A070997 a(n) = 8*a(n-1) - a(n-2), a(0)=1, a(-1)=1.

1, 7, 55, 433, 3409, 26839, 211303, 1663585, 13097377, 103115431, 811826071, 6391493137, 50320119025, 396169459063, 3119035553479, 24556114968769, 193329884196673, 1522082958604615, 11983333784640247, 94344587318517361

Offset: 0

Author

Joe Keane (jgk(AT)jgk.org), May 18 2002

Keywords

• nonn
• easy

Comments

A Pellian sequence.

In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005

a(n) = L(n,8), where L is defined as in A108299; see also A057080 for L(n,-8). - Reinhard Zumkeller, Jun 01 2005

Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7} which do not end in 0. - Tanya Khovanova, Jan 10 2007

Hankel transform of A158197. - Paul Barry, Mar 13 2009

For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(6)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011

Values of x (or y) in the solutions to x^2 - 8xy + y^2 + 6 = 0. - Colin Barker, Feb 05 2014

From Klaus Purath, May 06 2025: (Start)

Nonnegative solutions to the Diophantine equation 3*b(n)^2 - 5*a(n)^2 = -2. The corresponding b(n) are A057080(n). Note that (b(n)*b(n+2) - b(n+1)^2)/2 = -5 and (a(n)*a(n+2) - a(n+1)^2)/2 = 3.

(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A001090(n+1) = Lucas U(8,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.

a(n)=(t(i+2*n+1) + t(i))/(t(i+n+1) + t(i+n)) as long as t(i+n+1) + t(i+n) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 7*t(i+2) - 7*t(i+1) + t(i) or t(i+2) = 8*t(i+1) - t(i) regardless of initial values and including this sequence itself. (End)

Examples

			1 + 7*x + 55*x^2 + 433*x^3 + 3409*x^4 + 26839*x^5 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• A. Fink, R. K. Guy and M. Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), 76-114. See Section 13.
• Tanya Khovanova, Recursive Sequences
• J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
• H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
• H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (8,-1).

Crossrefs

a(n) = sqrt((3*A057080(n)^2+2)/5) (cf. Richardson comment).

Cf. A057080, A001090, A001091.

Row 8 of array A094954.

Cf. A001090.

Cf. similar sequences listed in A238379.

Cf. A041023.

Programs

Magma

I:=[1, 7]; [n le 2 select I[n] else 8*Self(n-1) - Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 26 2013

Mathematica

CoefficientList[Series[(1 - x)/(1 - 8*x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jan 26 2013 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
   ] (* Complement of A041023 *)
a[15, 20] (* Gerry Martens, Jun 07 2015 *)
LinearRecurrence[{8,-1},{1,7},20] (* Harvey P. Dale, Dec 04 2021 *)

PARI

{a(n) = subst( 9*poltchebi(n) - poltchebi(n-1), x, 4) / 5} /* Michael Somos, Jun 07 2005 */

PARI

{a(n) = if( n<0, n=-1-n); polcoeff( (1 - x) / (1 - 8*x + x^2) + x * O(x^n), n)} /* Michael Somos, Jun 07 2005 */

Sage

[lucas_number1(n,8,1)-lucas_number1(n-1,8,1) for n in range(1, 21)] # Zerinvary Lajos, Nov 10 2009

Formula

For all members x of the sequence, 15*x^2 - 6 is a square. Lim_{n->infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 12 2002

a(n) = (5+sqrt(15))/10 * (4+sqrt(15))^n + (5-sqrt(15))/10 * (4-sqrt(15))^n.

a(n) ~ 1/10*sqrt(10)*(1/2*(sqrt(10)+sqrt(6)))^(2*n+1)

a(n) = U(n, 4)-U(n-1, 4) = T(2*n+1, sqrt(5/2))/sqrt(5/2), with Chebyshev's U and T polynomials and U(-1, x) := 0. U(n, 4)=A001090(n+1), n>=-1.

Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 6) = a(n) - Benoit Cloitre, Nov 10 2002

a(n)*a(n+3) = 48 + a(n+1)a(n+2). - Ralf Stephan, May 29 2004

a(n) = (-1)^n*U(2n, i*sqrt(6)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005

G.f.: (1-x)/(1-8*x+x^2).

a(n) = a(-1-n).

a(n) = Jacobi_P(n,-1/2,1/2,4)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006

[a(n), A001090(n+1)] = [1,6; 1,7]^(n+1) * [1,0]. - Gary W. Adamson, Mar 21 2008

For n>0, a(n) is the numerator of the continued fraction [2,3,2,3,...,2,3] with n repetitions of 2,3. For the denominators see A136325. - Greg Dresden, Sep 12 2019

From Peter Bala, Apr 30 2025: (Start)

a(n) = (1/sqrt(5)) * sqrt(1 - T(2*n+1, -4)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.

a(n) divides a(3*n+1); a(n) divides a(5*n+2); in general, for k >= 0, a(n) divides a((2*k+1)*n + k).

The aerated sequence [b(n)]n>=1 = [1, 0, 7, 0, 55, 0, 433, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -10, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.

Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/6 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A291033(n-1) - 1/A291033(n).) (End)

In addition to the first formula above: In general, the following applies to all recurrences (a(n)) of the form (8,-1) with a(0) = 1 and arbitrary a(1): 15*a(n)^2 + y = b^2 where y = x^2 + 8*x + 1 and x = a(1) - 8. Also y = a(k+1)^2 - a(k)*a(k+1) for any k >=0. - Klaus Purath, May 06 2025