A006131 -id:A006131 - cp's OEIS frontend

A161007 a(n+1) = 2a(n) + 16a(n-1), a(0)=0, a(1)=1.

0, 1, 2, 20, 72, 464, 2080, 11584, 56448, 298240, 1499648, 7771136, 39536640, 203411456, 1039409152, 5333401600, 27297349632, 139929124864, 716615843840, 3672097685504, 18810048872448, 96373660712960, 493708103385088, 2529394778177536, 12958119210516480

Offset: 0

Sture Sjöstedt, Jun 02 2009

Colin Barker, Table of n, a(n) for n = 0..1000
J. Borowska, L. Lacinska, Recurrence form of determinant of a heptadiagonal symmetric Toeplitz matrix, J. Appl. Math. Comp. Mech. 13 (2014) 19-16, remark 2 for permanent of tridiagonal Toeplitz matrices a=2, b=4.
Index entries for linear recurrences with constant coefficients, signature (2,16).

Cf. A006131, A010473 (sqrt(17)).

[n le 2 select n-1 else 2*(Self(n-1) +8*Self(n-2)): n in [1..41]]; // G. C. Greubel, Oct 15 2022

LinearRecurrence[{2, 16}, {0, 1}, 50] (* T. D. Noe, Nov 07 2011 *)

concat(0, Vec(-x/(16*x^2+2*x-1) + O(x^40))) \\ Colin Barker, Jul 01 2015

A161007=BinaryRecurrenceSequence(2,16,0,1)
[A161007(n) for n in range(41)] # G. C. Greubel, Oct 15 2022

a(n) = ((1+sqrt(17))^n - (1-sqrt(17))^n)/(2*sqrt(17)).
Limit_{n -> oo} a(n+1)/a(n) = 1 + sqrt(17).
G.f.: x / (1 - 2*x - 16*x^2). - Colin Barker, Jul 01 2015
a(n) = 2^(n-1)*A006131(n-1). - R. J. Mathar, Mar 08 2021
a(n) = (4*i)^n*ChebyshevU(n, -i/4). - G. C. Greubel, Oct 15 2022

A193376 T(n,k) = number of ways to place any number of 2 X 1 tiles of k distinguishable colors into an n X 1 grid; array read by descending antidiagonals, with n, k >= 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 5, 5, 1, 5, 7, 11, 8, 1, 6, 9, 19, 21, 13, 1, 7, 11, 29, 40, 43, 21, 1, 8, 13, 41, 65, 97, 85, 34, 1, 9, 15, 55, 96, 181, 217, 171, 55, 1, 10, 17, 71, 133, 301, 441, 508, 341, 89, 1, 11, 19, 89, 176, 463, 781, 1165, 1159, 683, 144, 1, 12, 21, 109, 225, 673

Offset: 1

R. H. Hardin, Jul 24 2011

Transposed variant of A083856. - R. J. Mathar, Aug 23 2011

As to the sequences by columns beginning (1, N, ...), let m = (N-1). The g.f. for the sequence (1, N, ...) is 1/(1 - x - m*x^2). Alternatively, the corresponding matrix generator is [[1,1], [m,0]]. Another equivalency is simply: The sequence beginning (1, N, ...) is the INVERT transform of (1, m, 0, 0, 0, ...). Convergents to the sequences a(n)/a(n-1) are (1 + sqrt(4*m+1))/2. - Gary W. Adamson, Feb 25 2014

			Array T(n,k) (with rows n >= 1 and column k >= 1) begins as follows:
  ..1...1....1....1.....1.....1.....1......1......1......1......1......1...
  ..2...3....4....5.....6.....7.....8......9.....10.....11.....12.....13...
  ..3...5....7....9....11....13....15.....17.....19.....21.....23.....25...
  ..5..11...19...29....41....55....71.....89....109....131....155....181...
  ..8..21...40...65....96...133...176....225....280....341....408....481...
  .13..43...97..181...301...463...673....937...1261...1651...2113...2653...
  .21..85..217..441...781..1261..1905...2737...3781...5061...6601...8425...
  .34.171..508.1165..2286..4039..6616..10233..15130..21571..29844..40261...
  .55.341.1159.2929..6191.11605.19951..32129..49159..72181.102455.141361...
  .89.683.2683.7589.17621.35839.66263.113993.185329.287891.430739.624493...
  ...
Some solutions for n = 5 and k = 3 with colors = 1, 2, 3 and empty = 0:
..0....2....3....2....0....1....0....0....2....0....0....2....3....0....0....0
..0....2....3....2....2....1....2....3....2....1....0....2....3....1....1....1
..1....0....0....0....2....0....2....3....2....1....0....1....0....1....1....1
..1....2....2....0....3....2....2....3....2....0....3....1....3....3....2....1
..0....2....2....0....3....2....2....3....0....0....3....0....3....3....2....1

R. H. Hardin, Table of n, a(n) for n = 1..9999
Ron H. Hardin, Re: A193376 Tabl = 20 existing sequences, Sequence Fans mailing list, 2011.
Robert Israel, Re: A193376 Tabl = 20 existing sequences, Sequence Fans mailing list, 2011.

Column 1 is A000045(n+1), column 2 is A001045(n+1), column 3 is A006130, column 4 is A006131, column 5 is A015440, column 6 is A015441(n+1), column 7 is A015442(n+1), column 8 is A015443, column 9 is A015445, column 10 is A015446, column 11 is A015447, and column 12 is A053404,
Row 2 is A000027(n+1), row 3 is A004273(n+1), row 4 is A028387, row 5 is A000567(n+1), and row 6 is A106734(n+2).
Diagonal is A171180, superdiagonal 1 is A083859(n+1), and superdiagonal 2 is A083860(n+1).

T:= proc(n,k) option remember; `if`(n<0, 0,
      `if`(n<2 or k=0, 1, k*T(n-2, k) +T(n-1, k)))
    end;
seq(seq(T(n, d+1-n), n=1..d), d=1..12); # Alois P. Heinz, Jul 29 2011

T[n_, k_] := T[n, k] = If[n < 0, 0, If[n < 2 || k == 0, 1, k*T[n-2, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 12}] // Flatten (* Jean-François Alcover, Mar 04 2014, after Alois P. Heinz *)

With z X 1 tiles of k colors on an n X 1 grid (with n >= z), either there is a tile (of any of the k colors) on the first spot, followed by any configuration on the remaining (n-z) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n-1) X 1. Thus, T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n = 0, 1, ..., z-1.
The solution is T(n,k) = Sum_r r^(-n-1)/(1 + z*k*r^(z-1)), where the sum is over the roots r of the polynomial k*x^z + x - 1.
For z = 2, T(n,k) = ((2*k / (sqrt(1 + 4*k) - 1))^(n+1) - (-2*k/(sqrt(1 + 4*k) + 1))^(n+1)) / sqrt(1 + 4*k).
T(n,k) = Sum_{s=0..[n/2]} binomial(n-s,s) * k^s.
For z X 1 tiles, T(n,k,z) = Sum_{s = 0..[n/z]} binomial(n-(z-1)*s, s) * k^s. - R. H. Hardin, Jul 31 2011

Formula and proof from Robert Israel in the Sequence Fans mailing list.

A052923 Expansion of (1-x)/(1 - x - 4*x^2).

Original entry on oeis.org

1, 0, 4, 4, 20, 36, 116, 260, 724, 1764, 4660, 11716, 30356, 77220, 198644, 507524, 1302100, 3332196, 8540596, 21869380, 56031764, 143509284, 367636340, 941673476, 2412218836, 6178912740, 15827788084, 40543439044, 103854591380

Offset: 0

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

First differences of A006131.

This sequence {a(n)} appears in the formula for powers of c = (1 + sqrt(17))/2 = A222132, the fundamental (integer) algebraic number of Q(sqrt(17)): c^n = a(n) + A006131(n-1)*c. This is also valid for positive powers of 1/c = (-1 + sqrt(17)) /8. See the formula below and in A006131 in terms of Chebyshev or Fibonacci polynomials. - Wolfdieter Lang, Nov 27 2023

G. C. Greubel, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 908
Index entries for linear recurrences with constant coefficients, signature (1,4).
Index entries for sequences related to Chebyshev polynomials.

Cf. A006131, A026581, A049310, A222132.

a:=[1,0];; for n in [3..30] do a[n]:=a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, Oct 16 2019
a := n -> -(2*I)^n*ChebyshevU(n-2, -I/4):
seq(simplify(a(n)), n = 0..28);  # Peter Luschny, Dec 03 2023

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-x)/(1 -x -4*x^2) )); // G. C. Greubel, Oct 16 2019

spec := [S,{S=Sequence(Prod(Sequence(Z),Z,Union(Z,Z,Z,Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
seq(coeff(series((1-x)/(1 -x -4*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 16 2019

LinearRecurrence[{1,4}, {1,0}, 30] (* G. C. Greubel, Oct 16 2019 *)

my(x='x+O('x^30)); Vec((1-x)/(1 -x -4*x^2)) \\ G. C. Greubel, Oct 16 2019

def A052923_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((1-x)/(1 -x -4*x^2)).list()
A052923_list(30) # G. C. Greubel, Oct 16 2019

G.f.: (1-x)/(1 - x - 4*x^2).
a(n) = a(n-1) + 4*a(n-2), with a(0)=1, a(1)=0.
a(n) = Sum_{alpha=RootOf(-1+z+4*z^2)} (1/17)*(-1+9*alpha)*alpha^(-1-n).
If p[1]=0, and p[i]=4, ( i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
From Wolfdieter Lang, Nov 27 2023: (Start)
a(n) = 4*A006131(n-2), with A006131(-2) = 1/4 and A006131(-1) = 0.
a(n) = -(-2*i)^n*S(n-2, i/2), with i = sqrt(-1), and the S-Chebyshev polynomials (see A049310). S(-n, x) = -S(n-2, x). The Fibonacci polynomials are F(n, x) = (-i)^(n-1)*S(n-1, i*x). (End)

More terms from James Sellers, Jun 06 2000

A350470 Array read by ascending antidiagonals. T(n, k) = J(k, n) where J are the Jacobsthal polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 5, 1, 1, 1, 7, 9, 11, 1, 1, 1, 9, 13, 29, 21, 1, 1, 1, 11, 17, 55, 65, 43, 1, 1, 1, 13, 21, 89, 133, 181, 85, 1, 1, 1, 15, 25, 131, 225, 463, 441, 171, 1, 1, 1, 17, 29, 181, 341, 937, 1261, 1165, 341, 1

Offset: 0

Peter Luschny, Mar 19 2022

			Array starts:
n\k 0, 1,  2,  3,   4,    5,    6,     7,      8,      9, ...
---------------------------------------------------------------------
[0] 1, 1,  1,  1,   1,    1,    1,     1,      1,      1, ... A000012
[1] 1, 1,  3,  5,  11,   21,   43,    85,    171,    341, ... A001045
[2] 1, 1,  5,  9,  29,   65,  181,   441,   1165,   2929, ... A006131
[3] 1, 1,  7, 13,  55,  133,  463,  1261,   4039,  11605, ... A015441
[4] 1, 1,  9, 17,  89,  225,  937,  2737,  10233,  32129, ... A015443
[5] 1, 1, 11, 21, 131,  341, 1651,  5061,  21571,  72181, ... A015446
[6] 1, 1, 13, 25, 181,  481, 2653,  8425,  40261, 141361, ... A053404
[7] 1, 1, 15, 29, 239,  645, 3991, 13021,  68895, 251189, ... A350468
[8] 1, 1, 17, 33, 305,  833, 5713, 19041, 110449, 415105, ... A168579
[9] 1, 1, 19, 37, 379, 1045, 7867, 26677, 168283, 648469, ... A350469
      A005408 | A082108 |
           A016813   A014641

Rows: A000012, A001045, A006131, A015441, A015443, A015446, A053404, A350468, A168579, A350469.
Columns: A000012, A005408, A016813, A082108, A014641.
Cf. A350467 (main diagonal), A352361 (Fibonacci polynomials), A352362 (Lucas polynomials).

J := (n, x) -> add(2^k*binomial(n - k, k)*x^k, k = 0..n):
seq(seq(J(k, n-k), k = 0..n), n = 0..10);

T[n_, k_] := Hypergeometric2F1[(1 - k)/2, -k/2, -k, -8 n];
Table[T[n, k], {n, 0, 9}, {k, 0, 9}] // TableForm
(* or *)
T[n_, k_] := With[{s = Sqrt[8*n+1]}, ((1+s)^(k+1) - (1-s)^(k+1)) / (2^(k+1)*s)];
Table[Simplify[T[n, k]], {n, 0, 9}, {k, 0, 9}] // TableForm

T(n, k) = ([1, 2; k, 0]^n)[1, 1] ;
export(T)
for(k = 0, 9, print(parvector(10, n, T(n - 1, k))))

T(n, k) = Sum_{j=0..k} binomial(k - j, j)*(2*n)^j.
T(n, k) = ((1+s)^(k+1) - (1-s)^(k+1)) / (2^(k+1)*s) where s = sqrt(8*n + 1).
T(n, k) = [x^k] (1 / (1 - x - 2*n*x^2)).
T(n, k) = hypergeom([1/2 - k/2, -k/2], [-k], -8*n).

A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 6, 44, 312, 2224, 15840, 112832, 803712, 5724928, 40779264, 290475008, 2069084160, 14738305024, 104982503424, 747801460736, 5326668791808, 37942424436736, 270267896954880, 1925146777223168, 13713023838978048, 97679317251653632, 695780094221746176

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,8).

Sequences of the form a(n) = c*a(n-1) + d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A000045,A001045,A006130,A006131,A015440,A015441,A015442,A015443,A015445,A015446
.A000129,A002605,A015518,A063727,A002532,A083099,A015519,A003683,A002534,A083102
.A006190,A007482,A030195,A015521,A015523,A083858,A015524,A015525,A099012,A015528
.A001076,A090017,A015530,A057087,A015531,A085939,A015532,A180222,A015533,A180226
.A052918,A015535,A015536,A015537,A057088,A015540,A015541,A015544,A015545,A180250
.A005668,A135030,A090018,A135032,A015551,A057089,A015552,A189800,A189801,A190005
.A054413,A015555,A015559,A015561,A015562,A015564,A057090,A015565,A015566,A015568
.A041025,A190331,A015574,A190510,A015575,A190560,A015576,A057091,A015577,A190953
.A099371,A015579,A181353,A015580,A015581,A153191,A015583,A015584,A057092,A015585
A041041,A191014,A015588,A190954,A190955,A190956,A015589,A190957,A015591,A057093

I:=[0,1]; [n le 2 select I[n] else 6*Self(n-1)+8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2011

LinearRecurrence[{6, 8}, {0, 1}, 50]
CoefficientList[Series[-(x/(-1+6 x+8 x^2)),{x,0,50}],x] (* Harvey P. Dale, Jul 26 2011 *)

a(n)=([0,1; 8,6]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

G.f.: x/(1 - 2*x*(3+4*x)). - Harvey P. Dale, Jul 26 2011

A060959 Table by antidiagonals of generalized Fibonacci numbers: T(n,k) = T(n,k-1) + n*T(n,k-2) with T(n,0)=0 and T(n,1)=1.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 5, 5, 4, 1, 1, 0, 1, 8, 11, 7, 5, 1, 1, 0, 1, 13, 21, 19, 9, 6, 1, 1, 0, 1, 21, 43, 40, 29, 11, 7, 1, 1, 0, 1, 34, 85, 97, 65, 41, 13, 8, 1, 1, 0, 1, 55, 171, 217, 181, 96, 55, 15, 9, 1, 1, 0, 1, 89, 341, 508, 441, 301, 133, 71, 17, 10, 1, 1, 0

Offset: 0

Henry Bottomley, May 10 2001

			Square array begins as:
  0, 1, 1, 1,  1,  1,  1, ...
  0, 1, 1, 2,  3,  5,  8, ...
  0, 1, 1, 3,  5, 11, 21, ...
  0, 1, 1, 4,  7, 19, 40, ...
  0, 1, 1, 5,  9, 29, 65, ...
  0, 1, 1, 6, 11, 41, 96, ...

G. C. Greubel, Antidiagonals n = 0..100, flattened

Rows include A057427, A000045, A001045, A006130, A006131, A015440, A015441, A015442, A015443, A015445, A015446, A015447, A053404.

Columns include A000004, A000012 (twice), A000027, A000567 A005408, A028387.

Flat(List([0..12], n-> List([0..n], k-> (((1+Sqrt(1+4*k))/2)^(n-k) - ((1-Sqrt(1+4*k))/2)^(n-k))/Sqrt(1+4*k) ))); # G. C. Greubel, Jan 15 2020

[Round( (((1+Sqrt(1+4*k))/2)^(n-k) - ((1-Sqrt(1+4*k))/2)^(n-k) )/Sqrt(1+4*k) ): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 15 2020

seq(seq( round((((1+sqrt(1+4*k))/2)^(n-k) - ((1-sqrt(1+4*k))/2)^(n-k) )/sqrt(1+4*k)), k=0..n), n=0..12); # G. C. Greubel, Jan 15 2020

T[n_, k_]:= If[n==k==0, 0, Round[(((1+Sqrt[1+4n])/2)^k - ((1-Sqrt[1+4n])/2)^k)/Sqrt[1+4n]]]; Table[T[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 15 2020 *)

T(n,k) = ( ((1+sqrt(1+4*n))/2)^k - ((1-sqrt(1+4*n))/2)^k )/sqrt(1+4*n);
for(n=0,12, for(k=0,n, print1( round(T(k,n-k)), ", "))) \\ G. C. Greubel, Jan 15 2020

[[ round( (((1+sqrt(1+4*k))/2)^(n-k) - ((1-sqrt(1+4*k))/2)^(n-k) )/sqrt(1+4*k) ) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 15 2020

T(n, k) = ( ((1+sqrt(1+4*n))/2)^k - ((1-sqrt(1+4*n))/2)^k )/sqrt(1+4*n).

A026583 a(n) = Sum{T(i,j)}, 0<=j<=i, 0<=i<=2n, T given by A026568.

Original entry on oeis.org

1, 4, 11, 30, 77, 200, 511, 1314, 3361, 8620, 22067, 56550, 144821, 371024, 950311, 2434410, 6235657, 15973300, 40915931, 104809134, 268472861, 687709400, 1761600847, 4512438450, 11558841841, 29608595644, 75843963011

Offset: 0

Clark Kimberling

Amir Sapir, The Tower of Hanoi with Forbidden Moves, The Computer J. 47 (1) (2004) 20, case cyclic++, sequence d(n) (offset 1).
Index entries for linear recurrences with constant coefficients, signature (2,3,-4).

Cf. A006131, A026581 (first differences).

LinearRecurrence[{2,3,-4},{1,4,11},40] (* Harvey P. Dale, Apr 03 2024 *)

G.f.: (1+2x)/[(1-x)(1-x-4x^2)]. - Ralf Stephan, Feb 04 2004 (follows from first comment in A026581)

A097705 a(n) = 4a(n-1) + 17a(n-2), a(1)=1, a(2)=4.

Original entry on oeis.org

1, 4, 33, 200, 1361, 8844, 58513, 384400, 2532321, 16664084, 109705793, 722112600, 4753448881, 31289709724, 205967469873, 1355794944800, 8924626767041, 58747021129764, 386706739558753, 2545526317441000

Offset: 1

Ralf Stephan, Aug 27 2004

This is one of only two Lucas-type sequences whose 8th term is a square.

The other one is A006131. - Michel Marcus, Dec 07 2012

A. Bremner and N. Tzanakis, Lucas sequences whose 8th term is a square
Index entries for linear recurrences with constant coefficients, signature (4,17).

Cf. A006131.

f:= gfun:-rectoproc({a(n) = 4*a(n-1) + 17*a(n-2), a(1)=1, a(2)=4}, a(n), remember): map(f, [$0..20]); # Georg Fischer, Jun 18 2021

a[0]:0$
a[1]:1$
a[n]:=4*a[n-1] + 17*a[n-2]$
A097705(n):=a[n]$
makelist(A097705(n),n,1,30); /* Martin Ettl, Nov 03 2012 */

G.f.: 1/(1-4x-17x^2).

Definition adapted to offset by Georg Fischer, Jun 18 2021

A099580 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1) * 4^(k-1).

Original entry on oeis.org

0, 0, 1, 1, 9, 13, 65, 117, 441, 909, 2929, 6565, 19305, 45565, 126881, 309141, 833049, 2069613, 5467345, 13745797, 35877321, 90860509, 235418369, 598860405, 1544728185, 3940169805, 10135859761, 25896538981, 66507086889, 170093242813

Offset: 0

Paul Barry, Oct 23 2004

In general a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k-1) * r^(k-1) has g.f. x^2/((1-r*x^2)*(1-x-r*x^2)) and satisfies the recurrence a(n) = a(n-1) + 2*r*a(n-2) - r*a(n-3) - r^2*a(n-4).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,8,-4,-16).

Cf. A006131, A097038, A099579.

[n le 4 select Floor((n-1)/2) else Self(n-1) +8*Self(n-2) -4*Self(n-3) -16*Self(n-4): n in [1..41]]; // G. C. Greubel, Jul 24 2022

LinearRecurrence[{1,8,-4,-16}, {0,0,1,1}, 51] (* G. C. Greubel, Jul 24 2022 *)

@CachedFunction
def a(n): # a = A099580
    if (n<4): return (n//2)
    else: return a(n-1) +8*a(n-2) -4*a(n-3) -16*a(n-4)
[a(n) for n in (0..40)] # G. C. Greubel, Jul 24 2022

G.f.: x^2/((1-4*x^2) * (1-x-4*x^2)).
a(n) = a(n-1) + 8*a(n-2) - 4*a(n-3) - 16*a(n-4).
From G. C. Greubel, Jul 24 2022: (Start)
a(n) = (4*(2/i)^(n-1)*ChebyshevU(n-1, i/4) - 2^n*(1-(-1)^n))/4.
E.g.f.: ( 4*exp(x/2)*sinh(sqrt(17)*x/2) - sqrt(17)*sinh(2*x) )/(2*sqrt(17)). (End)

A102446 a(n) = a(n-1) + 4*a(n-2) for n>1, a(0) = a(1) = 2.

Original entry on oeis.org

2, 2, 10, 18, 58, 130, 362, 882, 2330, 5858, 15178, 38610, 99322, 253762, 651050, 1666098, 4270298, 10934690, 28015882, 71754642, 183818170, 470836738, 1206109418, 3089456370, 7913894042, 20271719522, 51927295690, 133014173778, 340723356538, 872780051650

Offset: 0

N. J. A. Sloane, based on a suggestion from R. K. Guy, Feb 23 2005

The continued fraction expansion c_0 = 0, c_n = 1/2 (n>0) (see a paper by Bremner & Tzanakis) has convergents 2/1, 2/5, 10/9, 18/29, 58/65, 130/181, ... where the numerators and denominators satisfy the recurrence a_n = a_{n-1} + 4a_{n-2}. The denominators are A006131 and the numerators are the present sequence.

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

Cf. A006131.

[n le 2 select 2 else Self(n-1) + 4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Dec 17 2015

a[0] = a[1] = 2; a[n_] := a[n] = a[n - 1] + 4a[n - 2]; Table[ a[n], {n, 0, 27}] (* Robert G. Wilson v, Feb 23 2005 *)
LinearRecurrence[{1, 4}, {2, 2}, 30] (* Vincenzo Librandi, Dec 17 2015 *)

Vec(-2 / (-1+x+4*x^2) + O(x^40)) \\ Colin Barker, Dec 22 2016

from sage.combinat.sloane_functions import recur_gen2b
it = recur_gen2b(2,2,1,4, lambda n: 0)
[next(it) for i in range(29)] # Zerinvary Lajos, Jul 09 2008

def A000129():
    x, y = 0, 1
    while True:
        x, y = (x + y) << 1, x - y
        yield x
a = A000129(); [next(a) for i in range(28)]  # Peter Luschny, Dec 17 2015

a(n) = 2 * A006131(n).
G.f.: Q(0)/x -1/x, where Q(k) = 1 + 4*x^2 + (2*k+3)*x - x*(2*k+1 + 4*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 05 2013
G.f.: -2 / ( -1+x+4*x^2 ). - R. J. Mathar, Feb 10 2016
a(n) = (2^(-n)*(-(1-sqrt(17))^(1+n) + (1+sqrt(17))^(1+n)))/sqrt(17). - Colin Barker, Dec 22 2016

More terms from Robert G. Wilson v, Feb 23 2005

cp's OEIS Frontend

A161007 a(n+1) = 2*a(n) + 16*a(n-1), a(0)=0, a(1)=1.

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A193376 T(n,k) = number of ways to place any number of 2 X 1 tiles of k distinguishable colors into an n X 1 grid; array read by descending antidiagonals, with n, k >= 1.

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A052923 Expansion of (1-x)/(1 - x - 4*x^2).

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A350470 Array read by ascending antidiagonals. T(n, k) = J(k, n) where J are the Jacobsthal polynomials.

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A189800 a(n) = 6*a(n-1) + 8*a(n-2), with a(0)=0, a(1)=1.

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A060959 Table by antidiagonals of generalized Fibonacci numbers: T(n,k) = T(n,k-1) + n*T(n,k-2) with T(n,0)=0 and T(n,1)=1.

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A026583 a(n) = Sum{T(i,j)}, 0<=j<=i, 0<=i<=2n, T given by A026568.

A161007 a(n+1) = 2a(n) + 16a(n-1), a(0)=0, a(1)=1.

A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.

A097705 a(n) = 4a(n-1) + 17a(n-2), a(1)=1, a(2)=4.