A058316 -id:A058316 - cp's OEIS frontend

A086902 a(n) = 7*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 7.

2, 7, 51, 364, 2599, 18557, 132498, 946043, 6754799, 48229636, 344362251, 2458765393, 17555720002, 125348805407, 894997357851, 6390330310364, 45627309530399, 325781497023157, 2326097788692498, 16608466017870643, 118585359913786999, 846705985414379636

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 18 2003

a(n+1)/a(n) converges to (7+sqrt(53))/2 = 7.14005... = A176439.
Lim a(n)/a(n+1) as n approaches infinity = 0.1400549... = 2/(7+sqrt(53)) = (sqrt(53)-7)/2 = 1/A176439 = A176439 - 7.
From Johannes W. Meijer, Jun 12 2010: (Start)
In general sequences with recurrence a(n) = k*a(n-1)+a(n-2) with a(0)=2 and a(1)=k [and a(-1)=0] have generating function (2-k*x)/(1-k*x-x^2). If k is odd (k>=3) they satisfy a(3n+1) = b(5n), a(3n+2)=b(5*n+3), a(3n+3)=2*b(5n+4) where b(n) is the sequence of numerators of continued fraction convergents to sqrt(k^2+4). [If k is even then a(n)/2, for n>=1, is the sequence of numerators of continued fraction convergents to sqrt(k^2/4+1).]
For the sequence given above k=7 which implies that it is associated with A041090.
For a similar statement about sequences with recurrence a(n) = k*a(n-1)+a(n-2) but with a(0)=1 [and a(-1)=0] see A054413; a sequence that is associated with A041091.
For more information follow the Khovanova link and see A087130, A140455 and A178765.
(End)

			a(4) = 7*a(3) + a(2) = 7*364 + 51 = 2599.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
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 (7,1).

Cf. A058316, A176439.

Cf. A000032 (k=1), A006497 (k=3), A087130 (k=5), A086902 (k=7), A087798 (k=9), A001946 (k=11), A088316 (k=13), A090301 (k=15), A090306 (k=17). - Johannes W. Meijer, Jun 12 2010

I:=[2,7]; [n le 2 select I[n] else 7*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016

RecurrenceTable[{a[0] == 2, a[1] == 7, a[n] == 7 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
LinearRecurrence[{7,1},{2,7},30] (* Harvey P. Dale, May 25 2023 *)

a(n)=([0,1; 1,7]^n*[2;7])[1,1] \\ Charles R Greathouse IV, Apr 06 2016

a(n) = ((7+sqrt(53))/2)^n + ((7-sqrt(53))/2)^n.
E.g.f. : 2exp(7x/2)cosh(sqrt(53)x/2); a(n)=2^(1-n)sum{k=0..floor(n/2), C(n, 2k)53^k7^(n-2k)}. a(n)=2T(n, 7i/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
G.f.: (2-7x)/(1-7x-x^2). - Philippe Deléham, Nov 16 2008
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 7*A097837(n), a(2n) = A099368(n).
a(3n+1) = A041090(5n), a(3n+2) = A041090(5*n+3), a(3n+3) = 2*A041090(5n+4).
Limit(a(n+k)/a(k), k=infinity) = (A086902(n) + A054413(n-1)*sqrt(53))/2.
Limit(A086902(n)/A054413(n-1), n=infinity) = sqrt(53). (End)

A086903 a(n) = 8*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 8.

Original entry on oeis.org

2, 8, 62, 488, 3842, 30248, 238142, 1874888, 14760962, 116212808, 914941502, 7203319208, 56711612162, 446489578088, 3515205012542, 27675150522248, 217885999165442, 1715412842801288, 13505416743244862, 106327921103157608

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 21 2003

a(n+1)/a(n) converges to (4+sqrt(15)) = 7.872983... a(0)/a(1)=2/8; a(1)/a(2)=8/62; a(2)/a(3)=62/488; a(3)/a(4)=488/3842; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.127016... = 1/(4+sqrt(15)) = (4-sqrt(15)).
Twice A001091. - John W. Layman, Sep 25 2003
Except for the first term, positive values of x (or y) satisfying x^2 - 8xy + y^2 + 60 = 0. - Colin Barker, Feb 13 2014

			a(4) = 3842 = 8*a(3) - a(2) = 8*488 - 62 = (4+sqrt(15))^4 + (4-sqrt(15))^4 = 3841.9997397 + 0.0002603 = 3842.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A086594, A058316, A006245, A009271, A005248, A174502.

I:=[2,8]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 15 2014

a[0] = 2; a[1] = 8; a[n_] := 8a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Jan 30 2004 *)
CoefficientList[Series[(2 - 8 x)/(1 - 8 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 15 2014 *)
LinearRecurrence[{8,-1},{2,8},30] (* Harvey P. Dale, Jan 18 2015 *)

[lucas_number2(n,8,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.
G.f.: (2-8*x)/(1-8*x+x^2). [Philippe Deléham, Nov 02 2008]
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = product {n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 4 - sqrt(15). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.12474 84992 41370 33639 ... = 2 + 1/(8 + 1/(62 + 1/(488 + ...))).  Cf. A174502 and A005248.
Also F(-alpha) = 0.87474 74663 84045 35032 ... has the continued fraction representation 1 - 1/(8 - 1/(62 - 1/(488 - ...))) and the simple continued fraction expansion 1/(1 + 1/((8-2) + 1/(1 + 1/((62-2) + 1/(1 + 1/((488-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((8^2-4) + 1/(1 + 1/((62^2-4) + 1/(1 + 1/((488^2-4) + 1/(1 + ...))))))).
(End)

cp's OEIS Frontend

A086902 a(n) = 7*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 7.

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