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 A098301 Member r=16 of the family of Chebyshev sequences S_r(n) defined in A092184. 11

%I

%S 0,1,16,225,3136,43681,608400,8473921,118026496,1643897025,

%T 22896531856,318907548961,4441809153600,61866420601441,

%U 861688079266576,12001766689130625,167163045568562176,2328280871270739841,32428769152221795600,451674487259834398561

%N Member r=16 of the family of Chebyshev sequences S_r(n) defined in A092184.

%C Also m such that (3*m^2 + m)/4 = m*(3*m + 1)/4 is a perfect square. - _Ctibor O. Zizka_, Oct 15 2010

%C Consequently A049451(k) is a square if and only if k = a(n). - _Bruno Berselli_, Oct 14 2011

%H Colin Barker, <a href="/A098301/b098301.txt">Table of n, a(n) for n = 0..874</a>

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).

%F a(n) = (T(n, 7)-1)/6 with Chebyshev's polynomials of the first kind evaluated at x=7: T(n, 7) = A011943(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n)/2; therefore: a(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n - 2)/12.

%F a(n) = A001353(n)^2 = S(n-1, 4)^2 with Chebyshev's polynomials of the second kind evaluated at x=4, S(n, 4):=U(n, 2).

%F a(n) = 14*a(n-1) - a(n-2) + 2, n >= 2, a(0)=0, a(1)=1.

%F a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3), n >= 3.

%F G.f.: x*(1+x)/((1-x)*(1 - 14*x + x^2)) = x*(1+x)/(1 - 15*x + 15*x^2 - x^3) (from the Stephan link, see A092184).

%F Conjecture: 4*A007655(n+1) + A046184(n) = A055793(n+2) + a(n+1). - _Creighton Dement_, Nov 01 2004

%F a(n) = (A001075(n)^2-1)/3. - _Parker Grootenhuis_, Nov 28 2017

%o (PARI) concat(0, Vec(x*(1+x)/((1-x)*(1-14*x+x^2)) + O(x^50))) \\ _Colin Barker_, Jun 15 2015

%Y Cf. A001075, A001353, A049451, A092184.

%K nonn,easy

%O 0,3

%A _Wolfdieter Lang_, Oct 18 2004

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Last modified March 31 19:16 EDT 2020. Contains 333151 sequences. (Running on oeis4.)