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

0, 1, 16, 225, 3136, 43681, 608400, 8473921, 118026496, 1643897025, 22896531856, 318907548961, 4441809153600, 61866420601441, 861688079266576, 12001766689130625, 167163045568562176, 2328280871270739841, 32428769152221795600, 451674487259834398561

Offset: 0

Author

Wolfdieter Lang, Oct 18 2004

Keywords

• nonn
• easy

Comments

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

Consequently A049451(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011

Links

• Colin Barker, Table of n, a(n) for n = 0..874
• S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Crossrefs

Cf. A001075, A001353, A049451, A092184.

Programs

Mathematica

LinearRecurrence[{# - 1, -# + 1, 1}, {0, 1, #}, 20] &[16] (* Michael De Vlieger, Feb 23 2021 *)

PARI

concat(0, Vec(x*(1+x)/((1-x)*(1-14*x+x^2)) + O(x^50))) \\ Colin Barker, Jun 15 2015

Formula

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.

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).

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

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

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).

Conjecture: 4*A007655(n+1) + A046184(n) = A055793(n+2) + a(n+1). - Creighton Dement, Nov 01 2004

a(n) = (A001075(n)^2-1)/3. - Parker Grootenhuis, Nov 28 2017