzo Librandi_, Nov 01 2014 *)
%t A002531 LinearRecurrence[{0, 4, 0, -1}, {1, 1, 2, 5}, 35] (* _Robert G. Wilson v_, Feb 11 2018 *)
%t A002531 a[ n_] := ChebyshevT[n, Sqrt[-1/2]]*Sqrt[2]^Mod[n,2]/I^n //Simplify; (* _Michael Somos_, Mar 22 2022 *)
%o A002531 (PARI) a(n)=contfracpnqn(vector(n,i,1+(i>1)*(i%2)))[1,1]
%o A002531 (PARI) apply( {A002531(n,w=quadgen(12))=real((2+w)^(n\/2)*if(bittest(n, 0), w-1, 1))}, [0..30]) \\ _M. F. Hasler_, Nov 04 2019
%o A002531 (Magma) m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1 +x-2*x^2+x^3)/(1-4*x^2+x^4))); // _G. C. Greubel_, Nov 16 2018
%o A002531 (Sage) s=((1+x-2*x^2+x^3)/(1-4*x^2+x^4)).series(x,40); s.coefficients(x, sparse=False) # _G. C. Greubel_, Nov 16 2018
%o A002531 (GAP) a:=[1,1,2,5];; for n in [5..40] do a[n]:=4*a[n-2]-a[n-4]; od; a; # _G. C. Greubel_, Nov 16 2018
%Y A002531 Bisections are A001075 and A001834.
%Y A002531 Cf. A002530 (denominators), A048788.
%Y A002531 Cf. A002316.
%Y A002531 Cf. A083332, A199710, A026150, A053120.
%K A002531 nonn,frac,easy,core,nice
%O A002531 0,3
%A A002531 _N. J. A. Sloane_
%E A002531 Name edited (as by discussion in A002530) by _M. F. Hasler_, Nov 04 2019
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