A006130 -id:A006130 - cp's OEIS frontend

A183584 T(n,k)=Half the number of nXk 0..3 arrays with each element equal to either the maximum or the minimum of its horizontal and vertical neighbors.

0, 2, 2, 2, 14, 2, 8, 78, 78, 8, 14, 407, 814, 407, 14, 38, 2216, 9667, 9667, 2216, 38, 80, 12024, 110674, 240358, 110674, 12024, 80, 194, 65277, 1282814, 5987795, 5987795, 1282814, 65277, 194, 434, 354615, 14823037, 149355515, 320224986, 149355515

Offset: 1

R. H. Hardin Jan 05 2011

Table starts
....0........2...........2..............8.................14
....2.......14..........78............407...............2216
....2.......78.........814...........9667.............110674
....8......407........9667.........240358............5987795
...14.....2216......110674........5987795..........320224986
...38....12024.....1282814......149355515........17195908538
...80....65277....14823037.....3724940988.......922710498351
..194...354615...171459557....92915045150.....49526167251643
..434..1926386..1982720806..2317648165893...2658141294593252
.1016.10465655.22929915669.57811518197886.142669482408162261

			Some solutions with a(1,1)<=1 for 4X3
..1..1..3....1..3..3....1..0..3....1..2..2....0..2..2....1..2..2....1..3..3
..1..3..3....1..1..1....1..0..3....1..1..3....0..0..0....1..1..3....1..0..0
..2..1..3....1..1..1....0..0..0....1..1..3....0..2..2....0..3..3....1..3..3
..2..1..1....2..2..2....0..1..1....1..1..1....2..2..2....0..0..3....3..3..3

R. H. Hardin, Table of n, a(n) for n = 1..111

Column 1 is 2*A006130(n-2)

A186168 T(n,k)=1/4 the number of nXk 0..3 arrays with every element equal to exactly one or two of its horizontal and vertical neighbors.

Original entry on oeis.org

0, 1, 1, 1, 7, 1, 4, 48, 48, 4, 7, 321, 702, 321, 7, 19, 2175, 14364, 14364, 2175, 19, 40, 14748, 253341, 751266, 253341, 14748, 40, 97, 99933, 4762206, 37402872, 37402872, 4762206, 99933, 97, 217, 677283, 87054174, 1899336597, 5033988714, 1899336597

Offset: 1

R. H. Hardin Feb 13 2011

Table starts
...0........1............1...............4.................7.................19
...1........7...........48.............321..............2175..............14748
...1.......48..........702...........14364............253341............4762206
...4......321........14364..........751266..........37402872.........1899336597
...7.....2175.......253341........37402872........5033988714.......704281652979
..19....14748......4762206......1899336597......704281652979....268995986029278
..40....99933.....87054174.....95752776009....96951738076992.101554003879403823
..97...677283...1610684397...4840082975532.13435825601048421
.217..4590168..29645381115.244420512852030
.508.31108893.546876640548

			Some solutions for 5X4 with a(1,1)=0
..0..0..1..0....0..0..1..1....0..0..1..1....0..0..1..0....0..0..1..1
..0..0..1..0....0..0..3..2....0..0..2..2....0..0..1..0....0..0..2..1
..1..1..3..3....1..1..3..2....1..1..0..0....1..1..0..0....1..1..2..2
..0..0..3..3....0..1..3..2....1..3..3..0....1..1..2..1....2..2..1..0
..1..1..0..0....0..0..0..0....0..0..2..2....0..0..2..1....3..3..1..0

R. H. Hardin, Table of n, a(n) for n = 1..84

Column 1 is A006130(n-2)

A206255 T(n,k) = Number of (n+1) X (k+1) 0..3 arrays with every 2 X 2 subblock having zero permanent.

Original entry on oeis.org

49, 361, 361, 1600, 8029, 1600, 9409, 99856, 99856, 9409, 47089, 1718209, 1364224, 1718209, 47089, 258064, 26512201, 49336576, 49336576, 26512201, 258064, 1343281, 434613664, 944701696, 4556324929, 944701696, 434613664, 1343281, 7198489

Offset: 1

R. H. Hardin, Feb 05 2012

Table starts
......49..........361...........1600..............9409................47089
.....361.........8029..........99856...........1718209.............26512201
....1600........99856........1364224..........49336576............944701696
....9409......1718209.......49336576........4556324929.........226637884225
...47089.....26512201......944701696......226637884225.......14139111560401
..258064....434613664....27285753856....16533087493120.....2261195747329024
.1343281...6990799321...603269103616..1006239668280961...186370092021300625
.7198489.113636628469.15860529270784.68438717405988481.24455901146017548025

			Some solutions for n=4, k=3
..2..0..3..3....0..0..1..1....0..3..3..0....0..0..0..0....0..2..0..3
..0..0..0..0....3..0..0..0....0..0..0..0....3..0..1..1....0..0..0..1
..2..1..1..0....3..0..1..1....1..0..1..0....2..0..0..0....1..0..0..1
..0..0..0..0....1..0..0..0....1..0..0..0....3..0..2..0....0..0..0..3
..2..1..1..0....3..0..0..2....2..0..2..3....2..0..3..0....1..0..0..1

R. H. Hardin, Table of n, a(n) for n = 1..611

Cf. A006130.

Column 1 is A006130(n+2)^2.

A083856 Square array T(n,k) of generalized Fibonacci numbers, read by antidiagonals upwards (n, k >= 0).

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 3, 1, 0, 1, 1, 4, 5, 5, 1, 0, 1, 1, 5, 7, 11, 8, 1, 0, 1, 1, 6, 9, 19, 21, 13, 1, 0, 1, 1, 7, 11, 29, 40, 43, 21, 1, 0, 1, 1, 8, 13, 41, 65, 97, 85, 34, 1, 0, 1, 1, 9, 15, 55, 96, 181, 217, 171, 55, 1

Offset: 0

Paul Barry, May 06 2003

Row n >= 0 of the array gives the solution to the recurrence b(k) = b(k-1) + n*b(k-2) for k >= 2 with b(0) = 0 and b(1) = 1.

			Array T(n,k) (with rows n >= 0 and columns k >= 0) begins as follows:
  0, 1, 1,  1,  1,   1,   1,    1,    1,     1, ... [A057427]
  0, 1, 1,  2,  3,   5,   8,   13,   21,    34, ... [A000045]
  0, 1, 1,  3,  5,  11,  21,   43,   85,   171, ... [A001045]
  0, 1, 1,  4,  7,  19,  40,   97,  217,   508, ... [A006130]
  0, 1, 1,  5,  9,  29,  65,  181,  441,  1165, ... [A006131]
  0, 1, 1,  6, 11,  41,  96,  301,  781,  2286, ... [A015440]
  0, 1, 1,  7, 13,  55, 133,  463, 1261,  4039, ... [A015441]
  0, 1, 1,  8, 15,  71, 176,  673, 1905,  6616, ... [A015442]
  0, 1, 1,  9, 17,  89, 225,  937, 2737, 10233, ... [A015443]
  0, 1, 1, 10, 19, 109, 280, 1261, 3781, 15130, ... [A015445]
  ...

A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, 21(2) (2015), 35-42.

Rows include A057427 (n=0), A000045 (n=1), A001045 (n=2), A006130 (n=3), A006131 (n=4), A015440 (n=5), A015441 (n=6), A015442 (n=7), A015443 (n=8), A015445 (n=9).
Columns include A000012 (k=1,2), A000027 (k=3), A005408 (k=4), A028387 (k=5), A000567 (k=6), A106734 (k=7).
Cf. A083857 (binomial transform), A083859 (main diagonal), A083860 (first subdiagonal), A083861 (second binomial transform), A110112, A110113 (diagonal sums), A193376 (transposed variant), A172237 (transposed variant).

function generalized_fibonacci(r, n)
   F = BigInt[1 r; 1 0]
   Fn = F^n
   Fn[2, 1]
end
for r in 0:6 println([generalized_fibonacci(r, n) for n in 0:9]) end # Peter Luschny, Mar 06 2017

A083856_row := proc(r, n) local R; R := proc(n) option remember;
if n<=1 then n else R(n-1)+r*R(n-2) fi end: R(n) end:
for r from 0 to 9 do seq(A083856_row(r, n), n=0..9) od; # Peter Luschny, Mar 06 2017

T[, 0] = 0; T[, 1|2] = 1; T[n_, k_] := T[n, k] = T[n, k-1] + n T[n, k-2];
Table[T[n-k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 22 2018 *)

T(n, k) = (((1 + sqrt(4*n + 1))/2)^k - ((1 - sqrt(4*n + 1))/2)^k)/sqrt(4*n + 1). [corrected by Michel Marcus, Jun 25 2018]
From Thomas Baruchel, Jun 25 2018: (Start)
The g.f. for row n >= 0 is x/(1 - x - n*x^2).
The g.f. for column k >= 1 is g(k,x) = 1/(1-x) + Sum_{m = 1..floor((k-1)/2)} (1 - x)^(-1 - m) * binomial(k - 1 - m, m) * Sum_{i = 0..m} x^i * Sum_{j = 0..i} (-1)^j * (i - j)^m * binomial(1 + m, j).
The g.f. for column k >= 1 is also g(k,x) = 1 + Sum_{m = 1..floor((k+1)/2)} ((1 - x)^(-m) * binomial(k-m, m-1) * Sum_{j = 0..m} (-1)^j * binomial(m, j) *  x^m * Phi(x, -m+1, -j+m)) + Sum_{s = 0..floor((k-1)/2)} binomial(k-s-1, s) * PolyLog(-s, x), where Phi is the Lerch transcendent function. (End)
T(n,k) = Sum_{i = 0..k} (-1)^(k+i) * binomial(k,i) * A083857(n,i). - Petros Hadjicostas, Dec 24 2019

Various sections edited by Petros Hadjicostas, Dec 24 2019

A143461 Square array A(n,k) of numbers of length n quaternary words with at least k 0-digits between any other digits (n,k >= 0), read by antidiagonals.

Original entry on oeis.org

1, 1, 4, 1, 4, 16, 1, 4, 7, 64, 1, 4, 7, 19, 256, 1, 4, 7, 10, 40, 1024, 1, 4, 7, 10, 22, 97, 4096, 1, 4, 7, 10, 13, 43, 217, 16384, 1, 4, 7, 10, 13, 25, 73, 508, 65536, 1, 4, 7, 10, 13, 16, 46, 139, 1159, 262144, 1, 4, 7, 10, 13, 16, 28, 76, 268, 2683, 1048576, 1, 4, 7, 10, 13, 16, 19, 49, 115, 487, 6160, 4194304

Offset: 0

Alois P. Heinz, Aug 16 2008

			A (3,1) = 19, because 19 quaternary words of length 3 have at least 1 0-digit between any other digits: 000, 001, 002, 003, 010, 020, 030, 100, 101, 102, 103, 200, 201, 202, 203, 300, 301, 301, 303.
Square array A(n,k) begins:
       1,   1,   1,  1,  1,  1,  1,  1,  ...
       4,   4,   4,  4,  4,  4,  4,  4,  ...
      16,   7,   7,  7,  7,  7,  7,  7,  ...
      64,  19,  10, 10, 10, 10, 10, 10,  ...
     256,  40,  22, 13, 13, 13, 13, 13,  ...
    1024,  97,  43, 25, 16, 16, 16, 16,  ...
    4096, 217,  73, 46, 28, 19, 19, 19,  ...
   16384, 508, 139, 76, 49, 31, 22, 22,  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0-9 give: A000302, A006130(n+1), A084386(n+2), A143454, A143455, A143456, A143457, A143458, A143459, A143460.

Main diagonal gives A016777.

A:= proc(n, k) option remember; if k=0 then 4^n elif n<=k+1 then 3*n+1 else A(n-1, k) +3*A(n-k-1, k) fi end: seq(seq(A(n, d-n), n=0..d), d=0..13);

a[n_, 0] := 4^n; a[n_, k_] /; n <= k+1 := 3*n+1; a[n_, k_] := a[n, k] = a[n-1, k] + 3*a[n-k-1, k]; Table[a[n-k, k], {n, 0, 13}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jan 15 2014, after Maple *)

G.f. of column k: 1/(x^k*(1-x-3*x^(k+1))).

A(n,k) = 4^n if k=0, else A(n,k) = 3*n+1 if n<=k+1, else A(n,k) = A(n-1,k) + 3*A(n-k-1,k).

A052533 Expansion of (1-x)/(1-x-3*x^2).

Original entry on oeis.org

1, 0, 3, 3, 12, 21, 57, 120, 291, 651, 1524, 3477, 8049, 18480, 42627, 98067, 225948, 520149, 1197993, 2758440, 6352419, 14627739, 33684996, 77568213, 178623201, 411327840, 947197443, 2181180963, 5022773292, 11566316181, 26634636057

Offset: 0

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

Form the graph with matrix A=[0,1,1,1;1,1,0,0;1,0,1,0;1,0,0,1]. A052533 counts closed walks of length n at the vertex without loop. - Paul Barry, Oct 02 2004
Let M = [0, sqrt(3); sqrt(3), 1] be a 2 X 2 matrix. Then A052533 = {[M^n](1,1)}. Note also that {[M^n](2,2)} = A006130. - L. Edson Jeffery, Nov 25 2011
Pisano period lengths:  1, 3, 1, 6, 24, 3, 24, 6, 1, 24,120, 6,156, 24, 24, 12, 16, 3, 90, 24, ... - R. J. Mathar, Aug 10 2012
a(n) appears in the formula for powers of the fundamental algebraic number c = (1 + sqrt(13))/2 = A209927 of the quadratic number field Q(sqrt(13)): c^n = a(n) + A006130(n-1), for n >=0, with  A006130(-1) = 0. The formulas given below and in A006130 in terms of S-Chebyshev polynomials are valid also for c^(-n), for n >= 0, with 1/c = (-1 + sqrt(13))/2 = A356033. - Wolfdieter Lang, Nov 26 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 463
Index entries for linear recurrences with constant coefficients, signature (1,3).

Cf. A006130, A049310, A209927, A274977, A356033.

a:=[1,0];; for n in [3..40] do a[n]:=a[n-1]+3*a[n-2]; od; a; # G. C. Greubel, May 09 2019

I:=[1,0]; [n le 2 select I[n] else Self(n-1)+3*Self(n-2): n in [1..40]]; // G. C. Greubel, May 09 2019

R:=PowerSeriesRing(Integers(), 33); Coefficients(R!( (1-x)/(1-x-3*x^2))); // Marius A. Burtea, Jan 15 2020

spec := [S,{S=Sequence(Prod(Z,Union(Z,Z,Z),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
seq(coeff(series((1-x)/(1-x-3*x^2), x, n+1), x, n), n = 0..40); # G. C. Greubel, Jan 15 2020

CoefficientList[Series[(1-x)/(1-x-3x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 07 2013 *)
LinearRecurrence[{1,3}, {1,0}, 40] (* G. C. Greubel, May 09 2019 *)

my(x='x+O('x^30)); Vec((1-x)/(1-x-3*x^2)) \\ G. C. Greubel, May 09 2019

[lucas_number1(n+1,1,-3) -lucas_number1(n,1,-3) for n in (0..40)] # G. C. Greubel, May 09 2019

G.f.: (1 - x)/(1 - x - 3*x^2).
a(n) = A006130(n) - A006130(n-1).
a(n) = a(n-1) + 3*a(n-2), with a(0)=1, a(1)=0.
a(n) = Sum_{alpha = RootOf(-1+x+3*x^2)} (1/13)*(-1 + 7*alpha)* alpha^(-n-1).
a(n) = Sum_{k=0..floor(n/2)} C(n-k-1,n-2*k)*3^k. - Paul Barry, Mar 16 2010
If p[1]=0, and p[i]=3, (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
G.f.: (Q(0) -1)*(1-x)/x, where Q(k) = 1 + 3*x^2 + (k+2)*x - x*(k+1 + 3*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
a(n) = 3^(n/2) * Fibonacci(n-1, 1/sqrt(3)). - G. C. Greubel, Jan 15 2020
From Wolfdieter Lang, Nov 27 2023: (Start)
a(n) = 3*A006130(n-2), with A006130(-2) = 1/3 and A006130(-1) = 0.
a(n) = 3*sqrt(-3)^(n-2)*S(n-2, 1/sqrt(-3)), with the S Chebyshev polynomials (see A049310), valid also for negative indices n, using S(-n, x) = - S(n-2, x), for n>= 2, and S(-1, x) = 0. (End)

More terms from James Sellers, Jun 06 2000

A108300 a(n+2) = 3*a(n+1) + a(n), with a(0) = 1, a(1) = 5.

Original entry on oeis.org

1, 5, 16, 53, 175, 578, 1909, 6305, 20824, 68777, 227155, 750242, 2477881, 8183885, 27029536, 89272493, 294847015, 973813538, 3216287629, 10622676425, 35084316904, 115875627137, 382711198315, 1264009222082, 4174738864561, 13788225815765, 45539416311856

Offset: 0

Creighton Dement, Jul 24 2005

Binomial transform is A109114.
Invert transform is A109115.
Inverse invert transform is A016777.
Inverse binomial transform is A006130.

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Sergio Falcon, The k-Fibonacci difference sequences, Chaos, Solitons & Fractals, Volume 87, June 2016, Pages 153-157.
Tanya Khovanova, Recursive Sequences
Vincent Vatter, Growth rates of permutation classes: from countable to uncountable, arXiv:1605.04297 [math.CO], 2016. (Mentions a signed version.)
Index entries for linear recurrences with constant coefficients, signature (3,1).

Row sums and main diagonal of A143972. - Gary W. Adamson, Sep 06 2008

Cf. A109114, A109115, A016777, A006130, A000004, A052924, A228916, A374439.

seriestolist(series((-2*x-1)/(x^2-1+3*x), x=0,25));

LinearRecurrence[{3,1},{1,5},40] (* Harvey P. Dale, Jul 04 2013 *)

Vec((1 + 2*x)/(1 - 3*x - x^2) + O(x^30)) \\ Andrew Howroyd, Jun 05 2021

G.f.: (1 + 2*x)/(1 - 3*x - x^2).
a(n) = A052924(n+1) - A052924(n).
a(n)*a(n-2) = a(n-1)^2 + 9*(-1)^n. - Roger L. Bagula, May 17 2010
a(n) = 3^n*Sum_{k=0..n} A374439(n, k)*(1/3)^k. - Peter Luschny, Jul 26 2024

A140167 a(n) = (-1)a(n-1) + 3a(n-2) with a(1)=-1 and a(2)=1.

Original entry on oeis.org

-1, 1, -4, 7, -19, 40, -97, 217, -508, 1159, -2683, 6160, -14209, 32689, -75316, 173383, -399331, 919480, -2117473, 4875913, -11228332, 25856071, -59541067, 137109280, -315732481, 727060321, -1674257764, 3855438727, -8878212019, 20444528200

Offset: 1

Gary W. Adamson, May 10 2008

A140165 is a companion sequence.

			a(5) = -19 = (-1)*7 + 3*(-4).
a(5) = -19 = term (1,2) of X^5 since X^5 = [ -2, -19; -19, -59].

Joerg Arndt, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (-1,3). [_R. J. Mathar_, Dec 12 2009]

Cf. A140165, A182228.

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

I:=[-1,1]; [n le 2 select I[n] else (-1)*Self(n-1) + 3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 31 2015

seq(coeff(series(-x/(1+x-3*x^2), x, n+1), x, n), n = 1..30); # G. C. Greubel, Dec 26 2019

RecurrenceTable[{a[n]== -a[n-1]+3*a[n-2], a[1]== -1, a[2]==1}, a, {n,30}] (* G. C. Greubel, Aug 30 2015 *)
Table[Round[-(-Sqrt[3])^(n-1)*(LucasL[n-1, 1/Sqrt[3]] + Fibonacci[n-1, 1/Sqrt[3] ]/Sqrt[3])/2], {n,30}] (* G. C. Greubel, Dec 26 2019 *)

first(m)=my(v=vector(m));v[1]=-1;v[2]=1;for(i=3,m,v[i]=-v[i-1] + 3*v[i-2]); v \\ Anders Hellström, Aug 30 2015

def A140167_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( -x/(1+x-3*x^2) ).list()
a=A140167_list(30); a[1:] # G. C. Greubel, Dec 26 2019

a(n) = (-1)*a(n-1) + 3*a(n-2), given a(1) = -1, a(2) = 1. a(n) = term (1,2) of X^n, where X = the 2x2 matrix [1,-1; -1,-2].
From R. J. Mathar, Dec 12 2009: (Start)
a(n) = (-1)^n*A006130(n-1).
G.f.: -x/(1+x-3*x^2). (End)
G.f.: -Q(0)/2 , where Q(k) = 1 + 1/(1 - x*(4*k-1 + 3*x)/( x*(4*k+1 + 3*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 08 2013
E.g.f.: (1/sqrt(13))*(exp(-(1+sqrt(13))*x/2) - exp(-(1-sqrt(13))*x/2)). G. C. Greubel, Aug 30 2015
a(n) = -(-sqrt(3))^(n-1)*(Lucas(n-1, 1/sqrt(3)) + Fibonacci(n-1, 1/sqrt(3) )/sqrt(3))/2. - G. C. Greubel, Dec 26 2019

A274977 a(n) = a(n-1) + 3*a(n-2) with n>1, a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 9, 27, 54, 135, 297, 702, 1593, 3699, 8478, 19575, 45009, 103734, 238761, 549963, 1266246, 2916135, 6714873, 15463278, 35607897, 81997731, 188821422, 434814615, 1001278881, 2305722726, 5309559369, 12226727547, 28155405654, 64835588295, 149301805257, 343808570142

Offset: 0

Bruno Berselli, Sep 13 2016

a(n)/a(n+1) converges to 1/A209927 as n approaches infinity.

			Table of similar sequences (not extendable on the left side) where this recurrence can be applied to the first two terms:
----------------------------------------------------------------------
(*)      -  -  1, -1,  2, -1,  5,   2,  17,  23,   74,  143,  365, ...
A052533: -  -  1,  0,  3,  3, 12,  21,  57, 120,  291,  651, 1524, ...
(^)      -  0, 1,  1,  4,  7, 19,  40,  97, 217,  508, 1159, 2683, ...
A006138: -  -  1,  2,  5, 11, 26,  59, 137, 314,  725, 1667, 3842, ...
A105476: -  -  1,  3,  6, 15, 33,  78, 177, 411,  942, 2175, 5001, ...
(^)      0, 1, 1,  4,  7, 19, 40,  97, 217, 508, 1159, 2683, 6160, ...
A105963: -  -  1,  5,  8, 23, 47, 116, 257, 605, 1376, 3191, 7319, ...
A274977: -  -  1,  6,  9, 27, 54, 135, 297, 702, 1593, 3699, 8478, ...
A075118: -  2, 1,  7, 10, 31, 61, 154, 337, 799, 1810, 4207, 9637, ...
----------------------------------------------------------------------
(*) see version A140165.
(^) see A006130 and the signed versions A140167, A182228.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,3).

Cf. A006130, A006138, A052533, A075118, A105476, A105963.

a:=[1,6];; for n in [3..40] do a[n]:=a[n-1]+3*a[n-2]; od; a; # G. C. Greubel, Jan 15 2020

[n le 2 select 5*n-4 else Self(n-1)+3*Self(n-2): n in [1..40]];

R:=PowerSeriesRing(Integers(), 32); Coefficients(R!((1 + 5*x)/(1- x-3*x^2))); // Marius A. Burtea, Jan 15 2020

seq(coeff(series((1+5*x)/(1-x-3*x^2), x, n+1), x, n), n = 0..40); # G. C. Greubel, Jan 15 2020

RecurrenceTable[{a[n]==a[n-1] +3a[n-2], a[0]==1, a[1]==6}, a, {n,0,40}]
Table[Round[Sqrt[3]^(n-1)*(Sqrt[3]*Fibonacci[n+1, 1/Sqrt[3]] + 5*Fibonacci[n, 1/Sqrt[3]])], {n,0,40}] (* G. C. Greubel, Jan 15 2020 *)
LinearRecurrence[{1,3},{1,6},40] (* Harvey P. Dale, Jul 11 2023 *)

v=vector(40); v[1]=1; v[2]=6; for(n=3, #v, v[n]=v[n-1]+3*v[n-2]); v

from sage.combinat.sloane_functions import recur_gen2
a = recur_gen2(1, 6, 1, 3)
[next(a) for n in range(40)]

G.f.: (1 + 5*x)/(1 - x - 3*x^2).
a(n) = ((13 + 11*sqrt(13))*(1 + sqrt(13))^n + (13 - 11*sqrt(13))*(1 - sqrt(13))^n)/(26*2^n).
3*a(n) + a(n+1) =  9*A105476(n+1).
3*a(n) - a(n+1) = 27*A006130(n-3) with n>1, A006130(-1) = 0.
a(n+1) - a(n)   = 27*A105476(n-3) with n>2.
a(n) = 3^((n-1)/2)*( sqrt(3)*Fibonacci(n+1, 1/sqrt(3)) + 5*Fibonacci(n, 1/sqrt(3)) ). - G. C. Greubel, Jan 15 2020
E.g.f.: (1/13)*exp(x/2)*(13*cosh((sqrt(13)*x)/2) + 11*sqrt(13)*sinh((sqrt(13)*x)/2)). - Stefano Spezia, Jan 15 2020

A074357 Coefficient of q^3 in nu(n), where nu(0)=1, nu(1)=b and, for n>=2, nu(n)=bnu(n-1)+lambda(1+q+q^2+...+q^(n-2))*nu(n-2) with (b,lambda)=(1,3).

Original entry on oeis.org

0, 0, 0, 0, 0, 30, 168, 639, 2415, 7872, 25542, 77727, 233547, 679410, 1949862, 5490132, 15276456, 41963844, 114153990, 307595853, 822263313, 2181777252, 5751280350, 15069310365, 39269077809, 101817186264, 262776963360

Offset: 0

Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002

nonn

Coefficient of q^0 is A006130.

			The first 6 nu polynomials are nu(0)=1, nu(1)=1, nu(2)=4, nu(3)=7+3q, nu(4)=19+15q+12q^2, nu(5)=40+45q+42q^2+30q^3+9q^4, so the coefficients of q^3 are 0,0,0,0,0,30.

M. Beattie, S. Dăscălescu and S. Raianu, Lifting of Nichols Algebras of Type B_2, arXiv:math/0204075 [math.QA], 2002.
Index entries for linear recurrences with constant coefficients, signature (4,6,-32,-19,96,54,-108,-81).

Coefficient of q^0, q^1 and q^2 are in A006130, A074355 and A074356. Related sequences with other values of b and lambda are in A074082-A074089, A074352-A074354, A074358-A074363.

nu := proc(b,lambda,n) global q; local qp,i ; if n = 0 then RETURN(1) ; elif n =1 then RETURN(b) ; fi ; qp:=0 ; for i from 0 to n-2 do qp := qp + q^i ; od ; RETURN( b*nu(b,lambda,n-1)+lambda*qp*nu(b,lambda,n-2)) ; end: A074357 := proc(n) RETURN( coeftayl(nu(1,3,n),q=0,3) ) ; end: for n from 0 to 30 do printf("%d,", A074357(n)) ; od ; # R. J. Mathar, Sep 20 2006

Join[{0, 0, 0}, LinearRecurrence[{4, 6, -32, -19, 96, 54, -108, -81}, {0, 0, 30, 168, 639, 2415, 7872, 25542}, 24]] (* Jean-François Alcover, Sep 22 2017 *)

Conjecture: O.g.f.: 3*x^5*(3*x+1)*(36*x^4+24*x^3-29*x^2-14*x+10)/(3*x^2+x-1)^4. - R. J. Mathar, Jul 22 2009

More terms from R. J. Mathar, Sep 20 2006

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