The OEIS is supported by the many generous donors to the OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A243470 Numerators of the rational convergents to the periodic continued fraction 1/(2 + 1/(7 + 1/(2 + 1/(7 + ...)))). 2
 1, 7, 15, 112, 239, 1785, 3809, 28448, 60705, 453383, 967471, 7225680, 15418831, 115157497, 245733825, 1835294272, 3916322369, 29249550855, 62415424079, 466157519408, 994730462895, 7429270759673, 15853271982241, 118402174635360, 252657621252961, 1887005523406087 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS The sequence of convergents to the simple periodic continued fraction 1/(2 + 1/(7 + 1/(2 + 1/(7 + ...)))) begins [0/1, 1/2, 7/15, 15/32, 112/239, 239/510, ...]. The present sequence is the sequence of numerators of the convergents. It is a strong divisibility sequence, that is gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. The sequence is closely related to A041111, the Lehmer numbers U_n(sqrt(R),Q)) with parameters R = 14 and Q = -1. See A243469 for the sequence of denominators to the convergents. LINKS G. C. Greubel, Table of n, a(n) for n = 1..1000 Peter Bala, Notes on 2-periodic continued fractions and Lehmer sequences L. Euler, Introductio in analysin infinitorum, Vol.1, Chapter 18, section 378. French and German translations. Eric W. Weisstein, MathWorld: Lehmer Number Index to divisibility sequences Index entries for linear recurrences with constant coefficients, signature (0,16,0,-1). FORMULA Let alpha = ( sqrt(14) + sqrt(18) )/2 and beta = ( sqrt(14) - sqrt(18) )/2 be the roots of the equation x^2 - sqrt(14)*x - 1 = 0. Then a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, while a(n) = 7*(alpha^n - beta^n)/(alpha^2 - beta^2) for n even. a(2*n + 1) = product {k = 1..n} (14 + 4*cos^2(k*Pi/(2*n+1)); a(2*n) = 7*product {k = 1..n-1} (14 + 4*cos^2(k*Pi/(2*n)). Recurrence equations: a(0) = 0, a(1) = 1 and for n >= 2, a(2*n) = 7*a(2*n - 1) + a(2*n - 2) and a(2*n + 1) = 2*a(2*n) + a(2*n - 1). Fourth-order recurrence: a(n) = 16*a(n - 2) - a(n - 4) for n >= 5. O.g.f.: x*(1 + 7*x - x^2)/(1 - 16*x^2 + x^4). a(2n-1) = A157456(n), a(2n) = 7*A077412(n-1). - Ralf Stephan, Jun 13 2014 a(n) = (1/2)*( 7*(1+(-1)^n)*ChebyshevU((n-2)/2, 8) + (1-(-1)^n)*(ChebyshevU((n- 1)/2, 8) - ChebyshevU((n-3)/2, 8)) ). - G. C. Greubel, May 21 2022 MATHEMATICA LinearRecurrence[{0, 16, 0, -1}, {1, 7, 15, 112}, 30] (* Harvey P. Dale, Nov 06 2017 *) PROG (PARI) Vec(x*(1+7*x-x^2)/(1-16*x^2+x^4)+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2015 (Magma) I:=[1, 7, 15, 112]; [n le 4 select I[n] else 16*Self(n-2) -Self(n-4): n in [1..31]]; // G. C. Greubel, May 21 2022 (SageMath) def b(n): return chebyshev_U(n, 8) # b=A077412 def A243470(n): return 7*((n-1)%2)*b(n//2 -1) +(n%2)*(b((n-1)//2) -b((n-1)//2 -1)) [A243470(n) for n in (1..30)] # G. C. Greubel, May 21 2022 CROSSREFS Cf. A041111, A077412, A243469. Sequence in context: A231466 A032004 A032018 * A068366 A156499 A137881 Adjacent sequences: A243467 A243468 A243469 * A243471 A243472 A243473 KEYWORD nonn,easy,frac AUTHOR Peter Bala, Jun 06 2014 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified April 22 07:09 EDT 2024. Contains 371888 sequences. (Running on oeis4.)