A027471 -id:A027471 - cp's OEIS frontend

A090018 a(n) = 6a(n-1) + 3a(n-2) for n > 2, a(0)=1, a(1)=6.

1, 6, 39, 252, 1629, 10530, 68067, 439992, 2844153, 18384894, 118841823, 768205620, 4965759189, 32099171994, 207492309531, 1341251373168, 8669985167601, 56043665125110, 362271946253463, 2341762672896108, 15137391876137037, 97849639275510546, 632510011281474387

Offset: 0

Paul Barry, Nov 19 2003

From Johannes W. Meijer, Aug 09 2010: (Start)

a(n) represents the number of n-move routes of a fairy chess piece starting in a given corner or side square on a 3 X 3 chessboard. This fairy chess piece behaves like a white queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen, see A180032. The central square leads to A180028. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,3).

Cf. A141041, A180028, A180032.

Sequences with g.f. of the form 1/(1 - 6*x - k*x^2): A106392 (k=-10), A027471 (k=-9), A006516 (k=-8), A081179 (k=-7), A030192 (k=-6), A003463 (k=-5), A084326 (k=-4), A138395 (k=-3), A154244 (k=-2), A001109 (k=-1), A000400 (k=0), A005668 (k=1), A135030 (k=2), this sequence (k=3), A135032 (k=4), A015551 (k=5), A057089 (k=6), A015552 (k=7), A189800 (k=8), A189801 (k=9), A190005 (k=10), A015553 (k=11).

[n le 2 select 6^(n-1) else 6*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 15 2011

a:= n-> (<<0|1>, <3|6>>^n. <<1,6>>)[1,1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 17 2011

Join[{a=1,b=6},Table[c=6*b+3*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 16 2011 *)
LinearRecurrence[{6,3}, {1,6}, 41] (* G. C. Greubel, Oct 10 2022 *)

my(x='x+O('x^30)); Vec(1/(1-6*x-3*x^2)) \\ G. C. Greubel, Jan 24 2018

[lucas_number1(n,6,-3) for n in range(1, 31)] # Zerinvary Lajos, Apr 24 2009

a(n) = (3+2*sqrt(3))^n*(sqrt(3)/4+1/2) + (1/2-sqrt(3)/4)*(3-2*sqrt(3))^n.
a(n) = (-i*sqrt(3))^n * ChebyshevU(n, isqrt(3)), i^2=-1.
From Johannes W. Meijer, Aug 09 2010: (Start)
G.f.: 1/(1 - 6*x - 3*x^2).
Limit_{k->oo} a(n+k)/a(k) = A141041(n) + A090018(n-1)*sqrt(12) for n >= 1.
Limit_{n->oo} A141041(n)/A090018(n-1) = sqrt(12). (End)
a(n) = Sum_{k=0..n} A099089(n,k)*3^k. - Philippe Deléham, Nov 21 2011
E.g.f.: exp(3*x)*(2*cosh(2*sqrt(3)*x) + sqrt(3)*sinh(2*sqrt(3)*x))/2. - Stefano Spezia, Apr 23 2025

Typo in Mathematica program corrected by Vincenzo Librandi, Nov 15 2011

A120908 Sum of the lengths of the drops in all ternary words of length n on {0,1,2}. The drops of a ternary word on {0,1,2} are the subwords 10,20 and 21, their lengths being the differences 1, 2 and 1, respectively.

Original entry on oeis.org

0, 4, 24, 108, 432, 1620, 5832, 20412, 69984, 236196, 787320, 2598156, 8503056, 27634932, 89282088, 286978140, 918330048, 2927177028, 9298091736, 29443957164, 92980917360, 292889889684, 920511081864, 2887057484028

Offset: 1

Emeric Deutsch, Jul 15 2006

a(n) = 4*A027471(n).

a(n) = Sum_{k>=0} k*A120907(n,k).

			a(2)=4 because the ternary words 00,01,02,11,12 and 22 have no drops, each of the words 10 and 21 has one drop of length 1 and the word 20 has one drop of length 2.

Vincenzo Librandi, Table of n, a(n) for n = 1..400
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Index entries for linear recurrences with constant coefficients, signature (6,-9).

Cf. A027471, A120906, A120907.

[4*(n-1)*3^(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 09 2011

Maple
```
seq(4*(n-1)*3^(n-2),n=1..27);
```

Table[4*(n-1)*3^(n-2), {n, 30}] (* Wesley Ivan Hurt, Jan 28 2014 *)
LinearRecurrence[{6,-9},{0,4},30] (* Harvey P. Dale, Jul 14 2023 *)

a(n) = 4*(n-1)*3^(n-2); \\ Altug Alkan, May 16 2018

a(n) = 4*(n-1)*3^(n-2).

G.f.: 4*z^2/(1-3*z)^2.

A064017 Number of ternary trees (A001764) with n nodes and maximal diameter.

Original entry on oeis.org

1, 3, 12, 45, 162, 567, 1944, 6561, 21870, 72171, 236196, 767637, 2480058, 7971615, 25509168, 81310473, 258280326, 817887699, 2582803260, 8135830269, 25569752274, 80196041223, 251048476872, 784526490225, 2447722649502

Offset: 1

Danail Bonchev (bonchevd(AT)aol.com), Sep 07 2001

A problem important for polymer science because it counts the trees having unbranched branches; they are called "combs".
Equals (1, 3, 9, 27, 81, ...) convolved with (1, 0, 3, 9, 27, 81, ...). Example: a(5) = 162 = (81, 27, 9, 3, 1) dot (1, 0, 3, 9, 27) = 81 + 3*27. - Gary W. Adamson, Jul 31 2010
Floretion Algebra Multiplication Program, FAMP Code: lesforseq[ - 'i + 'j - 'kk' - 'ki' - 'kj' ], vesforseq(n) = 3^n, tesforseq = A006234

			a(5) = 162 because we can write (5+1)*3^(5-2) = 6*3^3 = 6*27.

Harry J. Smith, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (6,-9).

Cf. A001764, A006234, A014915, A025192, A027261, A079272, A196389.

a:=n->ceil(sum(3^(n-2),j=0..n)): seq(a(n), n=1..26); # Zerinvary Lajos, Jun 05 2008

Join[{1},Table[(n+1)3^(n-2),{n,2,30}]] (* or *) Join[{1}, LinearRecurrence[ {6,-9},{3,12},30]] (* Harvey P. Dale, Feb 07 2012 *)

{ for (n=1, 200, if (n>1, a=(n + 1)*p; p*=3, a=p=1); write("b064017.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 06 2009

a(n)=if(n==1, 1, (n+1)*3^(n-2)); \\ Joerg Arndt, May 06 2013

@CachedFunction
def BB(n, k, x):  # modified cardinal B-splines
    if n == 1: return 0 if (x < 0) or (x >= k) else 1
    return x*BB(n-1, k, x) + (n*k-x)*BB(n-1, k, x-k)
def EulerianPolynomial(n, k, x):
    if n == 0: return 1
    return add(BB(n+1, k, k*m+1)*x^m for m in (0..n))
def A064017(n) : return 3^(n-1)*EulerianPolynomial(1,n-1,1/3) if n != 1 else 1
[A064017(n) for n in (1..25)]  # Peter Luschny, May 04 2013

a(n) = 3*a(n-1) + 3^(n-2).
a(n) = (n+1)*3^(n-2), for n > 1.
From Paul Barry, Sep 05 2003: (Start)
a(n) = (n+2)3^(n-1) + 0^n/3 (offset 0).
a(n) = A025192(n) + A027471(n). (End)
A006234(n+4) - a(n+2) = 3^n. - Creighton Dement, Mar 01 2005
a(n+1) = Sum_{k=0..n} A196389(n,k)*3^k. - Philippe Deléham, Oct 31 2011
G.f.: (1 - 3*x + 3*x^2)*x/(1 - 3*x)^2. - Philippe Deléham, Oct 31 2011
a(n) = 6*a(n-1) - 9*a(n-2), with a(1)=1, a(2)=3, a(3)=12. - Harvey P. Dale, Feb 07 2012
E.g.f.: (exp(3*x)*(1 + 3*x) - 1)/9. - Stefano Spezia, Mar 05 2020
From Amiram Eldar, Jan 18 2021: (Start)
Sum_{n>=1} 1/a(n) = 27*log(3/2) - 19/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = 17/2 - 27*log(4/3). (End)

A336951 E.g.f.: 1 / (1 - x * exp(3*x)).

Original entry on oeis.org

1, 1, 8, 69, 780, 11145, 191178, 3823785, 87406056, 2247785073, 64228084110, 2018771719569, 69221032558956, 2571290056399545, 102860527370221026, 4408690840306136505, 201557641172689004112, 9790792086366911655009, 503570143277542340304534

Offset: 0

Ilya Gutkovskiy, Aug 08 2020

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..383

Column k=3 of A351790.

Cf. A006153, A027471, A235328, A255927, A328182, A336948, A336950, A336952.

nmax = 18; CoefficientList[Series[1/(1 - x Exp[3 x]), {x, 0, nmax}], x] Range[0, nmax]!
Join[{1}, Table[n! Sum[(3 (n - k))^k/k!, {k, 0, n}], {n, 1, 18}]]
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] k 3^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 18}]

seq(n)={ Vec(serlaplace(1 / (1 - x*exp(3*x + O(x^n))))) } \\ Andrew Howroyd, Aug 08 2020

a(n) = n! * Sum_{k=0..n} (3 * (n-k))^k / k!.
a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * k * 3^(k-1) * a(n-k).
a(n) ~ n! * (3/LambertW(3))^n / (1 + LambertW(3)). - Vaclav Kotesovec, Aug 09 2021

A192015 Arithmetic derivative of prime powers: a(n) = A003415(A000961(n)).

Original entry on oeis.org

0, 1, 1, 4, 1, 1, 12, 6, 1, 1, 32, 1, 1, 1, 10, 27, 1, 1, 80, 1, 1, 1, 1, 14, 1, 1, 1, 192, 1, 1, 1, 1, 108, 1, 1, 1, 1, 1, 1, 1, 1, 22, 75, 1, 448, 1, 1, 1, 1, 1, 1, 1, 1, 26, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 405, 1, 1024, 1, 1, 1, 1, 1, 1, 1, 34

Offset: 1

Reinhard Zumkeller, Jun 26 2011

nonn

a(A000040(n)) = 1; a(A002808(n)) > 1;
A001787, A027471, A100484, A079705 and A051674 are subsequences;
A001787 and A024622 give record values and where they occur;
A192016(n) = A003415(a(n)).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Prime Power

a192015 = a003415 . a000961  -- Reinhard Zumkeller, Apr 16 2014

Join[{0}, Reap[For[n = 1, n <= 300, n++, f = FactorInteger[n]; If[Length[f] == 1, Sow[n*Total[Apply[#2/#1&, f, {1}]]]]]][[2, 1]]] (* Jean-François Alcover, Feb 21 2014 *)

from sympy import primepi, integer_nthroot, factorint
def A192015(n):
    if n == 1: return 0
    def f(x): return int(n+x-1-sum(primepi(integer_nthroot(x,k)[0]) for k in range(1,x.bit_length())))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return sum((m*e//p for p,e in factorint(m).items())) # Chai Wah Wu, Aug 15 2024

a(n) = A025474(n) * A025473(n)^(A025474(n) - 1).

A288834 a(n) = (n+1) * 3^(n-1).

Original entry on oeis.org

2, 9, 36, 135, 486, 1701, 5832, 19683, 65610, 216513, 708588, 2302911, 7440174, 23914845, 76527504, 243931419, 774840978, 2453663097, 7748409780, 24407490807, 76709256822, 240588123669, 753145430616, 2353579470675, 7343167948506, 22876792454961

Offset: 1

Gregory Gerard Wojnar, Jun 17 2017

nonn

Index entries for linear recurrences with constant coefficients, signature (6,-9).

Cf. A000244, A027471, A287768.

Table[(n + 1)*3^(n - 1), {n, 27}] (* Michael De Vlieger, Jun 23 2017 *)
LinearRecurrence[{6,-9},{2,9},40] (* Harvey P. Dale, Dec 16 2018 *)

a(n) = (n+1)*3^(n-1) \\ Felix Fröhlich, Jun 19 2017

Vec((z*(2-3*z)/(1-3*z)^2) + O(z^30)) \\ Felix Fröhlich, Jun 19 2017

O.g.f.: z*(2-3*z)/(1-3*z)^2.
a(n) = -A287768(n+1,2).
a(n) = (n+1)*A000244(n-1). - Felix Fröhlich, Jun 19 2017
a(n) = A027471(n)/3 for n >= 3. - Art Baker, Apr 12 2019
From Amiram Eldar, Jan 18 2021: (Start)
Sum_{n>=1} 1/a(n) = 9*log(3/2) - 3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3 - 9*log(4/3). (End)

A099097 Riordan array (1, 3+x).

Original entry on oeis.org

1, 0, 3, 0, 1, 9, 0, 0, 6, 27, 0, 0, 1, 27, 81, 0, 0, 0, 9, 108, 243, 0, 0, 0, 1, 54, 405, 729, 0, 0, 0, 0, 12, 270, 1458, 2187, 0, 0, 0, 0, 1, 90, 1215, 5103, 6561, 0, 0, 0, 0, 0, 15, 540, 5103, 17496, 19683, 0, 0, 0, 0, 0, 1, 135, 2835, 20412, 59049, 59049, 0, 0, 0, 0, 0, 0, 18, 945, 13608, 78732, 196830, 177147

Offset: 0

Paul Barry, Sep 25 2004

Row sums are A006190(n+1). Diagonal sums are A052931. The Riordan array (1, s+tx) defines T(n,k) = binomial(k,n-k)*s^k*(t/s)^(n-k). The row sums satisfy a(n) = s*a(n-1) + t*a(n-2) and the diagonal sums satisfy a(n) = s*a(n-2) + t*a(n-3).

Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1/3, -1/3, 0, 0, 0, 0, 0, ...] DELTA [3, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 10 2008

			Triangle begins:
  1;
  0, 3;
  0, 1, 9;
  0, 0, 6, 27;
  0, 0, 1, 27,  81;
  0, 0, 0,  9, 108, 243;
  ...

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A027465.

Diagonals are of the form 3^n*binomial(n+m, m): A000244 (m=0), A027471 (m=1), A027472 (m=2), A036216 (m=3), A036217 (m=4), A036219 (m=5), A036220 (m=6), A036221 (m=7), A036222 (m=8), A036223 (m=9), A172362 (m=10).

Table[3^(2*k-n)*Binomial[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 19 2021 *)

flatten([[3^(2*k-n)*binomial(k, n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 19 2021

Triangle: T(n, k) = binomial(k, n-k)*3^k*(1/3)^(n-k).
G.f. of column k: (3*x + x^2)^k.
G.f.: 1/(1 - 3*y*x - y*x^2). - Philippe Deléham, Nov 21 2011
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A006190(n+1), A135030(n+1), A181353(n+1) for x = 0,1,2,3 respectively. - Philippe Deléham, Nov 21 2011

A118441 Triangle L, read by rows, equal to the matrix log of A118435, with the property that L^2 consists of a single diagonal (two rows down from the main diagonal).

Original entry on oeis.org

0, 1, 0, -4, 2, 0, -12, 12, 3, 0, 32, -48, -24, 4, 0, 80, -160, -120, 40, 5, 0, -192, 480, 480, -240, -60, 6, 0, -448, 1344, 1680, -1120, -420, 84, 7, 0, 1024, -3584, -5376, 4480, 2240, -672, -112, 8, 0, 2304, -9216, -16128, 16128, 10080, -4032, -1008, 144, 9, 0

Offset: 0

Paul D. Hanna, Apr 28 2006

L = log(A118435) = log(H*[C^-1]*H], where C=Pascal's triangle and H=A118433 where H^2 = I (identity matrix).

			The matrix log, L = log(H*[C^-1]*H], begins:
     0;
     1,     0;
    -4,     2,      0;
   -12,    12,      3,     0;
    32,   -48,    -24,     4,     0;
    80,  -160,   -120,    40,     5,     0;
  -192,   480,    480,  -240,   -60,     6,     0;
  -448,  1344,   1680, -1120,  -420,    84,     7,   0;
  1024, -3584,  -5376,  4480,  2240,  -672,  -112,   8,  0;
  2304, -9216, -16128, 16128, 10080, -4032, -1008, 144,  9,  0;
  ...
The matrix square, L^2, is a single diagonal:
  0;
  0, 0;
  2, 0,  0;
  0, 6,  0,  0;
  0, 0, 12,  0,  0;
  0, 0,  0, 20,  0,  0;
  0, 0,  0,  0, 30,  0,  0;
  ...
From _Peter Luschny_, Apr 23 2020: (Start)
In unsigned form and without the main diagonal, as computed by the Maple script:
  [0], [0]
  [1], [1]
  [2], [4,   2]
  [3], [12,  12,   3]
  [4], [32,  48,   24,   4]
  [5], [80,  160,  120,  40,   5]
  [6], [192, 480,  480,  240,  60,  6]
  [7], [448, 1344, 1680, 1120, 420, 84, 7] (End)

Cf. A118435 (exp(L)), A118442 (column 0), A118443 (row sums), A027471 (unsigned row sums); A118433 (self-inverse triangle), A001815 (column 1?), A001789 (third of column 2?).

# Generalized Worpitzky transform of the harmonic numbers.
CL := p -> PolynomialTools:-CoefficientList(expand(p), x):
H := n -> add(1/k, k=1..n):
Trow := proc(n) local k,v; if n=0 then return [0] fi;
add(add((-1)^(n-v)*binomial(k,v)*H(k)*(-x+v-1)^n, v=0..k), k=0..n); CL(%) end:
for n from 0 to 7 do Trow(n) od; # Peter Luschny, Apr 23 2020

nmax = 12;
h[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2] - Quotient[k, 2]+n-k);
H = Table[h[n, k], {n, 0, nmax}, {k, 0, nmax}];
Cn = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, nmax}];
L = MatrixLog[H.Inverse[Cn].H ];
Table[L[[n+1, k+1]], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 08 2024 *)

/* From definition of L as matrix log of H*C^-1*H: */
{L(n,k)=local(H=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1)*(-1)^(r\2-(c-1)\2+r-c))),C=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1))),N=(H*C^-1*H)); Log=sum(p=1,n+1,-(N^0-N)^p/p);Log[n+1,k+1]}
for(n=0, 10, for(k=0, n, print1(L(n, k), ", ")); print(""))

/* The matrix power L^m is given by: */
{L(n,k,m)=if(m%2==0,if(n==k+m,n!/k!*2^(n-k-m)/(n-k-m)!), if(n>=k+m,n!/k!*2^(n-k-m)/(n-k-m)!*(-1)^(m\2+(n+1)\2-k\2+n-k)))}
for(n=0, 10, for(k=0, n, print1(L(n, k,1), ", ")); print(""))

For even exponents of L, L^(2m) is a single diagonal:
if n == k+2m, then [L^(2m)](n,k) = n!/k!*2^(n-k-2m)/(n-k-2m)!; else if n != k+2m: [L^(2m)](n,k) = 0.
For odd exponents of L:
if n >= k+2m+1, then [L^(2m+1)](n,k) = n!/k!*2^(n-k-2m-1)/(n-k-2m-1)!*(-1)^(m+[(n+1)/2]-[k/2]+n-k); else if n < k+2m+1: [L^(2m)](n,k) = 0.
Unsigned row sums equals A027471(n+1) = n*3^(n-1).

A224524 Table read by antidiagonals: T(n,k) is the number of idempotent n X n 0..k matrices of rank 1.

Original entry on oeis.org

1, 1, 6, 1, 10, 27, 1, 14, 69, 108, 1, 18, 123, 404, 405, 1, 22, 195, 892, 2155, 1458, 1, 26, 273, 1716, 5845, 10830, 5103, 1, 30, 375, 2732, 13525, 36042, 52241, 17496, 1, 34, 477, 4324, 24575, 99774, 213647, 244648, 59049, 1, 38, 603, 6060, 44545, 208146, 705215, 1232504, 1120599, 196830

Offset: 1

R. H. Hardin, Apr 09 2013

Table starts
      1,      1,      1,      1,      1,     1,    1,   1,  1, 1, ...
      6,     10,     14,     18,     22,    26,   30,  34, 38, ...
     27,     69,    123,    195,    273,   375,  477, 603, ...
    108,    404,    892,   1716,   2732,  4324, 6060, ...
    405,   2155,   5845,  13525,  24575, 44545, ...
   1458,  10830,  36042,  99774, 208146, ...
   5103,  52241, 213647, 705215, ...
  17496, 244648, ...
  59049, ...
  ...

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

Robert Israel, Table of n, a(n) for n = 1..10011

Column 1 is A027471(n+1).

f:= proc(n,k)
  local tot, a1, a0, a2, m,u;
  tot:= 0;
  for a1 from 1 to n do
    for a0 from 0 to n-a1 do
      a2:= n-a1-a0;
      if a0 = 0 then tot:= tot + n!/(a1!*a2!)*a1*(k-1)^a2
      elif a2 = 0 then tot:= tot + n!/(a0!*a1!)*a1*(k+1)^a0
      else
        u:= n!/(a0!*a1!*a2!)*a1;
        for m from 2 to k do
          tot:= tot + u*((m-1)^a2 - (m-2)^a2)*(floor(k/m)+1)^a0
        od
      fi
  od od;
  tot
end proc:
seq(seq(f(i,j-i),i=1..j-1),j=2..20); # Robert Israel, Dec 15 2019

Unprotect[Power]; 0^0 = 1; Protect[Power];
f[n_, k_] := Module[{tot, a1, a0, a2, m, u}, tot = 0; For[a1 = 1, a1 <= n, a1++, For[a0 = 0, a0 <= n - a1, a0++, a2 = n - a1 - a0; Which[a0 == 0, tot = tot + n!/(a1!*a2!)*a1*(k - 1)^a2, a2 == 0, tot = tot + n!/(a0!*a1!)*a1*(k + 1)^a0, True, u = n!/(a0!*a1!*a2!)*a1; For[m = 2, m <= k, m++, tot = tot + u*((m - 1)^a2 - (m - 2)^a2)*(Floor[k/m] + 1)^a0]]]]; tot];
Table[Table[f[i, j - i], {i, 1, j - 1}], {j, 2, 20}] // Flatten (* Jean-François Alcover, Feb 04 2023, after Robert Israel *)

A317497 Triangle T(n,k) = 3*T(n-1,k) + T(n-3,k-1) for k = 0..floor(n/3) with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

Original entry on oeis.org

1, 3, 9, 27, 1, 81, 6, 243, 27, 729, 108, 1, 2187, 405, 9, 6561, 1458, 54, 19683, 5103, 270, 1, 59049, 17496, 1215, 12, 177147, 59049, 5103, 90, 531441, 196830, 20412, 540, 1, 1594323, 649539, 78732, 2835, 15, 4782969, 2125764, 295245, 13608, 135, 14348907, 6908733, 1082565, 61236, 945, 1

Offset: 0

Zagros Lalo, Jul 31 2018

The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n) and  along a "second layer" of skew diagonals pointing top-right in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in triangles of coefficients in expansions of (1+3*x)^n and (3+x)^n are given in A304236 and A304249 respectively.)
The coefficients in the expansion of 1/(1-3x-x^3) are given by the sequence generated by the row sums.
The row sums give A052541.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.1038034027355..., when n approaches infinity.

			Triangle begins:
         1;
         3;
         9;
        27,        1;
        81,        6;
       243,       27;
       729,      108,       1;
      2187,      405,       9;
      6561,     1458,      54;
     19683,     5103,     270,      1;
     59049,    17496,    1215,     12;
    177147,    59049,    5103,     90;
    531441,   196830,   20412,    540,    1;
   1594323,   649539,   78732,   2835,   15;
   4782969,  2125764,  295245,  13608,  135;
  14348907,  6908733, 1082565,  61236,  945,  1;
  43046721, 22320522, 3897234, 262440, 5670, 18;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 364-366

G. C. Greubel, Rows n = 0..120 of the irregular triangle, flattened
Zagros Lalo, Second layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3x)^n
Zagros Lalo, Second layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n

Row sums give A052541.
Cf. A013610, A027465, A317496, A304236, A304249.
Cf. A000244 (column 0), A027471 (column 1), A027472 (column 2), A036216 (column 3), A036217 (column 4).
Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), A304249 (q=2), this sequence (q=3), A318773 (q=4).

Flat(List([0..16],n->List([0..Int(n/3)],k->3^(n-3*k)/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # Muniru A Asiru, Aug 01 2018

[3^(n-3*k)*Binomial(n-2*k,k): k in [0..Floor(n/3)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k] = 3^(n-3k)(n-2k)!/((n-3k)! k!); Table[T[n, k], {n, 0, 15}, {k, 0, Floor[n/3]} ]//Flatten
T[0, 0] = 1; T[n_, k_]:= T[n, k] = If[n<0 || k<0, 0, 3 T[n-1, k] + T[n-3, k-1]]; Table[T[n, k], {n, 0, 15}, {k, 0, Floor[n/3]}]//Flatten

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

T(n,k) = 3^(n-3*k) * (n-2*k)!/(k! * (n-3*k)!) where n is a nonnegative integer and k = 0..floor(n/3).

cp's OEIS Frontend

A090018 a(n) = 6*a(n-1) + 3*a(n-2) for n > 2, a(0)=1, a(1)=6.

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A120908 Sum of the lengths of the drops in all ternary words of length n on {0,1,2}. The drops of a ternary word on {0,1,2} are the subwords 10,20 and 21, their lengths being the differences 1, 2 and 1, respectively.

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A064017 Number of ternary trees (A001764) with n nodes and maximal diameter.

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A336951 E.g.f.: 1 / (1 - x * exp(3*x)).

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A192015 Arithmetic derivative of prime powers: a(n) = A003415(A000961(n)).

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A288834 a(n) = (n+1) * 3^(n-1).

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A099097 Riordan array (1, 3+x).

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A090018 a(n) = 6a(n-1) + 3a(n-2) for n > 2, a(0)=1, a(1)=6.