A201368 -id:A201368 - cp's OEIS frontend

A123125 Triangle of Eulerian numbers T(n,k), 0 <= k <= n, read by rows.

1, 0, 1, 0, 1, 1, 0, 1, 4, 1, 0, 1, 11, 11, 1, 0, 1, 26, 66, 26, 1, 0, 1, 57, 302, 302, 57, 1, 0, 1, 120, 1191, 2416, 1191, 120, 1, 0, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 0, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 0, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1

Offset: 0

Philippe Deléham, Sep 30 2006

The beginning of this sequence does not quite agree with the usual version, which is A173018. - N. J. A. Sloane, Nov 21 2010
Each row of A123125 is the reverse of the corresponding row in A173018. - Michael Somos, Mar 17 2011
A008292 (subtriangle for k>=1 and n>=1) is the main entry for these numbers.
Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,...] DELTA [1,0,2,0,3,0,4,0,5,0,6,...] where DELTA is the operator defined in A084938.
Row sums are the factorials. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008
If the initial zero column is deleted, the result is A008292. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008
This result gives an alternative method of calculating the Eulerian numbers by an Umbral Calculus expansion from Comtet. - Roger L. Bagula, Nov 21 2009
This function seems to be equivalent to the PolyLog expansion. - Roger L. Bagula, Nov 21 2009
A raising operator formed from the e.g.f. of this entry is the generator of a sequence of polynomials p(n,x;t) defined in A046802 that specialize to those for A119879 as p(n,x;-1), A007318 as p(n,x;0), A073107 as p(n,x;1), and A046802 as p(n,0;t). See Copeland link for more associations. - Tom Copeland, Oct 20 2015
The Eulerian numbers in this setup count the permutation trees of power n and width k (see the Luschny link). For the associated combinatorial statistic over permutations see the Sage program below and the example section. - Peter Luschny, Dec 09 2015 [See Elder et al. link. Peter Luschny, Jul 13 2022]
From Wolfdieter Lang, Apr 03 2017: (Start)
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k are the numerator polynomials of the o.g.f. G(n, x) of n-powers {m^n}_{m>=0} (with 0^0 = 1): G(n, x) = R(n, x)/(1-x)^(n+1). See the Aug 14 2008 formula, where f(x,n) = R(n, x). The e.g.f. of R(n, t) is given in Copeland's Oct 14 2015 formula below.
The first nine column sequences are A000007, A000012, A000295, A000460, A000498, A000505, A000514, A001243, A001244. (End)
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of this entry, A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
Let b(n) = (1/(n+1))*Sum_{k=0..n-1} (-1)^(n-k+1)*T(n, k+1) / binomial(n, k+1). Then b(n) = Bernoulli(n, 1) = -n*Zeta(1 - n) = Integral_{x=0..1} F_n(x) for n >= 1. Here F_n(x) are the signed Fubini polynomials (A278075). (See also Rzadkowski and Urlinska, example 1.) - Peter Luschny, Feb 15 2021
Patrick J. Burchell (see link) describes the following method: To get the k-th row of the triangle write the nonnegative integers with a fixed exponent k as a sequence, 0^k, 1^k, 2^k, ..., and then apply the first differences to them k + 1 times. - Peter Luschny, Apr 02 2023

			The triangle T(n, k) begins:
  n\k 0 1    2     3      4       5       6      7     8    9 10...
  0:  1
  1:  0 1
  2:  0 1    1
  3:  0 1    4     1
  4:  0 1   11    11      1
  5:  0 1   26    66     26       1
  6:  0 1   57   302    302      57       1
  7:  0 1  120  1191   2416    1191     120      1
  8:  0 1  247  4293  15619   15619    4293    247     1
  9:  0 1  502 14608  88234  156190   88234  14608   502    1
 10:  0 1 1013 47840 455192 1310354 1310354 455192 47840 1013  1
...  Reformatted. - _Wolfdieter Lang_, Feb 14 2015
------------------------------------------------------------------
The width statistic over permutations, n=4.
  [1, 2, 3, 4] => 3; [1, 2, 4, 3] => 2; [1, 3, 2, 4] => 2; [1, 3, 4, 2] => 2;
  [1, 4, 2, 3] => 2; [1, 4, 3, 2] => 1; [2, 1, 3, 4] => 3; [2, 1, 4, 3] => 2;
  [2, 3, 1, 4] => 2; [2, 3, 4, 1] => 3; [2, 4, 1, 3] => 2; [2, 4, 3, 1] => 2;
  [3, 1, 2, 4] => 3; [3, 1, 4, 2] => 3; [3, 2, 1, 4] => 2; [3, 2, 4, 1] => 3;
  [3, 4, 1, 2] => 3; [3, 4, 2, 1] => 2; [4, 1, 2, 3] => 4; [4, 1, 3, 2] => 3;
  [4, 2, 1, 3] => 3; [4, 2, 3, 1] => 3; [4, 3, 1, 2] => 3; [4, 3, 2, 1] => 2;
Gives row(4) = [0, 1, 11, 11, 1]. - _Peter Luschny_, Dec 09 2015
------------------------------------------------------------------
From _Wolfdieter Lang_, Apr 03 2017: (Start)
Recurrence: T(5, 3) = (6-3)*T(4, 2) + 3*T(4, 3) = 3*11 + 3*11 = 66.
O.g.f. column k=2: (x/(1 - 2*x))*E_x*(x/(1-x)) = (x/(1-x))^2/(1-2*x).
E.g.f. column k=2: A(2, x) = x*A(1, x) + x*E(1, x) = x*1 + x*(exp(x)-1) = x*exp(x), hence E(2, x) = (1 + int(x*exp(-x),x ))*exp(2*x) = exp(x)*(exp(x) - (1+x)). See A000295. (End)

L. Comtet, Advanced Combinatorics, Reidel, Holland, 1978, page 245. [Roger L. Bagula, Nov 21 2009]
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed.; Addison-Wesley, 1994, p. 268, Row reversed table 268. - Wolfdieter Lang, Apr 03 2017
Douglas C. Montgomery and Lynwood A. Johnson, Forecasting and Time Series Analysis, MaGraw-Hill, New York, 1976, page 91. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Paul Barry, Eulerian polynomials as moments, via exponential Riordan arrays, arXiv preprint arXiv:1105.3043 [math.CO], 2011, J. Int. Seq. 14 (2011) # 11.9.5
Paul Barry, Generalized Stirling Numbers, Exponential Riordan Arrays, and Toda Chain Equations, Journal of Integer Sequences, 17 (2014), #14.2.3.
Paul Barry, On a transformation of Riordan moment sequences, arXiv:1802.03443 [math.CO], 2018.
Paul Barry, Generalized Eulerian Triangles and Some Special Production Matrices, arXiv:1803.10297 [math.CO], 2018.
V. Batyrev and M. Blume, The functor of toric varieties associated with Weyl chambers and Losev-Manin moduli spaces, p. 11, arXiv:/0911.3607 [math.AG], 2009. [_Tom Copeland_, Oct 16 2015]
Anna Borowiec and Wojciech Mlotkowski, New Eulerian numbers of type D, arXiv:1509.03758 [math.CO], 2015.
Patrick J. Burchell, A Generalisation of Ramanujan's (back of the envelope) Method for Divergent Series, arXiv:2303.14045 math.NT, 2023.
A. Cohen, Eulerian polynomials of spherical type, Münster J. of Math. 1 (2008). [_Tom Copeland_, Oct 16 2015]
Tom Copeland, The Elliptic Lie Triad: Ricatti and KdV Equations, Infinigens, and Elliptic Genera
Jennifer Elder, Nadia Lafrenière, Erin McNicholas, Jessica Striker and Amanda Welch, Homomesies on permutations -- an analysis of maps and statistics in the FindStat database, arXiv:2206.13409 [math.CO], 2022-2023. (Def. 4.20 and Prop. 4.22.)
FindStat - Combinatorial Statistic Finder, The number of descents of a permutation.
F. Hirzebruch, Eulerian polynomials, Münster J. of Math. 1 (2008), pp. 9-12.
P. Hitczenko and S. Janson, Weighted random staircase tableaux, arXiv preprint arXiv:1212.5498 [math.CO], 2012.
Hsien-Kuei Hwang, Hua-Huai Chern, and Guan-Huei Duh, An asymptotic distribution theory for Eulerian recurrences with applications, arXiv:1807.01412 [math.CO], 2018.
Svante Janson, Euler-Frobenius numbers and rounding, arXiv preprint arXiv:1305.3512 [math.PR], 2013.
Katarzyna Kril and Wojciech Mlotkowski, Permutations of Type B with Fixed Number of Descents and Minus Signs, Volume 26(1) of The Electronic Journal of Combinatorics, 2019.
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli numbers, arXiv:1707.04451 [math.NT], 2017.
Huyile Liang, Yanni Pei, and Yi Wang, Analytic combinatorics of coordination numbers of cubic lattices, arXiv:2302.11856 [math.CO], 2023. See p. 22.
A. Losev and Y. Manin, New moduli spaces of pointed curves and pencils of flat connections, arXiv preprint arXiv:math/0001003 [math.AG], 2000 (p. 8). - _Tom Copeland_, Oct 16 2015
Peter Luschny, Permutation Trees
G. Rzadkowski and M. Urlinska, A Generalization of the Eulerian Numbers, arXiv:1612.06635 [math.CO], 2016.

See A008292 (subtriangle for k>=1 and n>=1), which is the main entry for these numbers. Another version has the zeros at the ends of the rows, as in Concrete Mathematics: see A173018.
Cf. A007318, A130850, A028246, A046802, A009006, A019538, A090582, A119879, A248727.
T(2n,n) gives A180056.

a123125 n k = a123125_tabl !! n !! k
a123125_row n = a123125_tabl !! n
a123125_tabl = [1] : zipWith (:) [0, 0 ..] a008292_tabl
-- Reinhard Zumkeller, Nov 06 2013

gf := 1/(1 - t*exp(x)): ser := series(gf, x, 12):
cx := n -> (-1)^(n + 1)*factor(n!*coeff(ser, x, n)*(t - 1)^(n + 1)):
seq(print(seq(coeff(cx(n), t, k), k = 0..n)), n = 0..9); # Peter Luschny, Feb 11 2021
A123125 := proc(n, k) option remember; if k = n then 1 elif k <= 0 or k > n then 0 else k*procname(n-1, k) + (n-k+1)*procname(n-1, k-1) fi end:
seq(print(seq(A123125(n, k), k=0..n)), n=0..10); # Peter Luschny, Mar 28 2021
# Alternative (Patrick J. Burchell):
t := a -> Statistics:-Difference([0, a]): Trow := k -> (t@@(k+1))([seq(n^k, n = 0..k)]):
seq(print(Trow(n)), n = 0..6); # Peter Luschny, Apr 02 2023

f[x_, n_] := f[x, n] = (1 - x)^(n + 1)*Sum[k^n*x^k, {k, 0, Infinity}];
Table[CoefficientList[f[x, n], x], {n,0,9}] // Flatten (* Roger L. Bagula, Aug 14 2008 *)
t[n_ /; n >= 0, 0] = 1; t[n_, k_] /; k<0 || k>n = 0; t[n_, k_] := t[n, k] = (n-k) t[n-1, k-1] + (k+1) t[n-1, k]; T[n_, k_] := t[n, n-k];
Table[T[n, k], {n,0,10}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 26 2019 *)
A123125[n_, k_] := Sum[(-1)^j*(n - j - k + 1)^n * Binomial[n + 1, j], {j, 0, n - k}];
Table[A123125[n, k], {n, 0, 9}, {k, 0, n}] // TableForm  (* Peter Luschny, Aug 12 2022 *)

from math import isqrt, comb
def A123125(n):
    a = (m:=isqrt(k:=n+1<<1))+(k>m*(m+1))
    b = comb(a+1,2)-n
    return sum(-(b-j)**(a-1)*comb(a,j) if j&1 else (b-j)**(a-1)*comb(a,j) for j in range(b)) # Chai Wah Wu, Nov 13 2024

def statistic_eulerian(pi):
    if not pi: return 0
    h, i, branch, next = 0, len(pi), [0], pi[0]
    while True:
        while next < branch[len(branch)-1]:
            del(branch[len(branch)-1])
        current = 0
        h += 1
        while next > current:
            i -= 1
            if i == 0: return h
            branch.append(next)
            current, next = next, pi[i]
def A123125_row(n):
    L = [0]*(n+1)
    for p in Permutations(n):
        L[statistic_eulerian(p)] += 1
    return L
[A123125_row(n) for n in range(7)] # Peter Luschny, Dec 09 2015

Sum_{k=0..n} T(n,k) = n! = A000142(n).
Sum_{k=0..n} 2^k*T(n,k) = A000629(n).
Sum_{k=0..n} 3^k*T(n,k) = abs(A009362(n+1)).
Sum_{k=0..n} 2^(n-k)*T(n,k) = A000670(n).
Sum_{k=0..n} T(n,k)*3^(n-k) = A122704(n). - Philippe Deléham, Nov 07 2007
G.f.: f(x,n) = (1 - x)^(n + 1)*Sum_{k>=0} k^n*x^k. - Roger L. Bagula and Gary W. Adamson, Aug 14 2008. f is not the g.f. of the triangle, it is the polynomial of row n. See an Apr 03 2017 comment above - Wolfdieter Lang, Apr 03 2017
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000629(n), A123227(n), A201355(n), A201368(n) for x = 0, 1, 2, 3, 4, 5 respectively. - Philippe Deléham, Dec 01 2011
E.g.f. (1-t)/(1-t*exp((1-t)x)). A123125 * A007318 = A130850 = unsigned A075263, related to reversed A028246. A007318 * A123125 = A046802. Evaluating the row polynomials at -1, giving the alternating-sign row sum, generates A009006. - Tom Copeland, Oct 14 2015
From Wolfdieter Lang, Apr 03 2017: (Start)
T(n, k) = A173018(n, n-k), 0 <= k <= n. Row reversed Euler's triangle. See Graham et al., p. 268.
Recurrence (from A173018): T(n, 0) = 1 if n=0 else 0; T(n, k) = 0 if n < k and T(n, k) = (n+1-k)*T(n-1, k-1) + k*T(n-1, k) else.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n-j, k-j)*S2(n, j)*j!, 0 <= k <= n, else 0. For S2(n, k)*k! see A131689.
The recurrence for the o.g.f. of the sequence of column k is
  G(k, x) = (x/(1 - k*x))*(E_x - (k-2))*G(k-1, x), with the Euler operator E_x  = x*d_x, for k >= 1, with G(0, x) = 1. (Proof from the recurrence of T(n, k)).
The e.g.f of the sequence of column k is found from E(k, x) = (1 + int(A(k, x),x)*exp(-k*x))*exp(k*x), k >= 1, with the recurrence
  A(k, x) = x*A(k-1, x) +(1 + (1-k)*(1-x))*E(k-1, x) for k >= 1, with  A(0,x)= 0. (Proof from the recurrence of T(n, k)). (End)
T(n, k) = Sum_{j=0..n-k} (-1)^j*(n-j-k+1)^n*binomial(n + 1, j). - Peter Luschny, Aug 12 2022
G.f.: Sum_{m >= 0} x^m/(1/(1-x)-m*t). - Mamuka Jibladze, Mar 12 2025

A201365 Expansion of e.g.f. exp(x) / (5 - 4*exp(x)).

Original entry on oeis.org

1, 5, 45, 605, 10845, 243005, 6534045, 204972605, 7348546845, 296387331005, 13282361478045, 654762261324605, 35211177242722845, 2051349014835939005, 128701394409842982045, 8651475271312083756605, 620334325261670875138845, 47259638324026516284867005

Offset: 0

Paul D. Hanna, Nov 30 2011

			E.g.f.: E(x) = 1 + 5*x + 45*x^2/2! + 605*x^3/3! + 10845*x^4/4! + 243005*x^5/5! + ...
O.g.f.: A(x) = 1 + 5*x + 45*x^2 + 605*x^3 + 10845*x^4 + 243005*x^5 + ...
where A(x) = 1 + 5*x/(1+x) + 2!*5^2*x^2/((1+x)*(1+2*x)) + 3!*5^3*x^3/((1+x)*(1+2*x)*(1+3*x)) + 4!*5^4*x^4/((1+x)*(1+2*x)*(1+3*x)*(1+4*x)) + ...

G. C. Greubel, Table of n, a(n) for n = 0..358

Cf. A027882, A032183, A050353.
Cf. A201366, A201367, A201368.
Cf. A000629, A201339, A201354.

R:=PowerSeriesRing(Rationals(), 20); Coefficients(R!(Laplace( 1/(5*Exp(-x) -4) ))); // G. C. Greubel, Jun 08 2020

seq(coeff(series(1/(5*exp(-x) - 4), x, n+1)*n!, x, n), n = 0..20); # G. C. Greubel, Jun 08 2020

Table[Sum[(-1)^(n-k)*5^k*StirlingS2[n,k]*k!,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 13 2013 *)
With[{nn=20},CoefficientList[Series[Exp[x]/(5-4Exp[x]),{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Jul 09 2015 *)
a[n_]:= If[n<0, 0, PolyLog[ -n, 4/5]/4]; (* Michael Somos, Apr 27 2019 *)

{a(n)=n!*polcoeff(exp(x+x*O(x^n))/(5 - 4*exp(x+x*O(x^n))), n)}

{a(n)=polcoeff(sum(m=0, n, 5^m*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)}

{a(n)=sum(k=0, n, (-1)^(n-k)*5^k*stirling(n, k, 2)*k!)}

[sum( (-1)^(n-j)*5^j*factorial(j)*stirling_number2(n,j) for j in (0..n)) for n in (0..20)] # G. C. Greubel, Jun 08 2020

O.g.f.: A(x) = Sum_{n>=0} n! * 5^n*x^n / Product_{k=0..n} (1+k*x).
O.g.f.: A(x) = 1/(1 - 5*x/(1-4*x/(1 - 10*x/(1-8*x/(1 - 15*x/(1-12*x/(1 - 20*x/(1-16*x/(1 - 25*x/(1-20*x/(1 - ...))))))))))), a continued fraction.
a(n) = Sum_{k=0..n} (-1)^(n-k) * 5^k * Stirling2(n,k) * k!.
a(n) = Sum_{k=0..n} A123125(n,k)*5^k*4^(n-k). - Philippe Deléham, Nov 30 2011
a(n) ~ n! / (4*(log(5/4))^(n+1)) . - Vaclav Kotesovec, Jun 13 2013
a(n) = log(5/4) * Integral_{x = 0..oo} (ceiling(x))^n * (5/4)^(-x) dx. - Peter Bala, Feb 14 2015
a(n) = (1/4) Sum_{k>=1} (4/5)^k * n^k. - Michael Somos, Apr 27 2019
a(n) = 1 + 4 * Sum_{k=0..n-1} binomial(n,k) * a(k). - Ilya Gutkovskiy, Jun 08 2020
From Seiichi Manyama, Nov 15 2023: (Start)
a(0) = 1; a(n) = -5*Sum_{k=1..n} (-1)^k * binomial(n,k) * a(n-k).
a(0) = 1; a(n) = 5*a(n-1) + 4*Sum_{k=1..n-1} binomial(n-1,k) * a(n-k). (End)
a(n) = (5/4)*A094417(n) - (1/4)*0^n. - Seiichi Manyama, Dec 21 2023

A201367 E.g.f.: 3exp(3x) / (5 - 2exp(3x)).

Original entry on oeis.org

1, 5, 35, 345, 4515, 73905, 1451835, 33273945, 871529715, 25681042305, 840815302635, 30281769805545, 1189735610250915, 50638609760802705, 2321120945531697435, 113992686944812385145, 5971520591679167948115, 332369999588147180115105, 19587647624733292373756235

Offset: 0

Paul D. Hanna, Nov 30 2011

			E.g.f.: E(x) = 1 + 5*x + 35*x^2/2! + 345*x^3/3! + 4515*x^4/4! + 73905*x^5/5! + ...
O.g.f.: A(x) = 1 + 5*x + 35*x^2 + 345*x^3 + 4515*x^4 + 73905*x^5 + ...
where A(x) = 1 + 5*x/(1+3*x) + 2!*5^2*x^2/((1+3*x)*(1+6*x)) + 3!*5^3*x^3/((1+3*x)*(1+6*x)*(1+9*x)) + 4!*5^4*x^4/((1+3*x)*(1+6*x)*(1+9*x)*(1+12*x)) + ...

Robert Israel, Table of n, a(n) for n = 0..250

Cf. A201365, A201366, A201368.

S:= series(3*exp(3*x)/(5-2*exp(3*x)),x,51):
seq(coeff(S,x,n)*n!,n=0..50); # Robert Israel, Nov 18 2019

Table[Sum[(-3)^(n-k)*5^k*StirlingS2[n,k]*k!,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 13 2013 *)
With[{nn=20},CoefficientList[Series[(3*Exp[3x])/(5-2*Exp[3x]),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Sep 07 2024 *)

{a(n)=n!*polcoeff(3*exp(3*x+x*O(x^n))/(5 - 2*exp(3*x+x*O(x^n))), n)}

{a(n)=polcoeff(sum(m=0, n, 5^m*m!*x^m/prod(k=1, m, 1+3*k*x+x*O(x^n))), n)}

{Stirling2(n, k)=if(k<0||k>n, 0, sum(i=0, k, (-1)^i*binomial(k, i)/k!*(k-i)^n))}
{a(n)=sum(k=0, n, (-3)^(n-k)*5^k*Stirling2(n, k)*k!)}

O.g.f.: A(x) = Sum_{n>=0} n! * 5^n*x^n / Product_{k=0..n} (1+3*k*x).
O.g.f.: A(x) = 1/(1 - 5*x/(1-2*x/(1 - 10*x/(1-4*x/(1 - 15*x/(1-6*x/(1 - 20*x/(1-8*x/(1 - 25*x/(1-10*x/(1 - ...))))))))))), a continued fraction.
a(n) = Sum_{k=0..n} (-3)^(n-k) * 5^k * Stirling2(n,k) * k!.
a(n) = Sum_{k=0..n} A123125(n,k)*5^k*2*(n-k). - Philippe Deléham, Nov 30 2011
a(n) ~ n! / (2*(log(5/2)/3)^(n+1)). - Vaclav Kotesovec, Jun 13 2013
a(n) = 3^n*log(5/2) * Integral_{x = 0..oo} (ceiling(x))^n * (5/2)^(-x) dx. - Peter Bala, Feb 06 2015

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A123125 Triangle of Eulerian numbers T(n,k), 0 <= k <= n, read by rows.

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A201365 Expansion of e.g.f. exp(x) / (5 - 4*exp(x)).

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A201367 E.g.f.: 3*exp(3*x) / (5 - 2*exp(3*x)).

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A201366 E.g.f.: 2*exp(2*x) / (5 - 3*exp(2*x)).

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A333983 a(0) = 0; a(n) = 4^(n-1) + (1/n) * Sum_{k=1..n-1} binomial(n,k)^2 * 4^(k-1) * (n-k) * a(n-k).

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Mathematica

Formula

A201367 E.g.f.: 3exp(3x) / (5 - 2exp(3x)).

A201366 E.g.f.: 2exp(2x) / (5 - 3exp(2x)).