A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

2, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, 384238402, 1995189565, 10360186227, 53796120700, 279340789727, 1450500069335, 7531841136402, 39109705751345, 203080369893127

Offset: 0

Author

Paul Barry, Aug 16 2003

Keywords

• nonn
• easy

Comments

Sequence is related to the fifth metallic mean [5;5,5,5,5,...] (see A098318).

The solution to the general recurrence b(n) = (2*k+1)*b(n-1)+b(n-2) with b(0)=2, b(1) = 2*k+1 is b(n) = ((2*k+1)+sqrt(4*k^2+4*k+5))^n+(2*k+1)-sqrt(4*k^2+4*k+5))^n)/2; b(n) = 2^(1-n)*Sum_{j=0..n} C(n, 2*j)*(4*k^2+4*k+5)^j*(2*k+1)^(n-2*j); b(n) = 2*T(n, (2*k+1)*x/2)(-1)^i with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003

Primes in this sequence include a(0) = 2; a(1) = 5; a(4) = 727; a(8) = 528527 (3) semiprimes in this sequence include a(7) = 101785; a(13) = 1995189565; a(16) = 279340789727; a(19) = 39109705751345; a(20) = 203080369893127 - Jonathan Vos Post, Feb 09 2005

a(n)^2 - 29*A052918(n-1)^2 = 4*(-1)^n, with n>0 - Gary W. Adamson, Oct 07 2008

For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010

Binomial transform of A072263. - Johannes W. Meijer, Aug 01 2010

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence L_{5,n}.
• Tanya Khovanova, Recursive Sequences
• Wikipedia, Metallic mean
• Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (5,1).

Crossrefs

Cf. A006497, A014448, A085447.

Cf. A086902, A000032, A052918.

Programs

Magma

I:=[2,5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

Mathematica

RecurrenceTable[{a[0] == 2, a[1] == 5, a[n] == 5 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)

PARI

{a(n) = if( n<0, (-1)^n * a(-n), polsym(x^2 - 5*x -1, n) [n + 1])} /* Michael Somos, Nov 04 2008 */

Sage

[lucas_number2(n,5,-1) for n in range(0, 21)] # Zerinvary Lajos, May 14 2009

Formula

a(n) = ((5+sqrt(29))/2)^n+((5-sqrt(29))/2)^n.

a(n) = A100236(n) + 1.

E.g.f. : 2*exp(5*x/2)*cosh(sqrt(29)*x/2); a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2k)*29^k*5^(n-2*k). a(n) = 2T(n, 5i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003

O.g.f.: (-2+5*x)/(-1+5*x+x^2). - R. J. Mathar, Dec 02 2007

a(-n) = (-1)^n * a(n). - Michael Somos, Nov 01 2008

A090248(n) = a(2*n). 5 * A097834(n) = a(2*n + 1). - Michael Somos, Nov 01 2008

Limit(a(n+k)/a(k), k=infinity) = (A087130(n) + A052918(n-1)*sqrt(29))/2. Limit(A087130(n)/A052918(n-1), n= infinity) = sqrt(29). - Johannes W. Meijer, Jun 12 2010

a(3n+1) = A041046(5n), a(3n+2) = A041046(5n+3) and a(3n+3) = 2*A041046 (5n+4). - Johannes W. Meijer, Jun 12 2010

a(n) = 2*A052918(n) - 5*A052918(n-1). - R. J. Mathar, Oct 02 2020

From Peter Bala, Jul 09 2025 : (Start)

The following series telescope (Cf. A000032):

For k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.

For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and

Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).

For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and

Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)