A179929 -id:A179929 - cp's OEIS frontend

A212846 Polylogarithm li(-n,-1/2) multiplied by (3^(n+1))/2.

1, -1, -1, 3, 15, -21, -441, -477, 19935, 101979, -1150281, -14838957, 60479055, 2328851979, 3529587879, -403992301437, -3333935426625, 72778393505979, 1413503392326039, -10851976875907917, -554279405351601105, -713848745428080021

Offset: 0

Stanislav Sykora, May 28 2012

sign

Apart from sign, same as A087674: a(n) = A087674*(-1)^n

Given integers n, p, q, 0=0, ((k^n)/(-p/q)^k) ) = s(n), multiplied by ((p+q)^(n+1))/q is an integer a(n). For this sequence set p=1 and q=2.

			a(5) = polylog(-5,-1/2)*3^6/2 = -21.
E.g.f.: A(x) = 1 - x - x^2/2! + 3*x^3/3! + 15*x^4/4! - 21*x^5/5! + ...
O.g.f.: G(x) = 1 - x - x^2 + 3*x^3 + 15*x^4 - 21*x^5 - 441*x^6 +...
where G(x) = 1 - x/(1-3*x) + 2!*x^2/((1-3*x)*(1-6*x)) - 3!*x^3/((1-3*x)*(1-6*x)*(1-9*x)) + 4!*x^4/((1-3*x)*(1-6*x)*(1-9*x)*(1-12*x)) +...

Seiichi Manyama, Table of n, a(n) for n = 0..455 (terms 0..100 from Stanislav Sykora)
OEIS-Wiki, Eulerian polynomials
S. Sykora, Finite and Infinite Sums of the Power Series (k^p)(x^k), Stan's Library Vol. I, April 2006, updated March 2012. See Eq.(29).
Eric W. Weisstein, MathWorld: Polylogarithm

Similar cases: A210246 (p=1,q=3), A212847 (p=2,q=3)
Cf. A210244 (similar).
Cf. A213127 through A213157.
Cf. A001045, A087674, A179929.

seq(add((-1)^(n-k)*combinat[eulerian1](n,k)*2^k,k=0..n),n=0..21); # Peter Luschny, Apr 21 2013

f[n_] := PolyLog[-n, -1/2] 3^(n + 1)/2; Array[f, 21] (* Robert G. Wilson v, May 28 2012 *)
a[ n_] := If[ n < 0, 0, n! 3/2 SeriesCoefficient[ 1 / (1 + Exp[3 x] / 2), {x, 0, n}]]; (* Michael Somos, Aug 27 2018 *)

/* for this sequence, run limnpq(nmax,1,2) */
limnpq(nmax,p,q) = {
  f=vector(nmax+1);f[1]=q/(p+q);r=-p/(p+q);
  for (i=2,nmax+1,p1=i-1;bc=1;m=p1;s=0;
    for(j=1,i-1,p2=j-1;if (p2,bc=bc*m/p2;m=m-1;);
    s=s+bc*f[j];);f[i]=r*s;);
fac=(p+q)/q;
for(i=1,nmax+1,f[i]=f[i]*fac;fac=(p+q)*fac;
  write("outputfile",i-1," ",f[i]););}

x='x+O('x^66); Vec(serlaplace(3/(2+exp(3*x)))) \\ Joerg Arndt, Apr 21 2013

/* O.g.f.: */
{a(n)=polcoeff(sum(m=0, n, m!*(-x)^m/prod(k=1, m, 1-3*k*x+x*O(x^n))), n)} \\ Paul D. Hanna, May 30 2013

a(n) = sum(k=0, n, k!*(-1)^k*3^(n-k)*stirling(n, k, 2)); \\ Seiichi Manyama, Mar 13 2022

General recurrence: s(n+1)=(-p/(p+q))*SUM(C(n+1,i)*s(i)), where i=0,1,2,...,n, C(n,m) are binomial coefficients, and the starting value is s(0)=SUM((-p/q)^k)=q/(p+q). For this sequence set p=1 and q=2.
From Peter Bala, Jun 24 2012: (Start)
E.g.f.: A(x) = 3/(2+exp(3*x)).
The compositional inverse (A(-x) - 1)^(-1) = x + x^2/2 + 3*x^3/3 + 5*x^4/4 + 11*x^5/5 + ... is the logarithmic generating function for A001045.
(End)
a(n+1) = -3*a(n) + 2*sum(k=0..n, binomial(n,k)*a(k)*a(n-k) ), with a(0) = 1. - Peter Bala, Mar 12 2013
Let A(x) be the g.f. of A212846, B(x) the g.f. of A087674, then A(x) = B(-x).
G.f.: 1/Q(0), where Q(k)= 1 + x*(k+1)/(1 - x*(2*k+2)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 20 2013
O.g.f.: Sum_{n>=0} n!*(-x)^n / Product_{k=0..n} (1-3*k*x). - Paul D. Hanna, May 30 2013
For n>0, a(n) = -A179929(n)/2. - Stanislav Sykora, May 15 2014
a(n) = Sum_{k=0..n} k! * (-1)^k * 3^(n-k) * Stirling2(n,k). - Seiichi Manyama, Mar 13 2022
a(n) ~ n! * (log(2) * cos(n*arctan(Pi/log(2))) - Pi * sin(n*arctan(Pi/log(2)))) * 3^(n+1) / (Pi^2 + log(2)^2)^(1 + n/2). - Vaclav Kotesovec, May 17 2022

A195205 Triangle of coefficients of a sequence of binomial type polynomials.

Original entry on oeis.org

3, 6, 9, 30, 54, 27, 222, 468, 324, 81, 2190, 5130, 4320, 1620, 243, 27006, 68400, 65610, 30780, 7290, 729, 399630, 1076166, 1135890, 618030, 187110, 30618, 2187, 6899262, 19532268, 22212792, 13471920, 4796820, 1020600, 122472, 6561

Offset: 1

Peter Bala, Sep 13 2011

Define a polynomial sequence P_n(x) by means of the recursion
P_(n+1)(x) = x*(P_n(x) + 2*P_n(x+1)), with P_0(x) = 1.
The first few values are
P_1(x) = 3*x, P_2(x) = 3*x*(3*x + 2),
P_3(x) = 3*x*(9*x^2 + 18*x + 10),
P_4(x) = 3*x*(27*x^3 + 108*x^2 + 156*x + 74).
The present table shows the coefficients of these polynomials (excluding P_0(x)) in ascending powers of x. Compare with A195204.
Triangle T(n,k) (1 <= k <= n), read by rows, given by (0, 2, 3, 4, 6, 6, 9, 8, 12, 10, 15, ...) DELTA (3, 0, 3, 0, 3, 0, 3, 0, 3, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 22 2011

			Triangle begins
n\k|.....1.......2......3......4......5......6
==============================================
..1|.....3
..2|.....6.......9
..3|....30......54.....27
..4|...222.....468....324.....81
..5|..2190....5130...4320...1620....243
..6|.27006...68400..65610..30780...7290....729
...
Triangle (0, 2, 3, 4, 6, 6, 9, ...) DELTA (3, 0, 3, 0, 3, 0, 3, 0, ...) begins:
1;
0,     3;
0,     6,     9;
0,    30,    54,    27;
0,   222,   468,   324,    81;
0,  2190,  5130,  4320,  1620,   243;
0, 27006, 68400, 65610, 30780,  7290,   729;
... - _Philippe Deléham_, Dec 22 2011

Wikipedia, Binomial type

Cf. A004123 ((1/3)*column 1), A050351 ((1/3)*row sums), A179929 (delta operator coeffs.), A195204.

# The function BellMatrix is defined in A264428.
# Adds (1,0,0,0, ..) as column 0.
BellMatrix(n -> `if`(n=0,3,polylog(-n, 2/3)), 10); # Peter Luschny, Jan 29 2016

BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
rows = 10;
M = BellMatrix[If[# == 0, 3, PolyLog[-#, 2/3]]&, rows];
Table[M[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 24 2018, after Peter Luschny *)

E.g.f.: F(x,z) := (exp(z)/(3 - 2*exp(z)))^x = 1 + 3*x*z + (6*x + 9*x^2)*z^2/2! + (30*x + 54*x^2 + 27*x^3)*z^3/3! + ....
The generating function F(x,z) = Sum_{n>=0} P_n(x)*z^n/n! satisfies the partial differential equation d/dz(F(x,z)) = x*F(x,z) + 2*x*F(x+1,z). Hence the row generating polynomials P_n(x) satisfy the recurrence P_(n+1)(x) = x*(P_n(x) + 2*P_n(x+1)), with P_0(x) = 1. The form of the e.g.f. shows that the polynomials P_n(x) are a sequence of binomial type. In what follows we denote P_n(x) by x^[n].
Relation with rising factorials
x^[n] = Sum_{k=1..n} (-1)^(n-k)*Stirling2(n,k)*3^k*x*(x+1)*...*(x+k-1),
and its inverse formula
3^n*x*(x+1)*...*(x+n-1) = Sum_{k=1..n} |Stirling1(n,k)|*x^[k].
The delta operator D*:
The row polynomials form a polynomial sequence of binomial type. If D denotes the derivative operator 1/3*d/dx then the associated delta operator D* is given by D* = D - 2*D^2/2! + 2*D^3/3! + 6*D^4/4! - 30*D^5/5! - ..., where the sequence of coefficients [1, -2, 2, 6, -30, -42, 882, ...] equals (-1)^n*A179929(n). D* is the lowering operator for the row polynomials, that is, (D*)x^[n] = n*x^[n-1].
Generalized Dobinski formula:
exp(-x)*Sum_{k >= 1} (-k)^[n]*x^k/k! = (-1)^n*Bell(n,3*x),
where the Bell (or exponential) polynomials are defined as
Bell(n,x) := Sum_{k = 1..n} Stirling2(n,k)*x^k.
Relation with the Bell polynomials:
The alternating n-th row entries (-1)^(n+k)*T(n,k) are the connection coefficients expressing the polynomial Bell(n,3*x) as a linear combination of Bell(k,x), 1 <= k <= n. For example for row 4:
Bell(4,3*x) = -222*Bell(1,x) + 468*Bell(2,x) - 324*Bell(3,x) + 81*Bell(4,x).
Generalized Bernoulli summation formula:
We have the following generalization of Bernoulli's formula for the sum of the powers of integers:
3*Sum_{k = 1..n} k^[p] = 1/(p+1)*Sum_{k = 0..p} (-1)^k * binomial(p+1,k)*B_k*n^[p+1-k], where B_k =[1, -1/2, 1/6, 0, -1/30, ...] denotes the sequence of Bernoulli numbers.
Relation with other sequences:
Row sums = 3*A050351(n) for n >= 1. Column 1 = 3*A004123.
T(n,k) = A185285(n,k)*3^k. - Philippe Deléham, Feb 17 2013
Also the Bell transform of 3*A004123. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 29 2016
Conjecture: o.g.f. as a continued fraction of Stieltjes type: 1/(1 - 3*x*z/(1 - 2*z/(1 - 3*(x + 1)*z/(1 - 4*z/(1 - 3*(x + 2)*z/(1 - 6*z/(1 - 3*(x + 3)*z/(1 - 8*z/(1 - ... ))))))))). - Peter Bala, Dec 12 2024

A383140 Triangle read by rows: the coefficients of polynomials (1/3^(m-n)) * Sum_{k=0..m} k^n * 2^(m-k) * binomial(m,k) in the variable m.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 2, 6, 1, 0, -6, 20, 12, 1, 0, -30, 10, 80, 20, 1, 0, 42, -320, 270, 220, 30, 1, 0, 882, -1386, -770, 1470, 490, 42, 1, 0, 954, 7308, -15064, 2800, 5180, 952, 56, 1, 0, -39870, 101826, -39340, -61992, 29820, 14364, 1680, 72, 1, 0, -203958, -40680, 841770, -666820, -86940, 139440, 34020, 2760, 90, 1

Offset: 0

Seiichi Manyama, Apr 17 2025

			f_n(m) = (1/3^(m-n)) * Sum_{k=0..m} k^n * 2^(m-k) * binomial(m,k).
f_0(m) = 1.
f_1(m) =    m.
f_2(m) =  2*m +   m^2.
f_3(m) =  2*m + 6*m^2 + m^3.
Triangle begins:
  1;
  0,   1;
  0,   2,    1;
  0,   2,    6,   1;
  0,  -6,   20,  12,   1;
  0, -30,   10,  80,  20,  1;
  0,  42, -320, 270, 220, 30, 1;
  ...

Eric Weisstein's World of Mathematics, Falling Factorial

Columns k=0..1 give A000007, A179929(n-1).
Row sums give A133494.
Alternating row sums give A212846.
Cf. A027471, A383136, A383137, A383138, A383139.
Cf. A129062, A209849.

T(n, k) = sum(j=k, n, 3^(n-j)*stirling(n, j, 2)*stirling(j, k, 1));

def a_row(n):
    s = sum(3^(n-k)*stirling_number2(n, k)*falling_factorial(x, k) for k in (0..n))
    return expand(s).list()
for n in (0..10): print(a_row(n))

T(n,k) = Sum_{j=k..n} 3^(n-j) * Stirling2(n,j) * Stirling1(j,k).
T(n,k) = [x^k] Sum_{k=0..n} 3^(n-k) * Stirling2(n,k) * FallingFactorial(x,k).
E.g.f. of column k (with leading zeros): g(x)^k / k! with g(x) = log(1 + (exp(3*x) - 1)/3).

cp's OEIS Frontend

A212846 Polylogarithm li(-n,-1/2) multiplied by (3^(n+1))/2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

PARI

Formula

A195205 Triangle of coefficients of a sequence of binomial type polynomials.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A383140 Triangle read by rows: the coefficients of polynomials (1/3^(m-n)) * Sum_{k=0..m} k^n * 2^(m-k) * binomial(m,k) in the variable m.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

Sage

Formula