A089817 a(n) = 5*a(n-1) - a(n-2) + 1 with a(0)=1, a(1)=6.

1, 6, 30, 145, 696, 3336, 15985, 76590, 366966, 1758241, 8424240, 40362960, 193390561, 926589846, 4439558670, 21271203505, 101916458856, 488311090776, 2339638995025, 11209883884350, 53709780426726, 257339018249281

Offset: 0

Author

Paul Barry, Nov 14 2003

Keywords

• nonn
• easy

Comments

Partial sums of Chebyshev sequence S(n,5) = U(n,5/2) = A004254(n) (Chebyshev's polynomials of the second kind, see A049310). - Wolfdieter Lang, Aug 31 2004

In this sequence 4*a(n)*a(n+2)+1 is a square. - Bruno Berselli, Jun 19 2012

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
• Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
• F. M. van Lamoen, Square wreaths around hexagons, Forum Geometricorum, 6 (2006) 311-325.
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (6,-6,1).

Crossrefs

Cf. A061278, A053142, A101368.

See. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

Programs

Magma

[Round((2/3 - Sqrt(21)/7)*(5/2 - Sqrt(21)/2)^n + (2/3 + Sqrt(21)/7)*(5/2 + Sqrt(21)/2)^n - 1/3): n in [0..30]]; // G. C. Greubel, Nov 20 2017

Mathematica

Join[{a=1,b=6},Table[c=5*b-a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011*)
CoefficientList[Series[1/(1 - 6*x + 6*x^2 - x^3), {x, 0, 50}], x] (* G. C. Greubel, Nov 20 2017 *)

PARI

a(n)=([0,1,0; 0,0,1; 1,-6,6]^n*[1;6;30])[1,1] \\ Charles R Greathouse IV, Nov 29 2016

PARI

x='x+O('x^50); Vec(1/(1-6*x+6*x^2-x^3)) \\ G. C. Greubel, Nov 20 2017

Formula

For n > 0, a(n-1) = Sum_{i=1..n} Sum_{j=1..i} b(n) with b(n) as in A004253.

a(n) = (2/3 - sqrt(21)/7)*(5/2 - sqrt(21)/2)^n + (2/3 + sqrt(21)/7)*(5/2 + sqrt(21)/2)^n - 1/3.

G.f.: 1/((1-x)*(1 - 5*x + x^2)) = 1/(1 - 6*x + 6*x^2 - x^3).

a(n) = 6*a(n-1) - 6*a(n-2) + a(n-3) for n >= 2, a(-1):=0, a(0)=1, a(1)=6.

a(n) = (S(n+1, 5) - S(n, 5) - 1)/3 for n >= 0.

a(n)*a(n-2) = a(n-1)*(a(n-1)-1) for n > 1. - Bruno Berselli, Nov 29 2016