%I #33 Nov 09 2024 06:18:48
%S 1,7,15,112,239,1785,3809,28448,60705,453383,967471,7225680,15418831,
%T 115157497,245733825,1835294272,3916322369,29249550855,62415424079,
%U 466157519408,994730462895,7429270759673,15853271982241,118402174635360,252657621252961,1887005523406087
%N Numerators of the rational convergents to the periodic continued fraction 1/(2 + 1/(7 + 1/(2 + 1/(7 + ...)))).
%C 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.
%C See A243469 for the sequence of denominators to the convergents.
%H G. C. Greubel, <a href="/A243470/b243470.txt">Table of n, a(n) for n = 1..1000</a>
%H Peter Bala, <a href="/A243469/a243469_1.pdf">Notes on 2-periodic continued fractions and Lehmer sequences</a>
%H L. Euler, <a href="http://eulerarchive.maa.org/pages/E101.html">Introductio in analysin infinitorum</a>, Vol.1, Chapter 18, section 378. French and German translations.
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/LehmerNumber.html">MathWorld: Lehmer Number</a>
%H <a href="/index/Di#divseq">Index to divisibility sequences</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,16,0,-1).
%F 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.
%F a(2*n + 1) = Product_{k = 1..n} (14 + 4*cos^2(k*Pi/(2*n+1)));
%F a(2*n) = 7*Product_{k = 1..n-1} (14 + 4*cos^2(k*Pi/(2*n))).
%F 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).
%F Fourth-order recurrence: a(n) = 16*a(n - 2) - a(n - 4) for n >= 5.
%F O.g.f.: x*(1 + 7*x - x^2)/(1 - 16*x^2 + x^4).
%F a(2n-1) = A157456(n), a(2n) = 7*A077412(n-1). - _Ralf Stephan_, Jun 13 2014
%F 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
%t LinearRecurrence[{0,16,0,-1},{1,7,15,112},30] (* _Harvey P. Dale_, Nov 06 2017 *)
%o (PARI) Vec(x*(1+7*x-x^2)/(1-16*x^2+x^4)+O(x^99)) \\ _Charles R Greathouse IV_, Nov 13 2015
%o (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
%o (SageMath)
%o def b(n): return chebyshev_U(n,8) # b=A077412
%o def A243470(n): return 7*((n-1)%2)*b(n//2 -1) +(n%2)*(b((n-1)//2) -b((n-1)//2 -1))
%o [A243470(n) for n in (1..30)] # _G. C. Greubel_, May 21 2022
%Y Cf. A041111, A077412, A243469.
%K nonn,easy,frac,changed
%O 1,2
%A _Peter Bala_, Jun 06 2014