A213127 -id:A213127 - cp's OEIS frontend

A015521 a(n) = 3a(n-1) + 4a(n-2), a(0) = 0, a(1) = 1.

0, 1, 3, 13, 51, 205, 819, 3277, 13107, 52429, 209715, 838861, 3355443, 13421773, 53687091, 214748365, 858993459, 3435973837, 13743895347, 54975581389, 219902325555, 879609302221, 3518437208883, 14073748835533

Offset: 0

Olivier Gérard

Inverse binomial transform of powers of 5 (A000351) preceded by 0. - Paul Barry, Apr 02 2003
Number of walks of length n between any two distinct vertices of the complete graph K_5. Example: a(2)=3 because the walks of length 2 between the vertices A and B of the complete graph ABCDE are: ACB, ADB, AEB. - Emeric Deutsch, Apr 01 2004
The terms of the sequence are the number of segments (sides) per iteration of the space-filling Peano-Hilbert curve. - Giorgio Balzarotti, Mar 16 2006
General form: k=4^n-k. Also: A001045, A078008, A097073, A115341, A015518, A054878. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
A further inverse binomial transform generates A015441. - Paul Curtz, Nov 01 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the central diagonal, and 2's along the subdiagonal and the superdiagonal. - John M. Campbell, Jul 19 2011
Pisano period lengths: 1, 1, 2, 2, 10, 2, 6, 2, 6, 10, 10, 2, 6, 6, 10, 2, 4, 6, 18, 10, ... - R. J. Mathar, Aug 10 2012
Sum_{i=0..m} (-1)^(m+i)*4^i, for m >= 0, gives the terms after 0. - Bruno Berselli, Aug 28 2013
The ratio a(n+1)/a(n) converges to 4 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
This is the Lucas sequence U(P=3,Q=-4), and hence for n>=0, a(n+2)/a(n+1) equals the continued fraction 3 + 4/(3 + 4/(3 + 4/(3 + ... + 4/3))) with n 4's. - Greg Dresden, Oct 07 2019
For n > 0, gcd(a(n), a(n+1)) = 1. - Kengbo Lu, Jul 27 2020

			G.f. = x + 3*x^2 + 13*x^3 + 51*x^4 + 205*x^5 + 819*x^6 + 3277*x^7 + 13107*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Jean-Paul Allouche, Jeffrey Shallit, Zhixiong Wen, Wen Wu, and Jiemeng Zhang, Sum-free sets generated by the period-k-folding sequences and some Sturmian sequences, arXiv:1911.01687 [math.CO], 2019.
J. Borowska and L. Lacinska, Recurrence form of determinant of a heptadiagonal symmetric Toeplitz matrix, J. Appl. Math. Comp. Mech. 13 (2014) 19-16, remark 2 for permanent of tridiagonal Toeplitz matrices a=3, b=2.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
E. M. García-Caballero, S. G. Moreno, and M. P. Prophet, A complete view of Viète-like infinite products with Fibonacci and Lucas numbers, Applied Mathematics and Computation 247 (2014) 703-711.
Dale Gerdemann, Fractal generated from (3,4) recursion A015521, YouTube Video, Dec 4, 2014.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
Index entries for linear recurrences with constant coefficients, signature (3,4).

Cf. A001045, A015518, A054878, A078008, A097073, A109200, A115341, A201455, A213127, A247281.

[Floor(4^n/5-(-1)^n/5): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011

seq(round(4^n/5),n=0..25) # Mircea Merca, Dec 28 2010

k=0;lst={k};Do[k=4^n-k;AppendTo[lst, k], {n, 0, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
LinearRecurrence[{3,4}, {0,1}, 30] (* Harvey P. Dale, Jun 26 2012 *)
CoefficientList[Series[x/((1 - 4 x) (1 + x)), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2014 *)

a(n) = 4^n/5-(-1)^n/5; \\ Altug Alkan, Jan 08 2016

first(n) = Vec(x/(1 - 3*x - 4*x^2) + O(x^n), -n) \\ Iain Fox, Dec 30 2017

def A015521(n): return ((1<<(n<<1))|1)//5 # Chai Wah Wu, Jun 28 2023

[lucas_number1(n,3,-4) for n in range(0, 24)] # Zerinvary Lajos, Apr 22 2009

From Paul Barry, Apr 02 2003: (Start)
a(n) = (4^n - (-1)^n)/5.
E.g.f.: (exp(4*x) - exp(-x))/5. (End)
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*5^(k-1). - Paul Barry, May 13 2003
a(2*n) = 4*a(2*n-1) - 1, a(2*n+1) = 4*a(2*n) + 1. In general this is true for all sequences of the type a(n) + a(n+1) = q^(n): i.e., a(2*n) = q*a(2n-1) - 1 and a(2*n+1) = q*a(2*n) + 1. - Amarnath Murthy, Jul 15 2003
From Emeric Deutsch, Apr 01 2004: (Start)
a(n) = 4^(n-1) - a(n-1).
G.f.: x/(1-3*x - 4*x^2). (End)
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*3^(n-2k)*4^k. - Paul Barry, Jul 29 2004
a(n) = 4*a(n-1) - (-1)^n, n > 0, a(0)=0. - Paul Barry, Aug 25 2004
a(n) = Sum_{k=0..n} A155161(n,k)*2^(n-k), n >= 1. - Philippe Deléham, Jan 27 2009
a(n) = round(4^n/5). - Mircea Merca, Dec 28 2010
The logarithmic generating function 1/5*log((1+x)/(1-4*x)) = x + 3*x^2/2 + 13*x^3/3 + 51*x^4/4 + ... has compositional inverse 5/(4+exp(-5*x)) - 1, the e.g.f. for a signed version of A213127. - Peter Bala, Jun 24 2012
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-5)^k = (4^n - (-1)^n)/5 = (-1)^(n-1)*Sum_{k=0..n-1} (-4)^k. Equals (-1)^(n-1)*Phi(n,-4), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
a(n+1) = 2^(2*n) - a(n), a(0) = 0. - Ben Paul Thurston, Dec 25 2015
a(n) = A247281(n)/5. - Altug Alkan, Jan 08 2016
From Kengbo Lu, Jul 27 2020: (Start)
a(n) = 3*Sum_{k=0..n-1} a(k) + 1 if n odd; a(n) = 3*Sum_{k=0..n-1} a(k) if n even.
a(n) = A030195(n) + Sum_{k=0..n-2} a(k)*A030195(n-k-1).
a(n) = A085449(n) + Sum_{k=0..n-1} a(k)*A085449(n-k).
a(n) = F(n) + 2*Sum_{k=0..n-1} a(k)*F(n-k) + 3*Sum_{k=0..n-2} a(k)*F(n-k-1), where F(n) denotes the Fibonacci numbers.
a(n) = F(n) + Sum_{k=0..n-1} a(k)*(L(n-k) + F(n-k+1)), where F(n) denotes the Fibonacci numbers and L(n) denotes the Lucas numbers.
a(n) = 3^(n-1) + 4*Sum_{k=0..n-2} 3^(n-k-2)*a(k).
a(m+n) = a(m)*a(n+1) + 4*a(m-1)*a(n).
a(2*n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)*3^(2n-2i-2j-1)*4^(i+j). (End)

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

Original entry on oeis.org

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

A210246 Polylogarithm li(-n,-1/3) multiplied by (4^(n+1))/3.

Original entry on oeis.org

1, -1, -2, 2, 40, 104, -1232, -13168, 16000, 1483904, 9695488, -151161088, -3287997440, 146760704, 866038110208, 10263094740992, -169941494497280, -6324725967978496, -15215847186563072, 2895126258819203072, 54295929047166484480

Offset: 0

Stanislav Sykora, Mar 19 2012

sign

Given n, consider the series s(n) = li(-n,-1/3) = SUM((-1)^k (k^n)/3^k) for k=0,1,2,... . Then a(n)=s(n)*(4^(n+1))/3. For more details, see A212846.

			a(5) = polylog(-5,-1/3)*4^6/3 = 104.

Seiichi Manyama, Table of n, a(n) for n = 0..436 (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 Weisstein's World of Mathematics, Polylogarithm

Similar to A210244. Cf. A210247 (sign changes).
Cf. A212846 (li(-n,-1/2)), A212847 (li(-n,-2/3)).
CF. A213127 through A213157.

seq(add((-1)^(n-k)*combinat[eulerian1](n,k)*3^k,k=0..n),n=0..20); # Peter Luschny, Apr 21 2013

Table[PolyLog[-n, -1/3] (4^(n+1))/3, {n, 30}] (* T. D. Noe, Mar 23 2012 *)
a[ n_] := If[ n < 1, Boole[n == 0], PolyLog[ -n, -1/3] 4^(n + 1) / 3]; (* Michael Somos, Nov 01 2014 *)

/* See in A212846, run limnpq(nmax,1,3) */

x='x+O('x^66); Vec(serlaplace( 4/(3+exp(4*x)) )) \\ Joerg Arndt, Apr 21 2013

a(n) = sum(k=0, n, k!*(-1)^k*4^(n-k)*stirling(n, k, 2)); \\ Seiichi Manyama, Mar 13 2022

Recurrence: s(n+1)=(-1/4)*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((-1/3)^k)=3/4.
From Peter Bala, Mar 12 2013: (Start)
E.g.f.: A(x) = 4/(3 + exp(4*x)) = 1 - x - 2*x^2/2! + 2*x^3/3! + 40*x^4/4! + ....
The compositional inverse (A(-x) - 1)^(-1) = x + 2*x^2/2 + 7*x^3/3 + 20*x^4/4 + 61*x^5/5 + ... is the logarithmic generating function for A015518.
Recurrence equation: a(n+1) = -4*a(n) + 3*sum {k = 0..n} binomial(n,k)*a(k)*a(n-k), with a(0) = 1.
(End)
G.f.: 1 + x/Q(0), where Q(k) = 2*x*(k+1) - 1 + 3*x^2*(k+1)*(k+2)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Sep 22 2013
G.f.: 1/Q(0), where Q(k) = 1 + x*(k+1)/( 1 - 3*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 17 2013
E.g.f.: 2 - W(0), where W(k) = 1 + x/( 4*k+1 - x/( 1 + 4*x/( 4*k+3 - 4*x/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2014
a(n) = Sum_{k=0..n} k! * (-1)^k * 4^(n-k) * Stirling2(n,k). - Seiichi Manyama, Mar 13 2022
a(n) ~ n! * cos((n+1)*arctan(Pi/log(3))) * 2^(2*n + 3) / (3 * (Pi^2 + log(3)^2)^((n+1)/2)). - Vaclav Kotesovec, May 17 2022

A212847 Polylogarithm li(-n,-2/3) multiplied by (5^(n+1))/3.

Original entry on oeis.org

1, -2, -2, 22, 94, -890, -9170, 67030, 1495870, -6581210, -362016050, 194447350, 120002960350, 554823694150, -51277487618450, -601106981110250, 26775789844186750, 591304973974171750, -16113120605399179250

Offset: 0

Stanislav Sykora, May 28 2012

sign

See the sequence A212846 which describes the general case of li(-n,-p/q). This sequence is obtained for p=2,q=3.

			polylog(-5,-2/3)*5^6/3 = -890.

Stanislav Sykora, Table of n, a(n) for n = 0..100

Cf. A212846 (li(-n,-1/2)), A210246 (li(-n,-1/3)).

Cf. A213127 through A213157.

f[n_] := PolyLog[-n, -2/3] 5^(n + 1)/3; f[0] = 1; Array[f, 20, 0] (* Robert G. Wilson v, Dec 25 2015 *)

PARI
```
\\ See A212846; run limnpq(nmax,2,3)
```

See formula in A212846, setting p=2,q=3.
From Vaclav Kotesovec, May 17 2022: (Start)
E.g.f.: 5/(3 + 2*exp(5*x)).
a(n) ~ n! * 2*(log(3/2) * cos(n*arctan(Pi/log(3/2))) - Pi * sin(n*arctan(Pi/log(3/2)))) * 5^(n+1) / (3 * (Pi^2 + log(3/2)^2)^(1 + n/2)). (End)

A213157 Polylogarithm li(-n,-99/100) multiplied by (199^(n+1))/100.

Original entry on oeis.org

1, -99, -99, 1960101, 7840701, -155226277899, -1319555674899, 26121225430931301, 381134689417943901, -7543761920163143670699, -168204228721945992219699, 3328727258163288077733522501

Offset: 0

Stanislav Sykora, Jun 06 2012

sign

See the sequence A212846 which describes the general case of li(-n,-p/q). This sequence is obtained for p=99,q=100.

			polylog(-5,-99/100)*199^6/100 = -155226277899.

Stanislav Sykora, Table of n, a(n) for n = 0..100

Cf. A210246, A212846, A212847, A213127-A213156.

PARI
```
in A212846; run limnpq(nmax, 99, 100)
```

See formula in A212846, setting p=99, q=100.

A213128 Polylogarithm li(-n,-1/5) multiplied by (6^(n+1))/5.

Original entry on oeis.org

1, -1, -4, -6, 96, 1104, 2016, -112176, -1718784, -642816, 437031936, 7656021504, -24274059264, -3939918299136, -72733516959744, 699443277686784, 67781787782086656, 1236409075147014144, -25430445045847425024

Offset: 0

Stanislav Sykora, Jun 06 2012

sign

See the sequence A212846 which describes the general case of li(-n,-p/q). This sequence is obtained for p=1,q=5.

			polylog(-5,-1/5)*6^6/5 = 1104.

Stanislav Sykora, Table of n, a(n) for n = 0..100
OEIS-Wiki, Eulerian polynomials

Cf. A212846, A210246, A212847, A213127
Cf. A213129 through A213157.
Cf. A015531.

seq(add((-1)^(n-k)*combinat[eulerian1](n,k)*5^k, k=0..n),n=0..18); # Peter Luschny, Apr 21 2013

Table[If[n == 0, 1, PolyLog[-n, -1/5] 6^(n+1)/5], {n, 0, 18}] (* Jean-François Alcover, Jun 29 2019 *)

PARI
```
/*See A212846; run limnpq(nmax,1,5) */
```

x='x+O('x^66); Vec(serlaplace( 6/(5+exp(6*x)) )) \\ Joerg Arndt, Apr 21 2013

a(n) = sum(k=0, n, k!*(-1)^k*6^(n-k)*stirling(n, k, 2)); \\ Seiichi Manyama, Mar 13 2022

See formula in A212846, setting p=1,q=5
From Peter Bala, Jun 24 2012: (Start)
E.g.f.: A(x) = 6/(5 + exp(6*x)) = 1 - x - 4*x^2/2! - 6 x^3/3! + 96*x^4/4! + ....
The compositional inverse (A(-x) - 1)^(-1) = x + 4*x^2/2 + 21*x^3/3 + 104*x^4/4 + 521*x^5/5 + ... is the logarithmic generating function for A015531.
(End)
G.f.: 1/Q(0), where Q(k) = 1 + x*(k+1)/( 1 - 5*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 17 2013
a(n) = Sum_{k=0..n} k! * (-1)^k * 6^(n-k) * Stirling2(n,k). - Seiichi Manyama, Mar 13 2022

A213129 Polylogarithm li(-n,-1/6) multiplied by (7^(n+1))/6.

Original entry on oeis.org

1, -1, -5, -13, 115, 2099, 11395, -177373, -5116685, -40481581, 948973795, 36701972867, 375364322515, -12090607539661, -580544884927805, -7188739235243293, 301374306966657715, 17150539711123411859, 246564346727945106595, -12988846468460187345853

Offset: 0

Stanislav Sykora, Jun 06 2012

sign

See the sequence A212846 which describes the general case of li(-n,-p/q). This sequence is obtained for p=1,q=6.

			polylog(-5,-1/6)*7^6/6 = 2099.

Stanislav Sykora, Table of n, a(n) for n = 0..100
OEIS-Wiki, Eulerian polynomials

Cf. A212846, A210246, A212847, A213127, A213128.

Cf. A213130 through A213157

seq(add((-1)^(n-k)*combinat[eulerian1](n,k)*6^k, k=0..n),n=0..17); # Peter Luschny, Apr 21 2013

Table[If[n == 0, 1, PolyLog[-n, -1/6] 7^(n+1)/6], {n, 0, 19}] (* Jean-François Alcover, Jun 27 2019 *)

/* See A212846; run limnpq(nmax,1,6) */

x='x+O('x^66); Vec(serlaplace( 7/(6+exp(7*x)) )) \\ Joerg Arndt, Apr 21 2013

a(n) = sum(k=0, n, k!*(-1)^k*7^(n-k)*stirling(n, k, 2)); \\ Seiichi Manyama, Mar 13 2022

See formula in A212846, setting p=1,q=6.
E.g.f.: 7/(6+exp(7*x)). [Joerg Arndt, Apr 21 2013]
a(n) = Sum_{k=0..n} k! * (-1)^k * 7^(n-k) * Stirling2(n,k). - Seiichi Manyama, Mar 13 2022

A213130 Polylogarithm li(-n,-1/7) multiplied by (8^(n+1))/7.

Original entry on oeis.org

1, -1, -6, -22, 120, 3464, 30864, -189232, -11564160, -173474176, 923222784, 112587838208, 2509094415360, -7947533372416, -2393798607108096, -74042111038461952, -8461127118520320, 94056121376877215744

Offset: 0

Stanislav Sykora, Jun 06 2012

sign

See the sequence A212846 which describes the general case of li(-n,-p/q). This sequence is obtained for p=1,q=7.

			polylog(-5,-1/7)*8^6/7 = 3464.

Stanislav Sykora, Table of n, a(n) for n = 0..100
OEIS-Wiki, Eulerian polynomials

Cf. A212846, A210246, A212847, A213127 to A213129.

Cf. A213131 through A213157.

seq(add((-1)^(n-k)*combinat[eulerian1](n,k)*7^k, k=0..n),n=0..17); # Peter Luschny, Apr 21 2013

f[n_] := PolyLog[-n, -1/7] 8^(n + 1)/7; f[0] = 1; Array[f, 20, 0] (* Robert G. Wilson v, Dec 25 2015 *)

PARI
```
\\ in A212846; run limnpq(nmax, 1, 7)
```

a(n) = sum(k=0, n, k!*(-1)^k*8^(n-k)*stirling(n, k, 2)); \\ Seiichi Manyama, Mar 13 2022

See formula in A212846, setting p=1,q=7.

a(n) = Sum_{k=0..n} k! * (-1)^k * 8^(n-k) * Stirling2(n,k). - Seiichi Manyama, Mar 13 2022

A213131 Polylogarithm li(-n,-1/8) multiplied by (9^(n+1))/8.

Original entry on oeis.org

1, -1, -7, -33, 105, 5199, 64953, -46593, -21769335, -497664081, -1941272487, 256114020447, 9566995408425, 99966666676239, -6245895772363527, -366865939437422913, -6924777575908002615, 259022993102904450159, 24387711970312991335833, 716398360186298080983327

Offset: 0

Stanislav Sykora, Jun 06 2012

sign

See the sequence A212846 which describes the general case of li(-n,-p/q). This sequence is obtained for p=1,q=8.

			polylog(-5,-1/8)*9^6/8 = 5199.

Stanislav Sykora, Table of n, a(n) for n = 0..100
OEIS-Wiki, Eulerian polynomials

Cf. A212846, A210246, A212847, A213127 to A213130

Cf. A213132 to A213157

seq(add((-1)^(n-k)*combinat[eulerian1](n,k)*8^k, k=0..n),n=0..17); # Peter Luschny, Apr 21 2013

Table[If[n == 0, 1, PolyLog[-n, -1/8] 9^(n+1)/8], {n, 0, 19}] (* Jean-François Alcover, Jun 27 2019 *)

/* See A212846; run limnpq(nmax, 1, 8) */

x='x+O('x^66); Vec(serlaplace( 9/(8+exp(9*x)) )) \\ Joerg Arndt, Apr 21 2013

a(n) = sum(k=0, n, k!*(-1)^k*9^(n-k)*stirling(n, k, 2)); \\ Seiichi Manyama, Mar 13 2022

See formula in A212846, setting p=1,q=8.
E.g.f.: 9/(8+exp(9*x)). [Joerg Arndt, Apr 21 2013]
a(n) = Sum_{k=0..n} k! * (-1)^k * 9^(n-k) * Stirling2(n,k). - Seiichi Manyama, Mar 13 2022

A213132 Polylogarithm li(-n,-1/9) multiplied by (10^(n+1))/9.

Original entry on oeis.org

1, -1, -8, -46, 64, 7280, 118720, 406160, -35578880, -1156775680, -12796467200, 444964083200, 27457634713600, 594958346547200, -9096689344716800, -1258068242084608000, -45330583283597312000, 24150498582339584000, 95678058298287259648000, 5379182782796767182848000

Offset: 0

Stanislav Sykora, Jun 06 2012

sign

See the sequence A212846 which describes the general case of li(-n,-p/q). This sequence is obtained for p=1,q=9.

			polylog(-5, -1/9)*10^6/9 = 7280.

Stanislav Sykora, Table of n, a(n) for n = 0..100
OEIS-Wiki, Eulerian polynomials

Cf. A212846, A210246, A212847, A213127 to A213131

Cf. A213133 to A213157

seq(add((-1)^(n-k)*combinat[eulerian1](n,k)*9^k, k=0..n),n=0..17); # Peter Luschny, Apr 21 2013

Table[If[n == 0, 1, PolyLog[-n, -1/9] 10^(n+1)/9], {n, 0, 19}] (* Jean-François Alcover, Jun 27 2019 *)

/* See A212846; run limnpq(nmax, 1, 9) */

x='x+O('x^66); Vec(serlaplace( 10/(9+exp(10*x)) )) \\ Joerg Arndt, Apr 21 2013

a(n) = sum(k=0, n, k!*(-1)^k*10^(n-k)*stirling(n, k, 2)); \\ Seiichi Manyama, Mar 13 2022

See formula in A212846, setting p=1,q=9.
E.g.f.: 10/(9+exp(10*x)). [Joerg Arndt, Apr 21 2013]
a(n) = Sum_{k=0..n} k! * (-1)^k * 10^(n-k) * Stirling2(n,k). - Seiichi Manyama, Mar 13 2022

cp's OEIS Frontend

A015521 a(n) = 3*a(n-1) + 4*a(n-2), a(0) = 0, a(1) = 1.

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

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A210246 Polylogarithm li(-n,-1/3) multiplied by (4^(n+1))/3.

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A212847 Polylogarithm li(-n,-2/3) multiplied by (5^(n+1))/3.

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A213157 Polylogarithm li(-n,-99/100) multiplied by (199^(n+1))/100.

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A213128 Polylogarithm li(-n,-1/5) multiplied by (6^(n+1))/5.

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A015521 a(n) = 3a(n-1) + 4a(n-2), a(0) = 0, a(1) = 1.