cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

Original entry on oeis.org

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401
Offset: 0

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Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 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 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

Examples

			a(1)^2 = 121 is the 5th star number (A003154).
		

Crossrefs

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A138281. Cf. A100047.
Cf. A142238.

Programs

  • GAP
    a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019
  • Magma
    I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015
    
  • Mathematica
    CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
    a[c_, n_] := Module[{},
       p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
       d := Numerator[Convergents[Sqrt[c], n p]];
       t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
       Return[t];
    ] (* Complement of A142238 *)
    a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)
  • PARI
    a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;
    

Formula

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Extensions

Chebyshev comments from Wolfdieter Lang, Oct 31 2002