A006130 -id:A006130 - cp's OEIS frontend

A123585 Triangle T(n,k), 0<=k<=n, given by [1, -1, 1, 0, 0, 0, 0, 0, ...] DELTA [1, 1, -1, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

1, 1, 1, 0, 2, 2, -1, 1, 5, 3, -1, -2, 4, 10, 5, 0, -4, -4, 12, 20, 8, 1, -2, -13, -4, 31, 38, 13, 1, 3, -11, -33, 3, 73, 71, 21, 0, 6, 6, -42, -74, 34, 162, 130, 34, -1, 3, 24, 0, -130, -146, 128, 344, 235, 55, -1, -4, 21, 72, -50, -352

Offset: 0

Philippe Deléham, Nov 13 2006

			Triangle begins:
1;
1, 1;
0, 2, 2;
-1, 1, 5, 3;
-1, -2, 4, 10, 5;
0, -4, -4, 12, 20, 8;
1, -2, -13, -4, 31, 38, 13;
1, 3, -11, -33, 3, 73, 71, 21;
0, 6, 6, -42, -74, 34, 162, 130, 34;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A010892, A099254, A000045, A000079 (row sums), A001629, A129707.

CoefficientList[CoefficientList[Series[1/(1 - (1 + y)*x + (1 - y^2)*x^2), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Oct 16 2017 *)

Sum_{k,0<=k<=n} T(n,k) = 2^n = A000079(n).
T(n,0) = A010892(n).
T(n,n) = Fibonacci(n+1) = A000045(n+1).
T(n+1,1) = A099254(n).
T(n+1,n) = A001629(n+2).
Sum_{k, 0<=k<=[n/2]} T(n-k,k) = A003269(n).
T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k-2) - T(n-2,k), n>0.
Sum_{k, 0<=k<=n} x^k*T(n,k) = (-1)^n*A003683(n+1), (-1)^n*A006130(n), A000007(n), A010892(n), A000079(n), A030195(n+1) for x=-3, -2, -1, 0, 1, 2 respectively . - Philippe Deléham, Dec 01 2006
T(n+2,n) = A129707(n+1).- Philippe Deléham, Dec 18 2011
G.f.: 1/(1-(1+y)*x+(1-y^2)*x^2). - Philippe Deléham, Dec 18 2011

A182228 a(n) = 3*a(n-2) - a(n-1) for n > 1, a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 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, 47079164257, -108412748857

Offset: 0

Alex Ratushnyak, Apr 19 2012

This is A006130 with minus signs on every other term. - T. D. Noe, Apr 23 2012

G. C. Greubel, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (-1,3).

Cf. A140167.

[n le 2 select n-1 else -Self(n-1) + 3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Aug 31 2015

RecurrenceTable[{a[n]== - a[n-1] + 3*a[n-2], a[0]== 0, a[1]== 1}, a, {n,0,200}] (* G. C. Greubel, Aug 30 2015 *)
LinearRecurrence[{-1,3},{0,1},40] (* Harvey P. Dale, Oct 23 2016 *)

prpr = 0
prev = 1
for i in range(2,55):
    current = prpr*3-prev
    print (current, end=',')
    prpr = prev
    prev = current

From R. J. Mathar, Apr 22 2013: (Start)
a(n) = -A140167(n).
G.f.: x/(1 + x - 3*x^2). (End)
G.f.: 1 - Q(0), where Q(k) = 1 + 3*x^2 - (k+2)*x + x*(k+1 - 3*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 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) = (-(-1)^n/sqrt(13))*(((1 + sqrt(13))/2)^n - ((1 - sqrt(13))/2)^n). - Taras Goy, Jul 17 2018

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.

A304236 Triangle T(n,k) = T(n-1,k) + 3*T(n-2,k-1) for k = 0..floor(n/2), with T(0,0) = 1, T(n,k) = 0 for n or k < 0, read by rows.

Original entry on oeis.org

1, 1, 1, 3, 1, 6, 1, 9, 9, 1, 12, 27, 1, 15, 54, 27, 1, 18, 90, 108, 1, 21, 135, 270, 81, 1, 24, 189, 540, 405, 1, 27, 252, 945, 1215, 243, 1, 30, 324, 1512, 2835, 1458, 1, 33, 405, 2268, 5670, 5103, 729, 1, 36, 495, 3240, 10206, 13608, 5103, 1, 39, 594, 4455, 17010, 30618, 20412, 2187

Offset: 0

Zagros Lalo, May 08 2018

The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A013610 ((1+3*x)^n).

The coefficients in the expansion of 1/(1-x-3x^2) are given by the sequence generated by the row sums.

			Triangle begins:
  1;
  1;
  1,  3;
  1,  6;
  1,  9,   9;
  1, 12,  27;
  1, 15,  54,   27;
  1, 18,  90,  108;
  1, 21, 135,  270,    81;
  1, 24, 189,  540,   405;
  1, 27, 252,  945,  1215,   243;
  1, 30, 324, 1512,  2835,  1458;
  1, 33, 405, 2268,  5670,  5103,   729;
  1, 36, 495, 3240, 10206, 13608,  5103;
  1, 39, 594, 4455, 17010, 30618, 20412, 2187;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 88, 363.

G. C. Greubel, Rows n = 0..100 of the irregular triangle, flattened
Zagros Lalo, Left-justified triangle
Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1+3x)^n

Row sums give A006130.
Cf. A013610.
Sequences of the form 3^k*binomial(n-(q-1)*k, k): A013610 (q=1), this sequence (q=2), A317496 (q=3), A318772 (q=4).

/* As triangle */ [[3^k*Binomial(n-k,k): k in [0..Floor(n/2)]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 05 2018

seq(seq( 3^k*binomial(n-k,k), k=0..floor(n/2)), n=0..24); # G. C. Greubel, May 12 2021

T[0, 0] = 1; T[n_, k_]:= If[n<0 || k<0, 0, T[n-1, k] + 3*T[n-2, k-1]]; Table[T[n, k], {n, 0, 14}, {k, 0, Floor[n/2]}]//Flatten
Table[3^k Binomial[n-k, k], {n, 0, 14}, {k, 0, Floor[n/2]}]//Flatten

T(n,k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, T(n-1,k) + 3*T(n-2,k-1)));
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 10 2018

flatten([[3^k*binomial(n-k,k) for k in (0..n//2)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = 3^k*binomial(n-k,k), n >= 0, 0 <= k <= floor(n/2).

A074324 a(2n+1) = 3^n, a(2n) = 4*3^(n-1) except for a(0) = 1.

Original entry on oeis.org

1, 1, 4, 3, 12, 9, 36, 27, 108, 81, 324, 243, 972, 729, 2916, 2187, 8748, 6561, 26244, 19683, 78732, 59049, 236196, 177147, 708588, 531441, 2125764, 1594323, 6377292, 4782969, 19131876, 14348907, 57395628, 43046721, 172186884, 129140163

Offset: 0

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

Also: Coefficient of the highest power of q in the expansion of nu(0)=1, nu(1)=b and for n>=2, nu(n)=b*nu(n-1)+lambda*(n-1)_q*nu(n-2) with (b,lambda)=(1,3), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity.
Instead of listing the coefficients of the highest power of q in each nu(n), if we list the coefficients of the smallest power of q (i.e., constant terms), we get a weighted Fibonacci numbers described by f(0)=1, f(1)=1, for n>=2, f(n)=f(n-1)+3f(n-2).
Sequences A162766, A166552 are essentially the same. - M. F. Hasler, Dec 03 2014

			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;
nu(6)=97+147q+180q^2+168q^3+147q^4+81q^5+36q^6;
by listing the coefficients of the highest power in each nu(n), we get, 1,1,4,3,12,9,36,....

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
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 (0,3).

Cf. A006130.

[1] cat [(1/6)*(7+(-1)^n)*3^Floor(n/2):n in [1..40]]; // Vincenzo Librandi, Jul 20 2013

CoefficientList[Series[-(1 + x + x^2) / (-1 + 3 x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
LinearRecurrence[{0,3},{1,1,4},40] (* Harvey P. Dale, Mar 13 2016 *)

a(n)=3^(n\2)\(3/4)^!bittest(n,0) \\ M. F. Hasler, Dec 03 2014

For given b and lambda, the recurrence relation is given by; t(0)=1, t(1)=b, t(2)=b^2+lambda and for n>=3, t(n) = lambda*t(n-2).
G.f.: -(1+x+x^2)/(-1+3*x^2). - R. J. Mathar, Dec 05 2007
a(n) = 3*a(n-2) for n>2. - Ralf Stephan, Jul 19 2013
a(n) = (1/6)*(7+(-1)^n)*3^floor(n/2) for n>0. - Ralf Stephan, Jul 19 2013

More terms from R. J. Mathar, Dec 05 2007

Simpler definition from M. F. Hasler, Dec 03 2014

A074355 Coefficient of q^1 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, 3, 15, 45, 147, 402, 1134, 2991, 7917, 20367, 52167, 131748, 330876, 824187, 2042763, 5035473, 12361755, 30226614, 73664298, 178971879, 433649769, 1048133619, 2527706127, 6083434824, 14613750648, 35045236083, 83909261319

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^1 are 0,0,0,3,15,45.

M. Beattie, S. Dăscălescu and S. Raianu, Lifting of Nichols Algebras of Type B_2, arXiv:math/0204075 [math.QA], 2002.

Coefficient of q^0, q^2 and q^3 are in A006130, A074356 and A074357. Related sequences with other values of b and lambda are in A074082-A074089, A074352-A074354, A074358-A074363.

nu := proc(n,b,lambda) option remember ; if n = 0 then 1 ; elif n = 1 then b ; else b*nu(n-1,b,lambda)+lambda*nu(n-2,b,lambda)*add(q^i,i=0..n-2) ; fi ; end:
A074355 := proc(n) local b,lambda,thisnu ; b := 1 ; lambda := 3 ; thisnu := nu(n,b,lambda) ; RETURN( coeftayl(thisnu,q=0,1) ) ; end: # R. J. Mathar, Mar 20 2007

nu[n_, b_, lambda_] := nu[n, b, lambda] = Which[ n == 0, 1, n == 1, b, True, b*nu[n - 1, b, lambda] + lambda*nu[n - 2, b, lambda]*Sum[q^i, {i, 0, n - 2}]];
a[n_] := a[n] = Coefficient[nu[n, 1, 3], q, 1];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 30}] (* Jean-François Alcover, Nov 23 2017, from 1st Maple program *)

G.f.: (9x^4+3x^3)/(1-3x-3x^2)^2 (conjectured). - Ralf Stephan, May 09 2004

More terms from R. J. Mathar, Mar 20 2007

A075118 Variant on Lucas numbers: a(n) = a(n-1) + 3*a(n-2) with a(0)=2 and a(1)=1.

Original entry on oeis.org

2, 1, 7, 10, 31, 61, 154, 337, 799, 1810, 4207, 9637, 22258, 51169, 117943, 271450, 625279, 1439629, 3315466, 7634353, 17580751, 40483810, 93226063, 214677493, 494355682, 1138388161, 2621455207, 6036619690, 13900985311, 32010844381, 73713800314, 169746333457

Offset: 0

Henry Bottomley, Sep 02 2002

The sequence 4,1,7,.. = 2*0^n+A075118(n) is given by trace(A^n) where A=[1,1,1,1;1,0,0,0;1,0,0,0;1,0,0,0]. - Paul Barry, Oct 01 2004

For n>2, a(n) is the numerator of the value of the continued fraction 1+3/(1+3/(1+...+3/7)) where there are n-2 1's. - Alexander Mark, Aug 16 2020

			a(4) = a(3)+3*a(2) = 10+3*7 = 31.

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", Wiley, 2001, p. 471.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Wikipedia, Lucas sequence
Index entries for linear recurrences with constant coefficients, signature (1,3).

Cf. A000032, A006130, A014551, A072265, A075117, A274977.

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

I:=[2,1]; [n le 2 select I[n] else Self(n-1)+3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jul 20 2013

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

a:= n-> (Matrix([[1,2]]). Matrix([[1,1], [3,0]])^n)[1,2]:
seq(a(n), n=0..35);  # Alois P. Heinz, Aug 15 2008

a[0]=2; a[1]=1; a[n_]:= a[n]= a[n-1] +3a[n-2]; Table[a[n], {n, 0, 30}]
CoefficientList[Series[(2-x)/(1-x-3x^2), {x,0,40}], x] (* Vincenzo Librandi, Jul 20 2013 *)
LinearRecurrence[{1,3},{2,1},40] (* Harvey P. Dale, Jun 18 2017 *)
Table[Round[Sqrt[3]^n*LucasL[n, 1/Sqrt[3]]], {n,0,40}] (* G. C. Greubel, Jan 15 2020 *)

my(x='x+O('x^30)); Vec((2-x)/(1-x-3*x^2)) \\ G. C. Greubel, Dec 21 2017

polsym(x^2-x-3, 44) \\ Joerg Arndt, Jan 22 2023

[lucas_number2(n,1,-3) for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009

a(n) = ((1+sqrt(13))/2)^n + ((1-sqrt(13))/2)^n.
a(n) = 2*A006130(n) - A006130(n-1) = A075117(3, n).
G.f.: (2-x)/(1-x-3*x^2). - Philippe Deléham, Nov 15 2008
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 13*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = 3^(n/2) * Lucas(n, 1/sqrt(3)). - G. C. Greubel, Jan 15 2020

A368152 Triangular array T(n,k), read by rows: coefficients of strong divisibility sequence of polynomials p(1,x) = 1, p(2,x) = 1 + 3x, p(n,x) = up(n-1,x) + v*p(n-2,x) for n >= 3, where u = p(2,x), v = 3 - x^2.

Original entry on oeis.org

1, 1, 3, 4, 6, 8, 7, 27, 25, 21, 19, 66, 126, 90, 55, 40, 204, 392, 504, 300, 144, 97, 522, 1363, 1884, 1851, 954, 377, 217, 1425, 4065, 7281, 8011, 6435, 2939, 987, 508, 3642, 12332, 24606, 34044, 31446, 21524, 8850, 2584, 1159, 9441, 35236, 82020, 127830

Offset: 1

Clark Kimberling, Jan 20 2024

Because (p(n,x)) is a strong divisibility sequence, for each integer k, the sequence (p(n,k)) is a strong divisibility sequence of integers.

			First eight rows:
    1
    1    3
    4    6    8
    7   27   25   21
   19   66  126   90   55
   40  204  392  504  300  144
   97  522 1363 1884 1851  954  377
  217 1425 4065 7281 8011 6435 2939 987
Row 4 represents the polynomial p(4,x) = 7 + 27*x + 25*x^2 + 21*x^3, so (T(4,k)) = (7,27,25,21), k=0..3.

Rigoberto Flórez, Robinson Higuita, and Antara Mukherjee, Characterization of the strong divisibility property for generalized Fibonacci polynomials, Integers, 18 (2018), 1-28, Paper No. A14.

Cf. A006130 (column 1); A001906 (p(n,n-1)); A090017 (row sums), (p(n,1)); A002605 (alternating row sums), (p(n,-1)); A004187, (p(n,2)); A004254, (p(n,-2)); A190988, (p(n,3)); A190978 (unsigned), (p(n,-3)); A094440, A367208, A367209, A367210, A367211, A367297, A367298, A367299, A367300, A367301, A368150, A368151.

p[1, x_] := 1; p[2, x_] := 1 + 3 x; u[x_] := p[2, x]; v[x_] := 3 - x^2;
p[n_, x_] := Expand[u[x]*p[n - 1, x] + v[x]*p[n - 2, x]]
Grid[Table[CoefficientList[p[n, x], x], {n, 1, 10}]]
Flatten[Table[CoefficientList[p[n, x], x], {n, 1, 10}]]

p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >= 3, where p(1,x) = 1, p(2,x) = 1 + 3*x, u = p(2,x), and v = 3 - x^2.

p(n,x) = k*(b^n - c^n), where k = -1/sqrt(13 + 6*x + 5*x^2), b = (1/2)*(3*x + 1 - 1/k), c = (1/2)*(3*x + 1 + 1/k).

A099232 a(n) = 2a(n-1) + 2a(n-2) - 3*a(n-3), with a(0) = 0, a(1) = 1, a(2) = 2.

Original entry on oeis.org

0, 1, 2, 6, 13, 32, 72, 169, 386, 894, 2053, 4736, 10896, 25105, 57794, 133110, 306493, 705824, 1625304, 3742777, 8618690, 19847022, 45703093, 105244160, 242353440, 558085921, 1285146242, 2959404006, 6814842733, 15693054752, 36137582952

Offset: 0

Paul Barry, Oct 08 2004

Partial sums of A006130 (with leading zero).

Specify a triangle by T(n,0) = T(n+1,1) = A001045(n) and T(n,k) = T(n-1,k-1) + T(n-1,k-2) + T(n-2,k-2) otherwise. Then T(n,n)= a(n-1). - J. M. Bergot, May 24 2013

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-3).

Cf. A001045, A006130.

LinearRecurrence[{2,2,-3},{0,1,2},40] (* Harvey P. Dale, Mar 02 2024 *)

G.f.: x/((1-x)*(1-x-3*x^2)).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k+1)*3^k.
a(n) = (1/2 + sqrt(13)/2)^n*(1/6 + 7*sqrt(13)/78) + (1/6 - 7*sqrt(13)/78)*(1/2 - sqrt(13)/2)^n - 1/3.
a(n+1) = Sum_{k=0..n} C(k+1,n-k+1)*3^(n-k). - Paul Barry, May 21 2006
a(n) = a(n-1) + 3*a(n-2) + 1, n > 1. - Gary Detlefs, Jun 21 2010
G.f.: Q(0)*x/(2-2*x), where Q(k) = 1 + 1/(1 - x*(4*k+1 + 3*x)/( x*(4*k+3 + 3*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 09 2013

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

cp's OEIS Frontend

A123585 Triangle T(n,k), 0<=k<=n, given by [1, -1, 1, 0, 0, 0, 0, 0, ...] DELTA [1, 1, -1, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

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A182228 a(n) = 3*a(n-2) - a(n-1) for 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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A304236 Triangle T(n,k) = T(n-1,k) + 3*T(n-2,k-1) for k = 0..floor(n/2), with T(0,0) = 1, T(n,k) = 0 for n or k < 0, read by rows.

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A074324 a(2n+1) = 3^n, a(2n) = 4*3^(n-1) except for a(0) = 1.

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A074355 Coefficient of q^1 in nu(n), where nu(0)=1, nu(1)=b and, for n>=2, nu(n)=b*nu(n-1)+lambda*(1+q+q^2+...+q^(n-2))*nu(n-2) with (b,lambda)=(1,3).

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A075118 Variant on Lucas numbers: a(n) = a(n-1) + 3*a(n-2) with a(0)=2 and a(1)=1.

A074355 Coefficient of q^1 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).

A368152 Triangular array T(n,k), read by rows: coefficients of strong divisibility sequence of polynomials p(1,x) = 1, p(2,x) = 1 + 3x, p(n,x) = up(n-1,x) + v*p(n-2,x) for n >= 3, where u = p(2,x), v = 3 - x^2.

A099232 a(n) = 2a(n-1) + 2a(n-2) - 3*a(n-3), with a(0) = 0, a(1) = 1, a(2) = 2.

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