A008544 -id:A008544 - cp's OEIS frontend

A303486 a(n) = n! * [x^n] 1/(1 - 3*x)^(n/3).

1, 1, 10, 162, 3640, 104720, 3674160, 152152000, 7264216960, 392841187200, 23734494784000, 1584471003315200, 115825295634048000, 9201578813819392000, 789383453851632640000, 72728093032166347776000, 7162140885524461957120000, 750766815289210771251200000

Offset: 0

Ilya Gutkovskiy, Apr 24 2018

nonn

			a(1) = 1;
a(2) = 2*5 = 10;
a(3) = 3*6*9 = 162;
a(4) = 4*7*10*13 = 3640;
a(5) = 5*8*11*14*17 = 104720, etc.

G. C. Greubel, Table of n, a(n) for n = 0..343
Index entries for sequences related to factorial numbers

Column k=3 of A303489.

Cf. A000407, A007559, A008544, A032031, A034000, A034001, A051604, A051605, A051606, A051607, A051608, A051609, A113551, A303487, A303488.

Table[n! SeriesCoefficient[1/(1 - 3 x)^(n/3), {x, 0, n}], {n, 0, 17}]
Table[Product[3 k + n, {k, 0, n - 1}], {n, 0, 17}]
Table[3^n Pochhammer[n/3, n], {n, 0, 17}]

a(n) = Product_{k=0..n-1} (3*k + n).
a(n) = 3^n*Gamma(4*n/3)/Gamma(n/3).
a(n) ~ 2^(8*n/3-1)*n^n/exp(n).

A107092 G.f. A(x) satisfies A(x)^3 = A(x^3) + 3*x.

Original entry on oeis.org

1, 1, -1, 2, -4, 9, -22, 55, -142, 376, -1011, 2758, -7614, 21220, -59630, 168759, -480533, 1375676, -3957075, 11430582, -33144264, 96434321, -281447954, 823734157, -2417092933, 7109265120, -20955593252, 61893804180, -183148075432, 542885589115, -1611809502764, 4792612539375

Offset: 0

Paul D. Hanna, May 11 2005

sign

Self-convolution cube is A107093.

			A(x)^3 = 1 + 3*x + x^3 - x^6 + 2*x^9 - 4*x^12 + 9*x^15 - 22*x^18 +...
A(x^3) = 1 + x^3 - x^6 + 2*x^9 - 4*x^12 + 9*x^15 - 22*x^18+...

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Ira M. Gessel, The Amdeberhan-Konvalinka Conjecture and Symmetric Functions, Séminaire Lotharingien Comb. (2024). See p. 83 of 109.

Cf. A107093, A008544, A352703, A352705, A361763.

{a(n)=local(A=1+x);for(i=1,n,A=(subst(A,x,x^3)+3*x+x*O(x^n))^(1/3)); polcoeff(A,n,x)}

A024395 a(n) = n-th elementary symmetric function of the first n+1 positive integers congruent to 2 mod 3.

Original entry on oeis.org

1, 7, 66, 806, 12164, 219108, 4591600, 109795600, 2951028000, 88084714400, 2891353030400, 103521905491200, 4015191638617600, 167714507921497600, 7506196028811110400, 358368551285791692800, 18180562447078051328000

Offset: 0

Clark Kimberling

Comment by R. J. Mathar, Oct 01 2016 (Start):
The k-th elementary symmetric functions of the integers 2+j*3, j=0..n-1, form a triangle T(n,k), 0<=k<=n, n>=0:
1
1 2
1 7 10
1 15 66 80
1 26 231 806 880
1 40 595 4040 12164 12320
1 57 1275 14155 80844 219108 209440
1 77 2415 39655 363944 1835988 4591600 4188800
1 100 4186 95200 1276009 10206700 46819324 109795600 96342400
This here is the first subdiagonal. The diagonal seems to be A008544. The first columns are A000012, A005449, A024391, A024392. (End)

			From _Gheorghe Coserea_, Dec 24 2015: (Start)
For n=1 we have a(1) = 2*5*(1/2 + 1/5) = 7.
For n=2 we have a(2) = 2*5*8*(1/2 + 1/5 + 1/8) = 66.
For n=3 we have a(3) = 2*5*8*11*(1/2 + 1/5 + 1/8 + 1/11) = 806.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A024216, A225470 (second column).

Table[ (-1)^(n+1)*Sum[(-3)^(n - k) k (-1)^(n - k) StirlingS1[n+1, k + 1], {k, 0, n}], {n, 1, 30}]
Join[{1},Table[Module[{c=NestList[3+#&,2,n+1]},Times@@c*Total[1/c]],{n,0,20}]] (* Harvey P. Dale, Jul 09 2019 *)

n = 16; a = vector(n); a[1] = 7; a[2] = 66;
for (k=2, n-1, a[k+1] = (6*k+7) * a[k] - (3*k+2)^2 * a[k-1]);
print(concat(1,a))  \\ Gheorghe Coserea, Aug 30 2015

E.g.f. (for offset 1): -(1/3)*log(1-3*x)/(1-3*x)^(2/3). - Vladeta Jovovic, Sep 26 2003
For n >= 1, a(n-1) = 3^(n-1)*n!*sum(binomial(k-1/3,k)/(n-k), k = 0..n-1). - Milan Janjic, Dec 14 2008, corrected by Peter Bala, Oct 08 2013
a(n) ~ (n+1)! * 3^n * (log(n) + gamma - Pi*sqrt(3)/6 + 3*log(3)/2) / (n^(1/3)*GAMMA(2/3)), where "GAMMA" is the Gamma function and "gamma" is the Euler-Mascheroni constant (A001620). - Vaclav Kotesovec, Oct 07 2013
a(n+1) = (6*n+7) * a(n) - (3*n+2)^2 * a(n-1). - Gheorghe Coserea, Aug 30 2015
a(n) = A225470(n+1, 1), n >= 0. - Wolfdieter Lang, May 29 2017

Formula (see Mathematica line), correction and more terms from Victor Adamchik (adamchik(AT)cs.cmu.edu), Jul 21 2001

A051608 a(n) = (3*n+8)!!!/8!!!.

Original entry on oeis.org

1, 11, 154, 2618, 52360, 1204280, 31311280, 908027120, 29056867840, 1016990374400, 38645634227200, 1584471003315200, 69716724145868800, 3276686034855833600, 163834301742791680000, 8683217992367959040000, 486260207572605706240000, 28689352246783736668160000

Offset: 0

Wolfdieter Lang

Related to A008544(n+1) ((3*n+2)!!! triple factorials).

Row m=8 of the array A(4; m,n) := ((3*n+m)(!^3))/m(!^3), m >= 0, n >= 0.

G. C. Greubel, Table of n, a(n) for n = 0..377

Cf. A032031, A007559(n+1), A034000(n+1), A034001(n+1), A051604, A051605, A051606, A051607, A051609 (rows m=0..9).

Cf. A008544.

m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(1/(1-3*x)^(11/3))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Aug 15 2018

With[{nn = 30}, CoefficientList[Series[1/(1 - 3*x)^(11/3), {x, 0, nn}], x]*Range[0, nn]!] (* G. C. Greubel, Aug 15 2018 *)

x='x+O('x^30); Vec(serlaplace(1/(1-3*x)^(11/3))) \\ G. C. Greubel, Aug 15 2018

a(n) = ((3*n+8)(!^3))/8(!^3).
E.g.f.: 1/(1-3*x)^(11/3).
Sum_{n>=0} 1/a(n) = 1 + 9*(9*e)^(1/3)*(Gamma(11/3) - Gamma(11/3, 1/3)). - Amiram Eldar, Dec 23 2022

A112333 An invertible triangle of ratios of triple factorials.

Original entry on oeis.org

1, 2, 1, 10, 5, 1, 80, 40, 8, 1, 880, 440, 88, 11, 1, 12320, 6160, 1232, 154, 14, 1, 209440, 104720, 20944, 2618, 238, 17, 1, 4188800, 2094400, 418880, 52360, 4760, 340, 20, 1, 96342400, 48171200, 9634240, 1204280, 109480, 7820, 460, 23, 1, 2504902400

Offset: 0

Paul Barry, Sep 04 2005

First column is A008544. Second column is A034000. Third column is A051605. As a square array read by antidiagonals, columns have e.g.f. (1/(1-3x)^(2/3)) * (1/(1-3x))^k.

			Triangle begins
      1;
      2,    1;
     10,    5,    1;
     80,   40,    8,   1;
    880,  440,   88,  11,  1;
  12320, 6160, 1232, 154, 14, 1;
Inverse triangle A112334 begins
   1;
  -2,  1;
   0, -5,  1;
   0,  0, -8,   1;
   0,  0,  0, -11,   1;
   0,  0,  0,   0, -14,   1;
   0,  0,  0,   0,   0, -17, 1;

nmax:=8: for n from 0 to nmax do for k from 0 to n do if k<=n then T(n, k) := mul(3*k1-1, k1=1..n)/ mul(3*j-1, j=1..k) else T(n, k) := 0: fi: od: od: for n from 0 to nmax do seq(T(n, k), k=0..n) od: seq(seq(T(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jul 04 2011, revised Nov 23 2012

Number triangle T(n, k)=if(k<=n, Product{k=1..n, 3k-1}/Product{j=1..k, 3j-1}, 0); T(n, k)=if(k<=n, 3^(n-k)*(n-1/3)!/(k-1/3)!, 0).

A144275 Lower triangular array called S2hat(-2) related to partition number array A144274.

Original entry on oeis.org

1, 2, 1, 10, 2, 1, 80, 14, 2, 1, 880, 100, 14, 2, 1, 12320, 1140, 108, 14, 2, 1, 209440, 14880, 1180, 108, 14, 2, 1, 4188800, 249280, 15400, 1196, 108, 14, 2, 1, 96342400, 4801280, 255400, 15480, 1196, 108, 14, 2, 1, 2504902400, 108574400, 4888960, 256440, 15512

Offset: 1

Wolfdieter Lang, Oct 09 2008

If in the partition array M32khat(-2)= A144274 entries with the same parts number m are summed one obtains this triangle of numbers S2hat(-2). In the same way the Stirling2 triangle A008277 is obtained from the partition array M_3 = A036040.

The first three columns are A008544, A144277, A144278.

			Triangle begins:
  [1];
  [2,1];
  [10,2,1];
  [80,14,2,1];
  [880,100,14,2,1];
  ...

Wolfdieter Lang, First 10 rows of the array and more.
Wolfdieter Lang, Combinatorial Interpretation of Generalized Stirling Numbers, J. Int. Seqs. Vol. 12 (2009) 09.3.3.

Row sums A144276.
A144270 (S2hat(-1)).
Cf. A004747, A008284, A008544.

a(n,m) = Sum_{q=1..p(n,m)} (Product_{j=1..n} |S2(-2;j,1)|^e(n,m,q,j)) if n>=m>=1, else 0. Here p(n,m) = A008284(n,m), the number of m parts partitions of n and e(n,m,q,j) is the exponent of j in the q-th m part partition of n. |S2(-2,n,1)|= A004747(n,1) = A008544(n-1) = (3*n-4)(!^3) (3-factorials) for n>=2 and 1 if n=1.

A346896 Expansion of e.g.f.: (1-12*x)^(-11/12).

Original entry on oeis.org

1, 11, 253, 8855, 416185, 24554915, 1743398965, 144702114095, 13746700839025, 1470896989775675, 175036741783305325, 22929813173612997575, 3278963283826658653225, 508239308993132091249875, 84875964601853059238729125, 15192797663731697603732513375

Offset: 0

Nikolaos Pantelidis, Aug 06 2021

G. C. Greubel, Table of n, a(n) for n = 0..300

Sequences of the form m^n*Pochhammer((m-1)/m, n): A000007 (m=1), A001147 (m=2), A008544 (m=3), A008545 (m=4), A008546 (m=5), A008543 (m=6), A049209 (m=7), A049210 (m=8), A049211 (m=9), A049212 (m=10), A254322 (m=11), this sequence (m=12).

m:=12; [Round(m^n*Gamma(n +(m-1)/m)/Gamma((m-1)/m)): n in [0..20]]; // G. C. Greubel, Feb 16 2022

CoefficientList[Series[(1-12*x)^(-11/12),{x,0,20}], x] * Range[0, 20]!
FullSimplify[Table[12^n Gamma[n+11/12]/Gamma[11/12],{n,0,15}]] (* Stefano Spezia, Aug 07 2021 *)

m=12; [m^n*rising_factorial((m-1)/m, n) for n in (0..20)] # G. C. Greubel, Feb 16 2022

G.f.: 1/(1-11*x/(1-12*x/(1-23*x/(1-24*x/(1-35*x/(1-36*x/(1-47*x/(1-48*x/(1-59*x/(1-60*x/(1-...))))))))))) (Stieltjes continued fraction).
G.f.: 1/Q(0) where Q(k) = 1 - x*(12*k+11)/(1 - x*(12*k+12)/Q(k+1) ) (continued fraction).
G.f.: 1/(1-11*x-132*x^2/(1-35*x-552*x^2/(1-59*x-1260*x^2/(1-83*x-2256*x^2/(1-107*x-3540*x^2/(1-...)))))) (Jacobi continued fraction).
G.f.: 1/G(0) where G(k) = 1 - x*(24*k+11) - 12*(k+1)*(12*k+11)*x^2/G(k+1) (continued fraction).
a(n) = 12^n*Gamma(n+11/12)/Gamma(11/12). - Stefano Spezia, Aug 07 2021
Sum_{n>=0} 1/a(n) = 1 + (e/12)^(1/12)*(Gamma(11/12) - Gamma(11/12, 1/12)). - Amiram Eldar, Dec 22 2022

A371077 Square array read by ascending antidiagonals: A(n, k) = 3^n*Pochhammer(k/3, n).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 4, 2, 1, 0, 28, 10, 3, 1, 0, 280, 80, 18, 4, 1, 0, 3640, 880, 162, 28, 5, 1, 0, 58240, 12320, 1944, 280, 40, 6, 1, 0, 1106560, 209440, 29160, 3640, 440, 54, 7, 1, 0, 24344320, 4188800, 524880, 58240, 6160, 648, 70, 8, 1

Offset: 0

Werner Schulte and Peter Luschny, Mar 10 2024

			The array starts:
  [0] 1,    1,     1,     1,     1,      1,      1,      1,      1, ...
  [1] 0,    1,     2,     3,     4,      5,      6,      7,      8, ...
  [2] 0,    4,    10,    18,    28,     40,     54,     70,     88, ...
  [3] 0,   28,    80,   162,   280,    440,    648,    910,   1232, ...
  [4] 0,  280,   880,  1944,  3640,   6160,   9720,  14560,  20944, ...
  [5] 0, 3640, 12320, 29160, 58240, 104720, 174960, 276640, 418880, ...
.
Seen as the triangle T(n, k) = A(n - k, k):
  [0] 1;
  [1] 0,       1;
  [2] 0,       1,      1;
  [3] 0,       4,      2,     1;
  [4] 0,      28,     10,     3,    1;
  [5] 0,     280,     80,    18,    4,   1;
  [6] 0,    3640,    880,   162,   28,   5,  1;
  [7] 0,   58240,  12320,  1944,  280,  40,  6, 1;
  [8] 0, 1106560, 209440, 29160, 3640, 440, 54, 7, 1;
.
Illustrating the LU decomposition of A:
    / 1                \   / 1 1 1 1 1 ... \   / 1   1   1    1    1 ... \
    | 0   1            |   |   1 2 3 4 ... |   | 0   1   2    3    4 ... |
    | 0   4   2        | * |     1 3 6 ... | = | 0   4  10   18   28 ... |
    | 0  28  24   6    |   |       1 4 ... |   | 0  28  80  162  280 ... |
    | 0 280 320 144 24 |   |         1 ... |   | 0 280 880 1944 3640 ... |
    | . . .            |   | . . .         |   | . . .                   |

Paolo Xausa, Table of n, a(n) for n = 0..11324 (first 150 antidiagonals, flattened).

Family m^n*Pochhammer(k/m, n): A094587 (m=1), A370419 (m=2), this sequence (m=3), A370915 (m=4).
Rows: A000012, A001477, A028552.
Columns: A000007, A007559, A008544, A032031, A007559 (shifted), A034000, A034001, A051604, A051605, A051606, A051607, A051608, A051609.
Cf. A303486 (main diagonal), A371079 (row sums of triangle), A371076, A371080.

A := (n, k) -> 3^n*pochhammer(k/3, n):
A := (n, k) -> local j; mul(3*j + k, j = 0..n-1):
# Read by antidiagonals:
T := (n, k) -> A(n - k, k): seq(seq(T(n, k), k = 0..n), n = 0..9);
seq(lprint([n], seq(T(n, k), k = 0..n)), n = 0..9);
# Using the generating polynomials of the rows:
P := (n, x) -> local k; add(Stirling1(n, k)*(-3)^(n - k)*x^k, k=0..n):
seq(lprint([n], seq(P(n, k), k = 0..9)), n = 0..5);
# Using the exponential generating functions of the columns:
EGFcol := proc(k, len) local egf, ser, n; egf := (1 - 3*x)^(-k/3);
ser := series(egf, x, len+2): seq(n!*coeff(ser, x, n), n = 0..len) end:
seq(lprint([k], EGFcol(k, 8)), k = 0..6);
# As a matrix product:
with(LinearAlgebra):
L := Matrix(7, 7, (n, k) -> A371076(n - 1,  k - 1)):
U := Matrix(7, 7, (n, k) -> binomial(n - 1, k - 1)):
MatrixMatrixMultiply(L, Transpose(U));

Table[3^(n-k)*Pochhammer[k/3, n-k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Mar 14 2024 *)

def A(n, k): return 3**n * rising_factorial(k/3, n)
def A(n, k): return (-3)**n * falling_factorial(-k/3, n)

A(n, k) = Product_{j=0..n-1} (3*j + k).
A(n, k) = A(n+1, k-3) / (k - 3) for k > 3.
A(n, k) = Sum_{j=0..n} Stirling1(n, j)*(-3)^(n - j)* k^j.
A(n, k) = k! * [x^k] (exp(x) * p(n, x)), where p(n, x) are the row polynomials of A371080.
E.g.f. of column k: (1 - 3*t)^(-k/3).
E.g.f. of row n: exp(x) * (Sum_{k=0..n} A371076(n, k) * x^k / (k!)).
Sum_{n>=0, k>=0} A(n, k) * x^k * t^n / (n!) = 1/(1 - x/(1 - 3*t)^(1/3)).
Sum_{n>=0, k>=0} A(n, k) * x^k * t^n /(n! * k!) = exp(x/(1 - 3*t)^(1/3)).
The LU decomposition of this array is given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L = A371076, i.e., A(n, k) = Sum_{i=0..k} A371076(n, i) * binomial(k, i).

A290596 Triangle read by rows. A generalization of unsigned Lah numbers, called L[3,1].

Original entry on oeis.org

1, 2, 1, 10, 10, 1, 80, 120, 24, 1, 880, 1760, 528, 44, 1, 12320, 30800, 12320, 1540, 70, 1, 209440, 628320, 314160, 52360, 3570, 102, 1, 4188800, 14660800, 8796480, 1832600, 166600, 7140, 140, 1, 96342400, 385369600, 269758720, 67439680, 7663600, 437920, 12880, 184, 1, 2504902400, 11272060800, 9017648640, 2630147520, 358656480, 25618320, 1004640, 21528, 234, 1, 72642169600, 363210848000, 326889763200, 108963254400, 17335063200, 1485862560, 72836400, 2081040, 33930, 290, 1

Offset: 0

Wolfdieter Lang, Sep 13 2017

For the general L[d,a] triangles see A286724, also for references.
This is the generalized signless Lah number triangle L[3,1], the Sheffer triangle ((1 - 3*t)^(-2/3), t/(1 - 3*t)). It is defined as transition matrix
risefac[3,1](x, n) = Sum_{m=0..n} L[3,1](n, m)*fallfac[3,1](x, m), where risefac[3,1](x, n):= Product_{0..n-1} (x  + (1 + 3*j)) for n >= 1 and risefac[3,1](x, 0) := 1, and  fallfac[3,1](x, n):= Product_{0..n-1} (x - (1 + 3*j)) for n >= 1 and fallfac[3,1](x, 0) := 1.
In matrix notation: L[3,1] = S1phat[3,1]*S2hat[3,1] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations A286718 and A111577 (but here with offsets 0), respectively.
The a- and z-sequences for this Sheffer matrix has e.g.f.s Ea(t) = 1 + 3*t and (Ez(t) = (1 + 3*t)*(1 - (1 + 3*t)^(-2/3))/t, respectively. That is, a = {1, 3, repeat(0)} and z(n) = A290597(n)/A038500(n+1). For the proof see the second W. Lang link. See also a W. Lang link under A006232 for Sheffer a- and z-sequences with references (in the Riordan case).
The inverse matrix T^(-1) = L^(-1)[3,1] is Sheffer ((1 + 3*t)^(-2/3), t/(1 + 3*t)). This means that  T^(-1)(n, m) = (-1)^(n-m)*T(n, m).
fallfac[3,1](x, n) = Sum_{m=0..n} (-1)^(n-m)*T(n, m)*risefac[3,1](x, m), n >= 0.

			The triangle T(n, m) begins:
n\m         0         1         2        3       4      5     6   7 8  ...
0:          1
1:          2         1
2:         10        10         1
3:         80       120        24        1
4:        880      1760       528       44       1
5:      12320     30800     12320     1540      70      1
6:     209440    628320    314160    52360    3570    102     1
7:    4188800  14660800   8796480  1832600  166600   7140   140   1
8:   96342400 385369600 269758720 67439680 7663600 437920 12880 184 1
...
n = 9: 2504902400 11272060800 9017648640 2630147520 358656480 25618320 1004640 21528 234 1,
n = 10: 72642169600 363210848000 326889763200 108963254400 17335063200 1485862560 72836400 2081040 33930 290 1.
...
Recurrence from a-sequence:  T(4, 2) = 2*T(3, 1) + 3*4*T(3, 2) = 2*120 + 12*24 = 528.
Recurrence from z-sequence: T(4, 0) = 4*(z(0)*T(3, 0) + z(1)*T(3, 1) + z(2)*T(3, 2) + z(3)*T(3, 3)) = 4*(2*80 + 1*120 - (10/3)*24 + 20*1) = 880.
Four term recurrence: T(4, 2) = T(3, 1) + 2*10*T(3, 2) - 3*3*8*T(2, 2) =  120 + 20*24 - 72*1 = 528.
Meixner type identity for n = 2: (D_x - 3*(D_x)^2)*(10 + 10*x + x^2 ) = (10 + 2*x) - 3*2 = 2*(2 + x).
Sheffer recurrence for R(3, x): [(2 + x) + 6*(1 + x)*D_x + 9*x*(D_x)^2] (10 + 10*x + x^2) = (2 + x)*(10 + 10*x + x^2) + 6*(1 + x)*(10 +2*x) + 9*2*x = 80 + 120*x + 24*x^2 + x^3 = R(3, x).
Boas-Buck recurrence for column m = 2 with n = 4: T(4, 2) = (4!*8/2)*(1*24/3! + 3*1/2!) = 528.

Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017.
Wolfdieter Lang, Note on a- and z-sequences of Sheffer number triangles for certain generalized Lah numbers.

Cf. A008544 (column m=0), A038500, A111577, A271703 L[1,0], A286718, A286724 L[2,1], A290597, A290598 L[3,2].

T(n, m) = L[3,1](n,m) = Sum_{k=m..n} A286718(n, k)*A111577(k+1, m+1), 0 <= m <= n.
E.g.f. of row polynomials R(n, x) := Sum_{m=0..n} T(n, m)*x^m:
(1 - 3*t)^(-2/3)*exp(x*t/(1 - 3*t)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - 3*t)^(-2/3)*(t/(1 - 3*t))^m/m!, m >= 0.
Three term recurrence for column entries m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 3*n*T(n-1, m) with T(n, m) = 0 for n < m, and for the column m = 0: T(n, 0) = n*Sum_{j=0}^(n-1) z(j)*T(n-1, j), from the a-sequence {1, 3 repeat(0)} and the z-sequence given above.
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(3*n - 2)*T(n-1, m) - 3*(n-1)*(3*n - 4)*T(n-2, m), n >= m >= 0, with T(0, 0) = 1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
Meixner type identity for (monic) row polynomials: (D_x/(1 + 3*D_x)) * R(n, x) = n*R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx. That is, Sum_{k=0..n-1} (-3)^k*(D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1.
General recurrence for Sheffer row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
R(n, x) = [(2 + x)*1 + 6*(1 + x)*D_x + 3^2*x*(D_x)^2]*R(n-1, x), n >= 1, with R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment in A286724 with references): T(n, m) = (n!/(n-m))*(2 + 3*m)*Sum_{p=0..n-1-m} 3^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, with input T(m, m) = 1.

A131178 Non-plane increasing unary binary (0-1-2) trees where the nodes of outdegree 1 come in 2 colors.

Original entry on oeis.org

1, 2, 5, 16, 64, 308, 1730, 11104, 80176, 643232, 5676560, 54650176, 569980384, 6401959328, 77042282000, 988949446144, 13488013248256, 194780492544512, 2969094574403840, 47640794742439936, 802644553810683904, 14166772337295285248, 261410917571703825920

Offset: 1

Wenjin Woan, Oct 31 2007

A labeled tree of size n is a rooted tree on n nodes that are labeled by distinct integers from the set {1,...,n}. An increasing tree is a labeled tree such that the sequence of labels along any branch starting at the root is increasing. Thus the root of an increasing tree will be labeled 1. In unary binary trees (sometimes called 0-1-2 trees) the outdegree of a node is either 0, 1 or 2. Here we are counting non-plane (where the subtrees stemming from a node are not ordered between themselves) increasing unary binary trees where the nodes of outdegree 1 come in two colors. An example is given below. - Peter Bala, Sep 01 2011
The number of plane increasing 0-1-2 trees on n nodes, where the nodes of outdegree 1 come in two colors, is equal to n!. Other examples of sequences counting increasing trees include A000111, A000670, A008544, A008545, A029768 and A080635. - Peter Bala, Sep 01 2011
Number of plane increasing 0-1-2 trees, where the nodes of outdegree 1 come in 2 colors, avoiding pattern T213. See A278679 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016

			G.f. = x + 2*x^2 + 5*x^3 + 16*x^4 + 64*x^5 + 308*x^6 + 1730*x^7 + 11104*x^8 + ...
a(3) = 5: Denoting the two types of node of outdegree 1 by the letters a or b, the 5 possible trees are
.
.  1a    1b    1a    1b      1
.  |     |     |     |      / \
.  2a    2b    2b    2a    2   3
.  |     |     |     |
.  3     3     3     3
- _Peter Bala_, Sep 01 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..100
Jean-Luc Baril, Sergey Kirgizov, Vincent Vajnovszki, Patterns in treeshelves, arXiv:1611.07793 [cs.DM], 2016.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48.
Lapo Cioni, Luca Ferrari, and Corentin Henriet. A direct bijection between two-stack sortable permutations and fighting fish, Euro. Conf. Comb., Graph Theory Appl. (2023) No. 12, 283-289.
D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA], 2005.

Cf. A000111, A000670, A008544, A008545, A029768, A080635, A278679

E:=  (2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x)):
S:= map(simplify,series(E,x,101)):
seq(coeff(S,x,j)*j!, j=1..100); # Robert Israel, Nov 23 2016

max = 25; f[x_] := (2*(Exp[Sqrt[2]*x] - 1))/((2 + Sqrt[2]) - (2 - Sqrt[2])*Exp[Sqrt[2]*x]); Drop[ Simplify[ CoefficientList[ Series[f[x], {x, 0, max}], x]*Range[0, max]!], 1] (* Jean-François Alcover, Oct 05 2011 *)

x='x+O('x^66); /* that many terms */
default(realprecision,1000); /* working with floats here */
egf=(2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x));
round(Vec(serlaplace(egf))) /* show terms */
/* Joerg Arndt, Sep 01 2011 */

/* the following program should be preferred. */
Vec( serlaplace( serreverse( intformal( 1/(1+2*x+1/2*x^2) + O(x^66) ) ) ) )
\\ Joerg Arndt, Mar 01 2014

{a(n) = if( n<1, 0, n! * polcoeff( 2 / (-2 + quadgen(8) * (-1 + 2 / (1 - exp(-quadgen(8) * x + x * O(x^n))))), n))};

E.g.f.: A(x) = (2*(exp(sqrt(2)*x)-1)) / ((2+sqrt(2))-(2-sqrt(2))*exp(sqrt(2)*x)) = x+2*x^2/2!+5*x^3/3!+16*x^4/4!+64*x^5/5!+....
From Peter Bala, Sep 01 2011: (Start)
The generating function A(x) satisfies the autonomous differential equation A' = 1+2*A+1/2*A^2 with A(0) = 0. It follows that the inverse function A(x)^-1 may be expressed as an integral A(x)^-1 = int {t = 0..x} 1/(1+2*t+1/2*t^2).
Applying [Dominici, Theorem 4.1] to invert the integral gives the following method for calculating the terms of the sequence: let f(x) = 1+2*x+1/2*x^2. Let D be the operator f(x)*d/dx. Then a(n) = D^n(f(x)) evaluated at x = 0. Compare with A000111(n+1) = D^n(1+x+x^2/2!) evaluated at x = 0.
(End)
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - m*x^2*(k+1)*(2*k+1)/( 1 - 2*x*(2*k+2) - m*x^2*(k+1)*(2*k+3)/Q(k+1) ) and m=1; (continued fraction). - Sergei N. Gladkovskii, Sep 24 2013
G.f.: 1/Q(0), where Q(k) = 1 - 2*x*(k+1) - 1/2*x^2*(k+1)*(k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 02 2013
a(n) ~ n! * 2^((n+3)/2) / log(3+2*sqrt(2))^(n+1). - Vaclav Kotesovec, Oct 08 2013
G.f.: conjecture: T(0)/(1-2*x) -1, where T(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 2*(1-2*x*(k+1))*(1-2*x*(k+2))/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2013
E.g.f.: x/(T(0)-x), where T(k) = 4*k + 1 + x^2/(8*k+6 + x^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2013

Terms >= 80176 from Peter Bala, Sep 01 2011

Changed offset to 1 to agree with name and example. - Michael Somos, Nov 23 2016

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