A123270 -id:A123270 - cp's OEIS frontend

A090390 Repeatedly multiply (1,0,0) by ([1,2,2],[2,1,2],[2,2,3]); sequence gives leading entry.

1, 1, 9, 49, 289, 1681, 9801, 57121, 332929, 1940449, 11309769, 65918161, 384199201, 2239277041, 13051463049, 76069501249, 443365544449, 2584123765441, 15061377048201, 87784138523761, 511643454094369, 2982076586042449, 17380816062160329, 101302819786919521, 590436102659356801

Offset: 0

Vim Wenders, Jan 30 2004

The values of a and b in (a,b,c)*A give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1; the values of c are A000129(2n)
Binomial transform of A086348. - Johannes W. Meijer, Aug 01 2010
All values of a(n) are squares.  sqrt(a(n+1)) = A001333(n). The ratio a(n+1)/a(n) converges to 3 + 2*sqrt(2). - Richard R. Forberg, Aug 14 2013

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 60 at p. 123.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A000129, A001333, A001541, A079291, A086348, A095344, A123270, A302946.

a090390 n = a090390_list !! n
a090390_list = 1 : 1 : 9 : zipWith (-) (map (* 5) $
   tail $ zipWith (+) (tail a090390_list) a090390_list) a090390_list
-- Reinhard Zumkeller, Aug 17 2013

[Evaluate(DicksonFirst(n,-1),2)^2/4: n in [0..40]]; // G. C. Greubel, Aug 21 2022

a:= n-> (<<1|0|0>>. <<1|2|2>, <2|1|2>, <2|2|3>>^n)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Aug 17 2013

CoefficientList[Series[(1-4x-x^2)/((1+x)(1-6x+x^2)),{x, 0, 30}], x] (* Harvey P. Dale, May 20 2012 *)
LinearRecurrence[{5,5,-1}, {1,1,9}, 30] (* Harvey P. Dale, May 20 2012 *)
Table[(ChebyshevT[n,3]+(-1)^n)/2, {n,0,30}] (* Eric W. Weisstein, Apr 17 2018 *)
(LucasL[Range[0, 40], 2]/2)^2 (* G. C. Greubel, Aug 21 2022 *)

a(n)=polcoeff((1-4*x-x^2)/((1+x)*(1-6*x+x^2))+x*O(x^n),n)

a(n)=if(n<0,0,([1,2,2;2,1,2;2,2,3]^n)[1,1])

Vec( (1-4*x-x^2)/((1+x)*(1-6*x+x^2)) + O(x^66) ) \\ Joerg Arndt, Aug 16 2013

use Math::Matrix; use Math::BigInt; $a = new Math::Matrix ([ 1, 2, 2], [ 2, 1, 2], [ 2, 2, 3]); $p = new Math::Matrix ([1, 0, 0]); $p->print(); for ($i=1; $i<20;$i++) { $p = $p->multiply($a); $p->print(); }

[lucas_number2(n,2,-1)^2/4 for n in (0..40)] # G. C. Greubel, Aug 21 2022

G.f.: (1-4*x-x^2)/((1+x)*(1-6*x+x^2)).
a(n) = A001333(n)^2
(a, b, c) = (1, 0, 0). Recursively multiply (a, b, c)*( [1, 2, 2], [2, 1, 2], [2, 2, 3] ).
M^n * [ 1 1 1] = [a(n+1) q a(n)], where M = the 3 X 3 matrix [4 4 1 / 2 1 0 / 1 0 0]. E.g. M^5 * [1 1 1] = [9801 4059 1681] where 9801 = a(6), 1681 = a(5). Similarly, M^n * [1 0 0] generates A079291 (Pell number squares). - Gary W. Adamson, Oct 31 2004
a(n) = (((1+sqrt(2))^(2*n) + (1-sqrt(2))^(2*n)) + 2*(-1)^n)/4 - Lambert Klasen (lambert.klasen(AT)gmx.net), Oct 09 2005
a(n) = (A001541(n) + (-1)^n)/2. - R. J. Mathar, Nov 20 2009
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3), with a(0)=1, a(1)=1, a(2)=9. - Harvey P. Dale, May 20 2012
(a(n)) = tesseq(- .5'j + .5'k - .5j' + .5k' - 2'ii' + 'jj' - 'kk' + .5'ij' + .5'ik' + .5'ji' + 'jk' + .5'ki' + 'kj' + e), apart from initial term. - Creighton Dement, Nov 16 2004
a(n) = A302946(n)/4. - Eric W. Weisstein, Apr 17 2018
E.g.f.: exp(-x)*(1 + exp(4*x)*cosh(2*sqrt(2)*x))/2. - Stefano Spezia, Aug 03 2024

A015537 Expansion of x/(1 - 5x - 4x^2).

Original entry on oeis.org

0, 1, 5, 29, 165, 941, 5365, 30589, 174405, 994381, 5669525, 32325149, 184303845, 1050819821, 5991314485, 34159851709, 194764516485, 1110461989261, 6331368012245, 36098688018269, 205818912140325, 1173489312774701, 6690722212434805

Offset: 0

Olivier Gérard

First differences give A122690(n) = {1, 4, 24, 136, 776, 4424, 25224, ...}. Partial sums of a(n) are {0, 1, 6, 35, 200, ...} = (A123270(n) - 1)/8. - Alexander Adamchuk, Nov 03 2006
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 5's along the main diagonal, and 2's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 19 2011
Pisano period lengths:  1, 1, 8, 1, 4, 8, 48, 1, 24, 4, 40, 8, 42, 48, 8, 2, 72, 24, 360, 4, ... - R. J. Mathar, Aug 10 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Lucyna Trojnar-Spelina and Iwona Włoch, On Generalized Pell and Pell-Lucas Numbers, Iranian Journal of Science and Technology, Transactions A: Science (2019), 1-7.
Index entries for linear recurrences with constant coefficients, signature (5,4).

Cf. A001076, A006190, A007482, A015520, A015521, A015523, A015524, A015525, A015528, A015529, A015530, A015531, A015532, A015533, A015534, A015535, A015536, A015443, A015447, A030195, A053404, A057087, A083858, A085939, A090017, A091914, A099012, A122690, A123270, A180222, A180226.

a:=[0,1];; for n in [3..30] do a[n]:=5*a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, Dec 26 2019

[n le 2 select n-1 else 5*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 12 2012

seq( simplify((2/I)^(n-1)*ChebyshevU(n-1, 5*I/4)), n=0..20); # G. C. Greubel, Dec 26 2019

LinearRecurrence[{5,4}, {0,1}, 30] (* Vincenzo Librandi, Nov 12 2012 *)
Table[2^(n-1)*Fibonacci[n, 5/2], {n, 0, 30}] (* G. C. Greubel, Dec 26 2019 *)

x='x+O('x^30); concat([0], Vec(x/(1-5*x-4*x^2))) \\ G. C. Greubel, Jan 01 2018

[lucas_number1(n,5,-4) for n in range(0, 22)] # Zerinvary Lajos, Apr 24 2009

a(n) = 5*a(n-1) + 4*a(n-2).
a(n) = Sum_{k=0..floor((n-1)/2)} C(n-k-1, k)*4^k*5^(n-2*k-1). - Paul Barry, Apr 23 2005
a(n) = Sum_{k=0..(n-1)} A122690(k). - Alexander Adamchuk, Nov 03 2006
a(n) = 2^(n-1)*Fibonacci(n, 5/2) = (2/i)^(n-1)*ChebyshevU(n-1, 5*i/4). - G. C. Greubel, Dec 26 2019

A052961 Expansion of (1 - 3x) / (1 - 5x + 3*x^2).

Original entry on oeis.org

1, 2, 7, 29, 124, 533, 2293, 9866, 42451, 182657, 785932, 3381689, 14550649, 62608178, 269388943, 1159120181, 4987434076, 21459809837, 92336746957, 397304305274, 1709511285499, 7355643511673, 31649683701868, 136181487974321, 585958388766001, 2521247479907042

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

a(n) is the number of tilings of a 2 X n rectangle using integer dimension tiles at least one of whose dimensions is 1, so allowable dimensions are 1 X 1, 1 X 2, 1 X 3, 1 X 4, ..., and 2 X 1. - David Callan, Aug 27 2014
a(n+1) counts closed walks on K_2 containing one loop on the index vertex and four loops on the other vertex. Equivalently the (1,1)entry of A^(n+1) where the adjacency matrix of digraph is A=(1,1;1,4). - _David Neil McGrath, Nov 05 2014
A production matrix for the sequence is M =
  1, 1, 0, 0, 0, 0, 0, ...
  1, 0, 4, 0, 0, 0, 0, ...
  1, 0, 0, 4, 0, 0, 0, ...
  1, 0, 0, 0, 4, 0, 0, ...
  1, 0, 0, 0, 0, 4, 0, ...
  1, 0, 0, 0, 0, 0, 4, ...
  ...
Take powers of M and extract the upper left term, getting the sequence starting (1, 1, 2, 7, 29, 124, ...). - Gary W. Adamson, Jul 22 2016
From Gary W. Adamson, Jul 29 2016: (Start)
The sequence is N=1 in an infinite set obtained from matrix powers of [(1,N); (1,4)], extracting the upper left terms.
The infinite set begins:
N=1  (A052961):  1,  2,  7,  29   124,  533,   2293, ...
N=2  (A052984):  1,  3, 13,  59,  269, 1227,   5597, ...
N=3  (A004253):  1,  4, 19,  91,  436, 2089,  10009, ...
N=4  (A000351):  1,  5, 25, 125,  625, 3125,  15625, ...
N=5  (A015449):  1,  6, 31, 161,  836, 4341,  22541, ...
N=6  (A124610):  1,  7, 37, 199, 1069, 5743,  30853, ...
N=7  (A111363):  1,  8, 43, 239, 1324, 7337,  40653, ...
N=8  (A123270):  1,  9, 49, 281, 1601, 9129,  52049, ...
N=9  (A188168):  1, 10, 55, 325, 1900, 11125, 65125, ...
N=10 (A092164):  1, 11, 61, 371, 2221, 13331, 79981, ...
... (End)

Karl V. Keller, Jr., Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1032
Kai Liang, Solving tiling enumeration problems by tensor network contractions, arXiv:2503.17698, March 2025.
Index entries for linear recurrences with constant coefficients, signature (5,-3).

Column k=2 of A254414.

Cf. A000351, A004253, A015449, A052984, A092164, A111363, A123270, A124610, A188168.

a:=[1,2];; for n in [3..30] do a[n]:=5*a[n-1]-3*a[n-2]; od; a; # G. C. Greubel, Oct 23 2019

I:=[1,2]; [n le 2 select I[n] else 5*Self(n-1)-3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 17 2014

spec:= [S,{S = Sequence(Union(Prod(Sequence(Union(Z,Z,Z)),Z),Z))}, unlabeled ]: seq(combstruct[count ](spec,size = n), n = 0..20);
seq(coeff(series((1-3*x)/(1-5*x+3*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 23 2019

CoefficientList[Series[(1-3x)/(1-5x+3x^2),{x,0,30}],x] (* or *) LinearRecurrence[{5,-3},{1,2},30] (* Harvey P. Dale, Nov 23 2013 *)

my(x='x+O('x^30)); Vec((1-3*x)/(1-5*x+3*x^2)) \\ G. C. Greubel, Oct 23 2019

def A052961_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((1-3*x)/(1-5*x+3*x^2)).list()
A052961_list(30) # G. C. Greubel, Oct 23 2019

G.f.: (1-3*x)/(1-5*x+3*x^2).
a(n) = 5*a(n-1) - 3*a(n-2), with a(0) = 1, a(1) = 2.
a(n) = Sum_{alpha=RootOf(1-5*z+3*z^2)} (-1 + 9*alpha)*alpha^(-1-n)/13.
E.g.f.: (1 + sqrt(13) + (sqrt(13)-1) * exp(sqrt(13)*x)) / (2*sqrt(13) * exp(((sqrt(13)-5)*x)/2)). - Vaclav Kotesovec, Feb 16 2015
a(n) = A116415(n) - 3*A116415(n-1). - R. J. Mathar, Feb 27 2019

A095344 Length of n-th string generated by a Kolakoski(9,1) rule starting with a(1)=1.

Original entry on oeis.org

1, 1, 9, 9, 49, 81, 281, 601, 1729, 4129, 11049, 27561, 71761, 182001, 469049, 1197049, 3073249, 7861441, 20154441, 51600201, 132217969, 338618769, 867490649, 2221965721, 5691928321, 14579791201, 37347504489, 95666669289, 245056687249, 627723364401

Offset: 1

Benoit Cloitre, Jun 03 2004

Each string is derived from the previous string using the Kolakoski(9,1) rule and the additional condition: "string begins with 1 if previous string ends with 9 and vice versa". The strings are 1 -> 9 -> 111111111 -> 919191919 -> 11111111191111111119... -> ... and each one contains 1,1,9,9,31,... elements.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,5,4).

Cf. A000002, A066983, A095342, A095343.

Cf. A123270, A090390.

a:=[1,1,9];; for n in [4..35] do a[n]:=5*a[n-2]+4*a[n-3]; od; a; # G. C. Greubel, Dec 26 2019

a095344 n = a095344_list !! (n-1)
a095344_list = tail xs where
   xs = 1 : 1 : 1 : zipWith (-) (map (* 5) $ zipWith (+) (tail xs) xs) xs
-- Reinhard Zumkeller, Aug 16 2013

R:=PowerSeriesRing(Integers(), 35); Coefficients(R!( x*(1+x+ 4*x^2)/((1+x)*(1-x-4*x^2)) )); // G. C. Greubel, Dec 26 2019

seq(simplify(2*(-1)^n -(2/I)^n*(ChebyshevU(n, I/4) -2*I*ChebyshevU(n-1, I/4)) ), n = 1..35); # G. C. Greubel, Dec 26 2019

Table[2*(-1)^n - 2^n*(Fibonacci[n+1, 1/2] - 2*Fibonacci[n, 1/2]), {n,35}] (* G. C. Greubel, Dec 26 2019 *)
LinearRecurrence[{0,5,4},{1,1,9},40] (* Harvey P. Dale, Oct 12 2022 *)

Vec(x*(1+x+4*x^2)/((1+x)*(1-x-4*x^2)) + O(x^50)) \\ Colin Barker, Apr 20 2016

vector(35, n, round( 2*(-1)^n - (2/I)^n*(polchebyshev(n, 2, I/4) -2*I*polchebyshev(n-1, 2, I/4)) )) \\ G. C. Greubel, Dec 26 2019

def A095344_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1+x+4*x^2)/((1+x)*(1-x-4*x^2)) ).list()
a=A095344_list(35); a[1:] # G. C. Greubel, Dec 26 2019

a(1) = a(2) = 1; for n>1, a(n) = a(n-1) + 4*a(n-2) - 4*(-1)^n.
G.f.: x*(1 + x + 4*x^2)/((1 + x)*(1 - x - 4*x^2)). - Colin Barker, Mar 25 2012
a(n) = 5*a(n-2) + 4*a(n-3). - Colin Barker, Mar 25 2012
a(n) = 2*(-1)^n + (2^(-1-n)*(-(-7+sqrt(17))*(1+sqrt(17))^n - (1-sqrt(17))^n*(7+sqrt(17))))/sqrt(17). - Colin Barker, Apr 20 2016
a(n) = 2*(-1)^n - 2^n*(Fibonacci(n+1, 1/2) - 2*Fibonacci(n, 1/2)) = 2*(-1)^n - (2/I)^n*(ChebyshevU(n, I/4) - 2*I*ChebyshevU(n-1, I/4)). - G. C. Greubel, Dec 26 2019

A188168 a(0) = a(1) = 1; a(n) = 5a(n-1) + 5a(n-2).

Original entry on oeis.org

1, 1, 10, 55, 325, 1900, 11125, 65125, 381250, 2231875, 13065625, 76487500, 447765625, 2621265625, 15345156250, 89832109375, 525886328125, 3078592187500, 18022392578125, 105504923828125, 617636582031250, 3615707529296875, 21166720556640625

Offset: 0

Harvey P. Dale, Mar 23 2011

First differences give {0,9,45,270,1575,9225,54000,316125,1850625,10833750,63421875,371278125, . . .}

Index entries for linear recurrences with constant coefficients, signature (5,5).

Cf. A123270.

Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{5,5},#]}]&, {1,1},40]][[1]]
LinearRecurrence[{5,5},{1,1},30] (* Harvey P. Dale, Nov 19 2024 *)

G.f.: (4x-1)/(5x^2+5x-1)

a(n) = 1/5 2^-n (-(5-3 Sqrt[5])^n (2+Sqrt[5])+(-2+Sqrt[5]) (5+3 Sqrt[5])^n)

cp's OEIS Frontend

A090390 Repeatedly multiply (1,0,0) by ([1,2,2],[2,1,2],[2,2,3]); sequence gives leading entry.

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A015537 Expansion of x/(1 - 5*x - 4*x^2).

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A052961 Expansion of (1 - 3*x) / (1 - 5*x + 3*x^2).

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A095344 Length of n-th string generated by a Kolakoski(9,1) rule starting with a(1)=1.

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A188168 a(0) = a(1) = 1; a(n) = 5*a(n-1) + 5*a(n-2).

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A015537 Expansion of x/(1 - 5x - 4x^2).

A052961 Expansion of (1 - 3x) / (1 - 5x + 3*x^2).

A188168 a(0) = a(1) = 1; a(n) = 5a(n-1) + 5a(n-2).