A027472 -id:A027472 - cp's OEIS frontend

A154372 Triangle T(n,k) = (k+1)^(n-k)*binomial(n,k).

1, 1, 1, 1, 4, 1, 1, 12, 9, 1, 1, 32, 54, 16, 1, 1, 80, 270, 160, 25, 1, 1, 192, 1215, 1280, 375, 36, 1, 1, 448, 5103, 8960, 4375, 756, 49, 1, 1, 1024, 20412, 57344, 43750, 12096, 1372, 64, 1

Offset: 0

Paul Curtz, Jan 08 2009

From A152650/A152656,coefficients of other exponential polynomials(*). a(n) is triangle A152818 where terms of each column is divided by the beginning one. See A000004, A001787(n+1), A006043=2*A027472, A006044=6*A038846.
(*) Not factorial as written in A006044. See A000110, Bell-Touchard. Second diagonal is 1,4,9,16,25, denominators of Lyman's spectrum of hydrogen, A000290(n+1) which has homogeneous indices for denominators series of Rydberg-Ritz spectrum of hydrogen.
The matrix inverse starts
      1;
     -1,      1;
      3,     -4,     1;
    -16,     24,    -9,     1;
    125,   -200,    90,   -16,   1;
  -1296,   2160, -1080,   240, -25,   1;
  16807, -28812, 15435, -3920, 525, -36, 1;
.. compare with A122525 (row reversed). - R. J. Mathar, Mar 22 2013
From Peter Bala, Jan 14 2015: (Start)
Exponential Riordan array [exp(z), z*exp(z)]. This triangle is the particular case a = 0, b = 1, c = 1 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. Cf. A059297.
This is the triangle of connection constants when expressing the monomials x^n as a linear combination of the basis polynomials (x - 1)*(x - k - 1)^(k-1), k = 0,1,2,.... For example, from row 3 we have x^3 = 1 + 12*(x - 1) + 9*(x - 1)*(x - 3) + (x - 1)*(x - 4)^2.
Let M be the infinite lower unit triangular array with (n,k)-th entry (k*(n - k + 1) + 1)/(k + 1)*binomial(n,k). M is the row reverse of A145033. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to the present triangle. See the Example section. (End)
T(n,k) is also the number of idempotent partial transformations of {1,2,...,n} having exactly k fixed points. - Geoffrey Critzer, Nov 25 2021

			With the array M(k) as defined in the Comments section, the infinite product M(0)*M(1)*M(2)*... begins
/1      \ /1        \ /1        \      /1        \
|1 1     ||0 1       ||0 1      |      |1  1      |
|1 3 1   ||0 1 1     ||0 0 1    |... = |1  4  1   |
|1 6 5 1 ||0 1 3 1   ||0 0 1 1  |      |1 12  9  1|
|...     ||0 1 6 5 1 ||0 0 1 3 1|      |...       |
|...     ||...       ||...      |      |          |
- _Peter Bala_, Jan 13 2015

G. C. Greubel, Table of n, a(n) for n = 0..1274
Peter Bala, A 3 parameter family of generalized Stirling numbers
Emanuele Munarini, Combinatorial identities involving the central coefficients of a Sheffer matrix, Applicable Analysis and Discrete Mathematics (2019) Vol. 13, 495-517.

Cf. A080108, A152818, A059297, A145033.

/* As triangle */ [[(k+1)^(n-k)*Binomial(n,k) : k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 15 2016

T[n_, k_] := (k + 1)^(n - k)*Binomial[n, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Sep 15 2016 *)

T(n,k) = (k+1)^(n-k)*binomial(n,k). k!*T(n,k) gives the entries for A152818 read as a triangular array.
E.g.f.: exp(x*(1+t*exp(x))) = 1 + (1+t)*x + (1+4*t+t^2)*x^2/2! + (1+12*t+9*t^2+t*3)*x^3/3! + .... O.g.f.: Sum_{k>=1} (t*x)^(k-1)/(1-k*x)^k = 1 + (1+t)*x + (1+4*t+t^2)*x^2 + .... Row sums are A080108. - Peter Bala, Oct 09 2011
From Peter Bala, Jan 14 2015: (Start)
Recurrence equation: T(n+1,k+1) = T(n,k+1) + Sum_{j = 0..n-k} (j + 1)*binomial(n,j)*T(n-j,k) with T(n,0) = 1 for all n.
Equals the matrix product A007318 * A059297. (End)

A217629 Triangle, read by rows, where T(n,k) = k!C(n, k)3^(n-k) for n>=0, k=0..n.

Original entry on oeis.org

1, 3, 1, 9, 6, 2, 27, 27, 18, 6, 81, 108, 108, 72, 24, 243, 405, 540, 540, 360, 120, 729, 1458, 2430, 3240, 3240, 2160, 720, 2187, 5103, 10206, 17010, 22680, 22680, 15120, 5040, 6561, 17496, 40824, 81648, 136080, 181440, 181440, 120960, 40320

Offset: 0

Vincenzo Librandi, Nov 10 2012

Triangle formed by the derivatives of x^n evaluated at x=3.
Sum(T(n,k), k=0..n) = A053486(n) (see the Formula section of A053486). Also:
first column:     A000244;
second column:    A027471;
third column:   2*A027472;
fourth column:  6*A036216;
fifth column:  24*A036217.

			Triangle begins:
1;
3,     1;
9,     6,     2;
27,    27,    18,     6;
81,    108,   108,    72,     24;
243,   405,   540,    540,    360,    120;
729,   1458,  2430,   3240,   3240,   2160,    720;
2187,  5103,  10206,  17010,  22680,  22680,   15120,   5040;
6561,  17496, 40824,  81648,  136080, 181440,  181440,  120960,  40320; etc.

Vincenzo Librandi, Rows n = 0..100, flattened

Cf. A000244, A027471, A027472, A036216, A036217, A053486, A090802, A218016, A218017.

[Factorial(n)/Factorial(n-k)*3^(n-k): k in [0..n], n in [0..10]];

Flatten[Table[n!/(n-k)!*3^(n-k), {n, 0, 10}, {k, 0, n}]]

T(n,k) = 3^(n-k)*n!/(n-k)! for n>=0, k=0..n.

E.g.f. (by columns): exp(3x)*x^k.

A318773 Triangle T(n,k) = 3*T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4), 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, 81, 1, 243, 6, 729, 27, 2187, 108, 6561, 405, 1, 19683, 1458, 9, 59049, 5103, 54, 177147, 17496, 270, 531441, 59049, 1215, 1, 1594323, 196830, 5103, 12, 4782969, 649539, 20412, 90, 14348907, 2125764, 78732, 540, 43046721, 6908733, 295245, 2835, 1, 129140163, 22320522, 1082565, 13608, 15

Offset: 0

Zagros Lalo, Sep 04 2018

The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n) and along a "third layer" skew diagonals pointing top-right in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in center-justified 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/(1-3*x-x^4) are given by the sequence generated by the row sums.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.035744112294..., when n approaches infinity.

			Triangle begins:
          1;
          3;
          9;
         27;
         81,        1;
        243,        6;
        729,       27;
       2187,      108;
       6561,      405,       1;
      19683,     1458,       9;
      59049,     5103,      54;
     177147,    17496,     270;
     531441,    59049,    1215,     1;
    1594323,   196830,    5103,    12;
    4782969,   649539,   20412,    90;
   14348907,  2125764,   78732,   540;
   43046721,  6908733,  295245,  2835,   1;
  129140163, 22320522, 1082565, 13608,  15;
  387420489, 71744535, 3897234, 61236, 135;
  ...

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

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

Row sums give A052917.
Cf. A013610, A027465.
Cf. A304236, A304249
Cf. A317496, A317497.
Cf. A000244 (column 0), A027471 (column 1), A027472 (column 2), A036216 (column 3).
Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), A304249 (q=2), A317497 (q=3), this sequence (q=4).

[3^(n-4*k)*Binomial(n-3*k,k): k in [0..Floor(n/4)], n in [0..24]]; // G. C. Greubel, May 12 2021

T[n_, k_]:= T[n, k] = 3^(n-4k)*(n-3k)!/((n-4k)! k!); Table[T[n, k], {n, 0, 17}, {k, 0, Floor[n/4]} ]//Flatten
T[0, 0] = 1; T[n_, k_]:= T[n, k] = If[n<0 || k<0, 0, 3T[n-1, k] + T[n-4, k-1]]; Table[T[n, k], {n, 0, 17}, {k, 0, Floor[n/4]}]//Flatten

flatten([[3^(n-4*k)*binomial(n-3*k,k) for k in (0..n//4)] for n in (0..24)]) # G. C. Greubel, May 12 2021

T(n,k) = 3^(n-4*k) * (n-3*k)!/(k! * (n-4*k)!) where n >= 0 and 0 <= k <= floor(n/4).

A094796 Triangle read by rows giving coefficients of polynomials arising in successive differences of central binomial numbers.

Original entry on oeis.org

1, 3, 1, 9, 15, 6, 27, 108, 135, 42, 81, 594, 1539, 1530, 456, 243, 2835, 12555, 25245, 22122, 6120, 729, 12393, 83835, 281475, 482436, 383292, 101520, 2187, 51030, 489888, 2466450, 6916833, 10546200, 7786692, 1980720

Offset: 0

Benoit Cloitre, Jun 11 2004

Let D_0(n)=binomial(2*n,n) and D_{k+1}(n)=D_{k}(n+1)-D_{k}(n); then D_{k}(n)*(n+1)*(n+2)*...*(n+k) = binomial(2*n,n)*P_{k}(n) where P_{k} is a polynomial with integer coefficients of degree k.

			The third differences of the central binomial numbers are given by D_3(n) = binomial(2*n,n)*(n+1)*(n+2)*(n+3)*(27*n^3 + 108*n^2 + 135*n + 42) and the fourth row of the triangle is 27, 108, 135, 42.
From _M. F. Hasler_, Nov 15 2019: (Start)
The table reads:
  n  |  row(n)
  0  |    1
  1  |    3      1
  2  |    9     15       6
  3  |   27    108     135       42
  4  |   81    594    1539     1530      456
  5  |  243   2835   12555    25245    22122      6120
  6  |  729  12393   83835   281475   482436    383292    101520
  7  | 2187  51030  489888  2466450  6916833  10546200   7786692   1980720
  8  | 6561 201204 2602530 18329976 75981969 186899076 260520300 181218384 44634240
(End)

Cf. A000984 (central binomial coefficients), A163771 (square array of central binomial coefficients and higher differences), A000244 (column k=0).

Main diagonal gives A098461.

Dnk := proc(n,k)
    option remember;
    if k < 0 then
        0 ;
    elif k = 0 then
        binomial(2*n,n) ;
    else
        procname(n+1,k-1)-procname(n,k-1) ;
    end if;
end proc:
A094796 := proc(n,k)
    local xyvec,i,x ;
    xyvec := [] ;
    for i from 0 to n do
        xyvec := [op(xyvec),[i,Dnk(i,n)*mul(i+j,j=1..n)/Dnk(i,0)]] ;
    end do:
    CurveFitting[PolynomialInterpolation](xyvec,x) ;
    coeff(%,x,n-k) ;
end proc: # R. J. Mathar, Nov 19 2019

Dnk[n_, k_] := Dnk[n, k] = Which[k < 0, 0, k == 0, Binomial[2*n, n], True, Dnk[n + 1, k - 1] - Dnk[n, k - 1]];
T[n_, k_] := Module[{xyvec, i, x , ip}, xyvec = {}; For[i = 0, i <= n, i++, AppendTo[xyvec, {i, Dnk[i, n]*Product[i + j, {j, 1, n}]/Dnk[i, 0]}]]; ip = InterpolatingPolynomial[xyvec, x]; Coefficient[ip, x, n - k]];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 01 2024, after R. J. Mathar *)

apply( {A094796_row(n,D(n,k)=if(k,D(n+1,k-1)-D(n,k-1),binomial(2*n,n)))=Vec(polinterpolate([0..n],vector(n+1,k,D(k--,n)*(n+k)!/k!/binomial(2*k,k))))}, [0..8]) \\ M. F. Hasler, Nov 15 2019

T(n,0) = 3^n. T(n,1) = A027472(n+2) + 6*A027472(n+1). T(n,2) = 3*(2*A036217(n-2) + 15*A036217(n-3) + 18*A036217(n-4)). - R. J. Mathar, Nov 19 2019

Corrected and edited by M. F. Hasler, following observations by R. J. Mathar and Don Reble, Nov 15 2019

More terms from Don Reble, Nov 15 2019

A082149 A transform of C(n,2).

Original entry on oeis.org

0, 0, 1, 6, 30, 140, 615, 2562, 10220, 39384, 147645, 541310, 1948650, 6908772, 24180611, 83702010, 286978200, 975725744, 3293074233, 11041484022, 36804946550, 122037454140, 402723598431, 1323234680306, 4330586226180

Offset: 0

Paul Barry, Apr 07 2003

Represents the mean of C(n,2) with its second binomial transform. Binomial transform of A080929 (preceded by two zeros).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (12,-57,136,-171,108,-27).

Cf. A082150, A000217, A027472.

[Binomial(n,2)*(1 + 3^(n-2))/2: n in [0..30]]; // G. C. Greubel, Feb 10 2018

CoefficientList[Series[(x^2/(1-3*x)^3 + x^2/(1-x)^3)/2, {x,0,50}], x] (* or *) Table[Binomial[n,2]*(1 + 3^(n-2))/2, {n,0,30}] (* G. C. Greubel, Feb 10 2018 *)
LinearRecurrence[{12,-57,136,-171,108,-27},{0,0,1,6,30,140},30] (* Harvey P. Dale, Aug 11 2021 *)

for(n=0,30, print1(binomial(n,2)*(1 + 3^(n-2))/2, ", ")) \\ G. C. Greubel, Feb 10 2018

a(n) = C(n, 2)*(3^(n-2) + 1)/2.
G.f.: (x^2/(1-3x)^3+x^2/(1-x)^3)/2.
G.f.: x^2(14*x^3-15*x^2+6*x-1)/((1-x)^3*(3*x-1)^3).
E.g.f.: x^2*exp(2*x)*cosh(x)/2.

A082151 A transform of C(n,2).

Original entry on oeis.org

0, 0, 1, 12, 102, 760, 5295, 35364, 228956, 1445616, 8936685, 54252220, 324214242, 1911205608, 11132579003, 64170616020, 366497915640, 2076171038176, 11676266706969, 65242364726124, 362433045180830, 2002838101907160, 11015341078090503, 60321223747375492

Offset: 0

Paul Barry, Apr 07 2003

Represents the mean of the second and fourth binomial transforms of C(n,2). Binomial transform of A082150

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (24,-237,1232,-3555,5400,-3375).

Cf. A027472, A081135, A000217.

[Binomial(n,2)*(3^(n-2) + 5^(n-2))/2: n in [0..30]]; // G. C. Greubel, Feb 10 2018

CoefficientList[Series[(x^2/(1-5*x)^3 + x^2/(1-3*x)^3)/2, {x,0,50}], x] (* or *) Table[Binomial[n,2]*(3^(n-2) + 5^(n-2))/2, {n,0,30}] (* G. C. Greubel, Feb 10 2018 *)
LinearRecurrence[{24,-237,1232,-3555,5400,-3375},{0,0,1,12,102,760},30] (* Harvey P. Dale, Apr 10 2023 *)

for(n=0,30, print1(binomial(n,2)*(3^(n-2) + 5^(n-2))/2, ", ")) \\ G. C. Greubel, Feb 10 2018

a(n) = C(n, 2)*(3^(n-2) + 5^(n-2))/2.
G.f.: (x^2/(1-5*x)^3 + x^2/(1-3*x)^3)/2.
a(n) = x^2*(76*x^3 - 51*x^2 + 12*x - 1)/((1-3*x)^3*(5*x-1)^3).
E.g.f.: x^2*exp(4*x)*cosh(x)/2.

A116138 a(n) = 3^n * n*(n + 1).

Original entry on oeis.org

0, 6, 54, 324, 1620, 7290, 30618, 122472, 472392, 1771470, 6495390, 23383404, 82904796, 290166786, 1004423490, 3443737680, 11708708112, 39516889878, 132497807238, 441659357460, 1464449448420, 4832683179786, 15878816162154

Offset: 0

Mohammad K. Azarian, Apr 08 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-27,27).

Cf. A007758, A036289, A128796, A027472.

List([0..30], n-> 3^n*n*(n+1)); # G. C. Greubel, May 10 2019

[(n^2+n)*3^n: n in [0..30]]; // Vincenzo Librandi, Feb 28 2013

I:=[0,6,54]; [n le 3 select I[n] else 9*Self(n-1)-27*Self(n-2)+27*Self(n-3): n in [1..30]]; // Vincenzo Librandi, Feb 28 2013

Table[(n^2 + n) 3^n, {n, 0, 30}] (* Vincenzo Librandi, Feb 28 2013 *)

a(n)=(n^2+n)*3^n \\ Charles R Greathouse IV, Feb 28 2013

[3^n*n*(n+1) for n in (0..30)] # G. C. Greubel, May 10 2019

From R. J. Mathar, Dec 19 2008: (Start)
G.f.: 6*x/(1-3*x)^3.
a(n) = 6 * A027472(n+2). (End)
a(n) = 9*a(n-1) -27*a(n-2) +27*a(n-3). - Vincenzo Librandi, Feb 28 2013
E.g.f.: 3*x*(2 + 3*x)*exp(3*x). - G. C. Greubel, May 10 2019
From Amiram Eldar, Jul 20 2020: (Start)
Sum_{n>=1} 1/a(n) = 1 - 2*log(3/2).
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(4/3) - 1. (End)

A153703 Partial sums of A069996.

Original entry on oeis.org

1, 13, 94, 526, 2551, 11299, 47020, 186988, 718429, 2686729, 9831658, 35340826, 125154355, 437641663, 1513809688, 5187129880, 17627632249, 59469045061, 199327841590, 664232428390, 2201904349231, 7264715299483, 23865295832644, 78091766836996

Offset: 1

Bruno Berselli, Dec 12 2010

The first differences are in the third row of the square array of A072590.

The general formula for the partial sums of the sequence 1, 4*m, 9*m^2, 16*m^3, 25*m^4,...,n^2*m^(n-1),... is (n^2*m^(n+2)-(2*n*(n+1)-1)*m^(n+1)+(n+1)^2*m^n-m-1)/(m-1)^3 with m>1 (see also References).

"Supplemento al Periodico di Matematica", Raffaello Giusti Editore (Livorno) - Apr / May, 1913 - p. 99 (Problem 1277, case x=3).

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (10,-36,54,-27).

Cf. A069996, A072590, A027472, A003462.

[(3^n*(n^2-n+1)-1)/2: n in [1..25]]; // Vincenzo Librandi, Aug 19 2013

CoefficientList[Series[(1 + 3 x) / ((1 - x) (1 - 3 x)^3), {x, 0, 25}], x] (* Vincenzo Librandi, Aug 19 2013 *)

a(n) = (3^n*(n^2-n+1)-1)/2 \\ Michel Marcus, Jun 07 2013

a(n) = (3^n*(n^2 - n + 1) - 1)/2.
G.f.: x*(1+3*x)/((1-x)*(1-3*x)^3).
a(n) = 10*a(n-1) - 36*a(n-2) + 54*a(n-3) - 27a(n-4) for n>4.
a(n) = 9*A027472(n+1) + A003462(n) for n>2.
E.g.f.: (1/2)*((1 + 9*x^2)*exp(x) - exp(-x))*exp(2*x). - G. C. Greubel, Aug 24 2016

A367591 Expansion of 1/((1-x) * (1-3*x)^3).

Original entry on oeis.org

1, 10, 64, 334, 1549, 6652, 27064, 105796, 401041, 1483606, 5380840, 19198306, 67559437, 234963352, 808919632, 2760370984, 9346519297, 31429487170, 105039380080, 349114288150, 1154561484781, 3801030845140, 12462203297224, 40705156945324, 132494756301649

Offset: 0

Seiichi Manyama, Nov 24 2023

Index entries for linear recurrences with constant coefficients, signature (10,-36,54,-27).

Partial sums of A027472.

Cf. A055580, A367592.

PARI
```
a(n) = ((2*n^2+4*n+3)*3^(n+1)-1)/8;
```

a(n) = ((2*n^2+4*n+3) * 3^(n+1) - 1)/8.

A126177 Triangle read by rows: T(n,k) is number of hex trees with n edges and k leaves (n >= 1, 1 <= k <= 1 + floor(n/2)).

Original entry on oeis.org

3, 9, 1, 27, 9, 81, 54, 2, 243, 270, 30, 729, 1215, 270, 5, 2187, 5103, 1890, 105, 6561, 20412, 11340, 1260, 14, 19683, 78732, 61236, 11340, 378, 59049, 295245, 306180, 85050, 5670, 42, 177147, 1082565, 1443420, 561330, 62370, 1386, 531441, 3897234

Offset: 1

Emeric Deutsch, Dec 19 2006

A hex tree is a rooted tree where each vertex has 0, 1, or 2 children and, when only one child is present, it is either a left child, or a middle child, or a right child (name due to an obvious bijection with certain tree-like polyhexes; see the Harary-Read paper).
Also number of hex trees with n edges and k-1 nodes of outdegree 2.
Row n has 1 + floor(n/2) terms.
Sum of terms in row n = A002212(n+1).
T(n,1) = 3^n (A000244).
T(n,2) = A027472(n+1).
Sum_{k=1..1+floor(n/2)} k*T(n,k) = A026375(n).

			Triangle starts:
    3;
    9,   1;
   27,   9;
   81,  54,   2;
  243, 270,  30;

F. Harary and R. C. Read, The enumeration of tree-like polyhexes, Proc. Edinburgh Math. Soc. (2) 17 (1970), 1-13.

Cf. A002212, A000244, A027472, A026375.

T:=(n,k)->3^(n-2*k+2)*binomial(2*k-2,k-1)*binomial(n,2*k-2)/k: for n from 1 to 13 do seq(T(n,k),k=1..1+floor(n/2)) od; # yields sequence in triangular form

T(n,k) = 3^(n-2k+2)*binomial(2k-2,k-1)*binomial(n,2k-2)/k. Proof: There are Catalan(k-1) full binary trees with k leaves. Each of them has 2k-2 edges. Additional n-2k+2 edges can be inserted as paths at the existing 2k-1 vertices in 3^(n-2k+2)*binomial(n,2k-2) ways.

G.f.: G=G(t,z) satisfies z^2*G^2-(1-3z-2tz^2)G+tz(3+tz)=0.

cp's OEIS Frontend

A154372 Triangle T(n,k) = (k+1)^(n-k)*binomial(n,k).

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A217629 Triangle, read by rows, where T(n,k) = k!*C(n, k)*3^(n-k) for n>=0, k=0..n.

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A318773 Triangle T(n,k) = 3*T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4), with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

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A094796 Triangle read by rows giving coefficients of polynomials arising in successive differences of central binomial numbers.

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A082149 A transform of C(n,2).

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A082151 A transform of C(n,2).

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A116138 a(n) = 3^n * n*(n + 1).

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A217629 Triangle, read by rows, where T(n,k) = k!C(n, k)3^(n-k) for n>=0, k=0..n.