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A002531
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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.
(Formerly M1340 N0513)
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32
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1, 1, 2, 5, 7, 19, 26, 71, 97, 265, 362, 989, 1351, 3691, 5042, 13775, 18817, 51409, 70226, 191861, 262087, 716035, 978122, 2672279, 3650401, 9973081, 13623482, 37220045, 50843527, 138907099, 189750626, 518408351, 708158977, 1934726305
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OFFSET
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0,3
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COMMENTS
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Numerators of continued fraction convergents to sqrt(3), for n >= 1.
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
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
If signs are disregarded, all terms of A002316 appear to be elements of this sequence. - Creighton Dement, Jun 11 2007
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
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
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REFERENCES
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I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 181.
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).
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.
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LINKS
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FORMULA
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G.f.: (1 + x - 2*x^2 + x^3)/(1 - 4*x^2 + x^4).
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.
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
a(n) = ((1 + sqrt(3))^n + (1 - sqrt(3))^n)/(2*2^floor(n/2)). - Bruno Berselli, Nov 10 2011
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
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
a(n) = ((1 - sqrt(2))*(-1)^n + 1 + sqrt(2))*(((sqrt(2) - sqrt(6))/2)^n + ((sqrt(6) + sqrt(2))/2)^n)/4.
E.g.f.: cosh(sqrt(3/2)*x)*(sqrt(2)*sinh(x/sqrt(2)) + cosh(x/sqrt(2))). (End)
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EXAMPLE
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1 + 1/(1 + 1/(2 + 1/(1 + 1/2))) = 19/11 so a(5) = 19.
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.
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
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MAPLE
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with(numtheory): tp := cfrac (tan(Pi/3), 100): seq(nthnumer(tp, i), i=-1..32 ); # Zerinvary Lajos, Feb 07 2007
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MATHEMATICA
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Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[3], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 01 2006 *)
Join[{1}, Numerator[Convergents[Sqrt[3], 40]]] (* Harvey P. Dale, Jan 23 2012 *)
CoefficientList[Series[(1 + x - 2 x^2 + x^3)/(1 - 4 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Nov 01 2014 *)
a[ n_] := ChebyshevT[n, Sqrt[-1/2]]*Sqrt[2]^Mod[n, 2]/I^n //Simplify; (* Michael Somos, Mar 22 2022 *)
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PROG
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(PARI) a(n)=contfracpnqn(vector(n, i, 1+(i>1)*(i%2)))[1, 1]
(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
(Magma) m:=40; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1 +x-2*x^2+x^3)/(1-4*x^2+x^4))); // G. C. Greubel, Nov 16 2018
(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
(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
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CROSSREFS
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KEYWORD
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nonn,frac,easy,core,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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