cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A052849 a(0) = 0; a(n) = 2*n! (n >= 1).

Original entry on oeis.org

0, 2, 4, 12, 48, 240, 1440, 10080, 80640, 725760, 7257600, 79833600, 958003200, 12454041600, 174356582400, 2615348736000, 41845579776000, 711374856192000, 12804747411456000, 243290200817664000, 4865804016353280000, 102181884343418880000, 2248001455555215360000
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

For n >= 1 a(n) is the size of the centralizer of a transposition in the symmetric group S_(n+1). - Ahmed Fares (ahmedfares(AT)my-deja.com), May 12 2001
For n > 0, a(n) = n! - A062119(n-1) = number of permutations of length n that have two specified elements adjacent. For example, a(4) = 12 as of the 24 permutations, 12 have say 1 and 2 adjacent: 1234, 2134, 1243, 2143, 3124, 3214, 4123, 4213, 3412, 3421, 4312, 4321. - Jon Perry, Jun 08 2003
With different offset, denominators of certain sums computed by Ramanujan.
From Michael Somos, Mar 04 2004: (Start)
Stirling transform of a(n) = [2, 4, 12, 48, 240, ...] is A000629(n) = [2, 6, 26, 150, 1082, ...].
Stirling transform of a(n-1) = [1, 2, 4, 12, 48, ...] is A007047(n-1) = [1, 3, 11, 51, 299, ...].
Stirling transform of a(n) = [1, 4, 12, 48, 240, ...] is A002050(n) = [1, 5, 25, 149, 1081, ...].
Stirling transform of 2*A006252(n) = [2, 2, 4, 8, 28, ...] is a(n) = [2, 4, 12, 48, 240, ...].
Stirling transform of a(n+1) = [4, 12, 48, 240, ...] is 2*A005649(n) = [4, 16, 88, 616, ...].
Stirling transform of a(n+1) = [4, 12, 48, 240, ...] is 4*A083410(n) = [4, 16, 88, 616, ...]. (End)
Number of {12, 12*, 21, 21*}-avoiding signed permutations in the hyperoctahedral group.
Permanent of the (0, 1)-matrices with (i, j)-th entry equal to 0 if and only if it is in the border but not the corners. The border of a matrix is defined the be the first and the last row, together with the first and the last column. The corners of a matrix are the entries (i = 1, j = 1), (i = 1, j = n), (i = n, j = 1) and (i = n, j = n). - Simone Severini, Oct 17 2004

References

  • B. C. Berndt, Ramanujan's Notebooks Part V, Springer-Verlag, see p. 520.

Crossrefs

Essentially the same sequence as A098558.
Row 3 of A276955 (from term a(2)=4 onward).

Programs

  • Haskell
    a052849 n = if n == 0 then 0 else 2 * a000142 n
    a052849_list = 0 : fs where fs = 2 : zipWith (*) [2..] fs
    -- Reinhard Zumkeller, Aug 31 2014
    
  • Magma
    [0] cat [2*Factorial(n-1): n in [2..25]]; // Vincenzo Librandi, Nov 03 2014
  • Maple
    spec := [S,{B=Cycle(Z),C=Cycle(Z),S=Union(B,C)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    Join[{0}, 2Range[20]!] (* Harvey P. Dale, Jul 13 2013 *)
  • PARI
    a(n)=if(n<1,0,n!*2)
    

Formula

a(n) = T(n, 2) for n>1, where T is defined as in A080046.
D-finite with recurrence: {a(0) = 0, a(1) = 2, (-1 - n)*a(n+1) + a(n+2)=0}.
E.g.f.: 2*x/(1-x).
a(n) = A090802(n, n - 1) for n > 0. - Ross La Haye, Sep 26 2005
For n >= 1, a(n) = (n+3)!*Sum_{k=0..n+2} (-1)^k*binomial(2, k)/(n + 3 - k). - Milan Janjic, Dec 14 2008
G.f.: 2/Q(0) - 2, where Q(k) = 1 - x*(k + 1)/(1 - x*(k + 1)/Q(k+1) ); (continued fraction ). - Sergei N. Gladkovskii, Apr 01 2013
G.f.: -2 + 2/Q(0), where Q(k) = 1 + k*x - x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
G.f.: W(0) - 2 , where W(k) = 1 + 1/( 1 - x*(k+1)/( x*(k+1) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 21 2013
a(n) = A245334(n, n-1), n > 0. - Reinhard Zumkeller, Aug 31 2014
From Amiram Eldar, Jan 15 2023: (Start)
Sum_{n>=1} 1/a(n) = (e-1)/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = (e-1)/(2*e). (End)

Extensions

More terms from Ross La Haye, Sep 26 2005

A024167 a(n) = n!*(1 - 1/2 + 1/3 - ... + c/n), where c = (-1)^(n+1).

Original entry on oeis.org

1, 1, 5, 14, 94, 444, 3828, 25584, 270576, 2342880, 29400480, 312888960, 4546558080, 57424792320, 948550176000, 13869128448000, 256697973504000, 4264876094976000, 87435019510272000, 1627055289796608000, 36601063093905408000, 754132445894209536000
Offset: 1

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Author

Keywords

Comments

Stirling transform of (-1)^n*a(n-1) = [0, 1, -1, 5, -14, 94, ...] is A000629(n-2) = [0, 1, 2, 6, 26, ...]. - Michael Somos, Mar 04 2004
Stirling transform of a(n) = [1, 1, 5, 14, 94, ...] is A052882(n) = [1, 2, 9, 52, 375, ...]. - Michael Somos, Mar 04 2004
a(n) is the number of n-permutations that have a cycle with length greater than n/2. - Geoffrey Critzer, May 28 2009
From Jens Voß, May 07 2010: (Start)
a(4n) is divisible by 6*n + 1 for all n >= 1; the quotient of a(4*n) and 6*n+1 is A177188(n).
a(4*n+3) is divisible by 6*n + 5 for all n >= 0; the quotient of a(4*n+3) and 6*n + 5 is A177174(n). (End)

Examples

			G.f. = x + x^2 + 5*x^3 + 14*x^4 + 94*x^5 + 444*x^6 + 3828*x^7 + 25584*x^8 + ...
		

Crossrefs

Programs

  • Maple
    a := n -> n!*(log(2) - (-1)^n*LerchPhi(-1, 1, n+1));
    seq(simplify(a(n)), n=1..20); # Peter Luschny, Dec 27 2018
  • Mathematica
    f[k_] := k (-1)^(k + 1)
    t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 18}]    (* A024167 signed *)
    (* Clark Kimberling, Dec 30 2011 *)
    a[ n_] := If[ n < 0, 0, n! Sum[ -(-1)^k / k, {k, n}]]; (* Michael Somos, Nov 28 2013 *)
    a[ n_] := If[ n < 0, 0, n! (PolyGamma[n + 1] - PolyGamma[(n + Mod[n, 2, 1]) / 2])]; (* Michael Somos, Nov 28 2013 *)
    a[ n_] := If[ n < 1, 0, (-1)^Quotient[n, 2] SymmetricPolynomial[ n - 1, Table[ -(-1)^k k, {k, n}]]]; (* Michael Somos, Nov 28 2013 *)
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( log(1 + x + x * O(x^n)) / (1 - x), n))}; /* Michael Somos, Mar 02 2004 */
    
  • PARI
    x='x+O('x^33); Vec(serlaplace(log(1+x)/(1-x))) \\ Joerg Arndt, Dec 27 2018
    
  • Python
    def A():
        a, b, n = 1, 1, 2
        yield(a)
        while True:
            yield(a)
            b, a = a, a + b * n * n
            n += 1
    a = A(); print([next(a) for  in range(20)]) # _Peter Luschny, May 19 2020

Formula

E.g.f.: log(1 + x)/(1 - x). - Vladeta Jovovic, Aug 25 2002
a(n) = a(n-1) + a(n-2) * (n-1)^2, n > 1. - Michael Somos, Oct 29 2002
b(n) = n! satisfies the above recurrence with b(1) = 1, b(2) = 2. This gives the finite continued fraction expansion a(n)/n! = 1/(1 + 1^2/(1 + 2^2/(1 + 3^2/(1 + ... + (n-1)^2/1)))). Cf. A142979. - Peter Bala, Jul 17 2008
a(n) = A081358(n) - A092691(n). - Gary Detlefs, Jul 09 2010
E.g.f.: (x/(x-1))/G(0) where G(k) = -1 + (x-1)*k + x*(k+1)^2/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 18 2012
a(n) ~ log(2)*n!. - Daniel Suteu, Dec 03 2016
a(n) = (1/2)*n!*((-1)^n*(digamma((n+1)/2) - digamma((n+2)/2)) + log(4)). - Daniel Suteu, Dec 03 2016
a(n) = n!*(log(2) - (-1)^n*LerchPhi(-1, 1, n+1)). - Peter Luschny, Dec 27 2018
a(n) = A054651(n,n-1). - Pontus von Brömssen, Oct 25 2020
a(n) = Sum_{k=0..n} (-1)^k*k!*A094587(n, k+1). - Mélika Tebni, Jun 20 2022
a(n) = n * a(n-1) - (-1)^n * (n-1)! for n > 1. - Werner Schulte, Oct 20 2024

Extensions

More terms from Benoit Cloitre, Jan 27 2002
a(21)-a(22) from Pontus von Brömssen, Oct 25 2020

A163626 Triangle read by rows: The n-th derivative of the logistic function written in terms of y, where y = 1/(1 + exp(-x)).

Original entry on oeis.org

1, 1, -1, 1, -3, 2, 1, -7, 12, -6, 1, -15, 50, -60, 24, 1, -31, 180, -390, 360, -120, 1, -63, 602, -2100, 3360, -2520, 720, 1, -127, 1932, -10206, 25200, -31920, 20160, -5040, 1, -255, 6050, -46620, 166824, -317520, 332640, -181440, 40320, 1, -511, 18660
Offset: 0

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Author

Keywords

Comments

Apart from signs and offset, same as A028246. - Joerg Arndt, Nov 06 2016
Triangle T(n,k), read by rows, given by (1,0,2,0,3,0,4,0,5,0,6,0,7,0,8,0,9,...) DELTA (-1,-1,-2,-2,-3,-3,-4,-4,-5,-5,-6,-6,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 05 2011
The "Stirling-Bernoulli transform" maps a sequence b_0, b_1, b_2, ... to a sequence c_0, c_1, c_2, ..., where if B has o.g.f. B(x), c has e.g.f. exp(x)*B(1 - exp(x)). More explicity, c_n = Sum_{k = 0..n} A163626(n,k)*b_k. - Philippe Deléham, May 26 2015
Row sums of absolute values of terms give A000629. - Yahia DJEMMADA, Aug 16 2016
This is the triangle of connection constants for expressing the monomial polynomials (-x)^n as a linear combination of the basis polynomials {binomial(x+n,n)}n>=0, that is, (-x)^n = Sum_{k = 0..n} T(n,k)*binomial(x+k,k). Cf. A145901. - Peter Bala, Jun 06 2019
Row sums for n > 0 are zero. - Shel Kaphan, May 14 2024
The Akiyama-Tanigawa algorithm applied to a sequence yields the same result as the Stirling-Bernoulli Transform applied to the same sequence. See Philippe Deléham's comment of May 26 2015. - Shel Kaphan, May 16 2024

Examples

			y = 1/(1+exp(-x))
y^(0) = y
y^(1) = y-y^2
y^(2) = y-3*y^2+2*y^3
y^(3) = y-7*y^2+12*y^3-6*y^4
Triangle begins :
n\k 0     1     2     3     4     5    6
----------------------------------------
0:  1
1:  1    -1
2:  1    -3     2
3:  1    -7    12    -6
4:  1   -15    50   -60    24
5:  1   -31   180  -390   360  -120
6:  1   -63   602 -2100  3360 -2520  720
7:  1  -127 ... - Reformatted by _Philippe Deléham_, May 26 2015
Change of basis constants: x^4 = 1 - 15*binomial(x+1,1) + 50*binomial(x+2,2) - 60*binomial(x+3,3) + 24*binomial(x+4,4). - _Peter Bala_, Jun 06 2019
		

Crossrefs

Programs

  • Maple
    A163626 := (n, k) -> add((-1)^j*binomial(k, j)*(j+1)^n, j = 0..k):
    for n from 0 to 6 do seq(A163626(n, k), k = 0..n) od; # Peter Luschny, Sep 21 2017
  • Mathematica
    Derivative[0][y][x] = y[x]; Derivative[1][y][x] = y[x]*(1-y[x]);
    Derivative[n_][y][x] := Derivative[n][y][x] = D[Derivative[n-1][y][x], x];
    row[n_] := CoefficientList[Derivative[n][y][x], y[x]] // Rest;
    Table[row[n], {n, 0, 9}] // Flatten
    (* or *) Table[(-1)^k*k!*StirlingS2[n+1, k+1], {n, 0, 9}, {k, 0, n}] // Flatten
    (* Jean-François Alcover, Dec 16 2014 *)
  • Python
    from sympy.core.cache import cacheit
    @cacheit
    def T(n, k):return 1 if n==0 and k==0 else 0 if k>n or k<0 else (k + 1)*T(n - 1, k) - k*T(n - 1, k - 1)
    for n in range(51): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Sep 11 2017

Formula

T(n, k) = (-1)^k*k!*Stirling2(n+1, k+1). - Jean-François Alcover, Dec 16 2014
T(n, k) = (k+1)*T(n-1,k) - k*T(n-1,k-1), T(0,0) = 1, T(n,k) = 0 if k>n or if k<0. - Philippe Deléham, May 29 2015
Worpitzky's representation of the Bernoulli numbers B(n, 1) = Sum_{k = 0..n} T(n,k)/(k+1) = A164555(n)/A027642(n) (Bernoulli numbers). - Philippe Deléham, May 29 2015
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(k, j)*(j+1)^n. - Peter Luschny, Sep 21 2017
Let W_n(x) be the row polynomials of this sequence and F_n(x) the row polynomials of A278075. Then W_n(1 - x) = F_n(x). Also Integral_{x=0..1} U_n(x) = Bernoulli(n, 1) for U in {W, F}. - Peter Luschny, Aug 10 2021

A007047 Number of chains in power set of n-set.

Original entry on oeis.org

1, 3, 11, 51, 299, 2163, 18731, 189171, 2183339, 28349043, 408990251, 6490530291, 112366270379, 2107433393523, 42565371881771, 921132763911411, 21262618727925419, 521483068116543603, 13542138653027381291, 371206349277313644531
Offset: 0

Views

Author

N. J. A. Sloane, Roger B. Nelsen

Keywords

Comments

Stirling transform of A052849(n-1) = [1,2,4,12,48,...] is a(n-1) =[1,3,11,51,299,...]. - Michael Somos, Mar 04 2004
It is interesting to note that a chain in the power set of a set X can be thought of as a fuzzy subset of X and conversely. Chains originating with empty set are fuzzy subsets with empty core and those chains not ending with the whole set are with support strictly contained in X. - Venkat Murali (v.murali(AT)ru.ac.za), May 18 2005
Equals the binomial transform of A000629: (1, 2, 6, 26, 150, 1082, ...) and the double binomial transform of A000670: (1, 1, 3, 13, 75, 541, ...). - Gary W. Adamson, Aug 04 2009
Row sums of A038719. - Peter Bala, Jul 09 2014
Also the number of restricted barred preferential arrangements of an n-set having two bars, where one fixed section is a free section and the other two sections are restricted sections. - Sithembele Nkonkobe, Jun 16 2015
The number of all predictable outcomes of a race between a given number registered competitors, where clean finishes, dead heats (ties), disqualifications, cancellations and their combinations are all counted. (11 outcomes for two competitors, 51 for three, 299 for four, etc.. Example for two competitors shown below.) - Gergely Földvári, Jul 28 2024

Examples

			If there are two registered competitors, A and B, in a race, the total number of predictable outcomes counting all possibilities of clean finishes (f), dead heats (t), disqualifications (d), cancellations (c) and their combinations is 11 (eleven). Here is the breakdown: AfBf, BfAf, AtBt, AfBd, AfBc, BfAd, BfAc, AdBd, AcBc, AdBc, AcBd. - _Gergely Földvári_, Jul 28 2024
		

References

  • V. Murali, Counting fuzzy subsets of a finite set, preprint, Rhodes University, Grahamstown 6140, South Africa, 2003.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000629, A000629, A000670. Row sums of A038719.

Programs

  • Haskell
    a007047 = sum . a038719_row  -- Reinhard Zumkeller, Feb 05 2014
  • Maple
    P := proc(n,x) option remember; if n = 0 then 1 else
       (n*x+2*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x);
       expand(%) fi end:
    A007047 := n -> 2^n*subs(x=1/2, P(n,x)):
    seq(A007047(n), n=0..19);  # Peter Luschny, Mar 07 2014
    # second Maple program:
    b:= proc(n) option remember; `if`(n=0, 4,
          add(b(n-j)*binomial(n, j), j=1..n))
        end:
    a:= n-> `if`(n=0, 1, b(n)-1):
    seq(a(n), n=0..21);  # Alois P. Heinz, Feb 07 2020
  • Mathematica
    Table[LerchPhi[1/2, -n, 2]/2, {n, 0, 10}] (* Vladimir Reshetnikov, Feb 16 2011 *)
    Table[2*PolyLog[-n, 1/2] - 1 , {n, 0, 19}] (* Jean-François Alcover, Aug 14 2013 *)
    With[{nn=20},CoefficientList[Series[Exp[2x]/(2-Exp[x]),{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Dec 08 2015 *)
    Table[-(-1)^k HurwitzLerchPhi[2, -k, -1], {k, 0, 30}] (* Federico Provvedi,Sep 05 2020 *)
    a[n_]:= a[n] = 2^n + Sum[Binomial[n, k]*a[k], {k, 0, n-1}]; Table[a[n], {n, 0, 20}] (* Rajesh Kumar Mohapatra, Jul 02 2025 *)
    a[0] = 1; a[n_]:= a[n] = 2^n + Sum[Binomial[n, k]*a[n-k], {k, 1, n}]; Table[a[n], {n, 0, 20}] (* Rajesh Kumar Mohapatra, Jul 02 2025 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(subst((y+1)^2/(1-y),y,exp(x+x*O(x^n))-1),n));
    
  • PARI
    my(x='x+O('x^66)); Vec(serlaplace(exp(2*x)/(2-exp(x)))) \\ Joerg Arndt, Aug 14 2013
    

Formula

E.g.f.: exp(2*x)/(2-exp(x)).
a(n) = Sum_{k>=1} (k+1)^n/2^k = 2*A000629(n)-1. - Benoit Cloitre, Sep 08 2002
a(n) = one less than sum of quotients with numerator 4 times (n!)((k_1 + k_2 + ... + k_n)!) and with denominator (k_1!k_2!...k_n!)(1!^k_1 2!^k_2...n!^k_n) where the sum is taken over all partitions 1*k_1 + 2*k_2 + ... + n*k_n = n. E.g. a(1) = 3 because the membership value of x to {x} is either 1, alpha with 0 < alpha < 1 or 0. a(2) = 11 since the membership values x and y to {x, y} are 1 >= alpha >= beta >= 0 for {empty set, x, y} in that order or {empty set, y, x} exercising all possible > or = . - Venkat Murali (v.murali(AT)ru.ac.za), May 18 2005
G.f.: 1/Q(0), where Q(k) = 1 - 3*x*(k+1) - 2*x^2*(k+1)*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 02 2013
a(n) ~ 2*n! / (log(2))^(n+1). - Vaclav Kotesovec, Sep 27 2017
a(n) = 4 * A000670(n) - 1 for n > 0. - Alois P. Heinz, Feb 07 2020
a(n) = -(-1)^n Phi(2,-n,-1), where Phi(z,s,a) is the Lerch zeta function. - Federico Provvedi, Sep 05 2020
a(n) = 1 + 2 * Sum_{k=0..n-1} (-1)^(n-k-1) * binomial(n,k) * a(k). - Ilya Gutkovskiy, Apr 28 2021
From Rajesh Kumar Mohapatra, Jul 02 2025: (Start)
a(n) = 2*A000629(n) - 1.
a(n) = 2^n + Sum_{k=0..n-1} binomial(n,k) * a(k).
a(n) = 2^n + Sum_{k=1..n} binomial(n,k) * a(n-k), a(0) = 1. (End)

A002050 Number of simplices in barycentric subdivision of n-simplex.

Original entry on oeis.org

0, 1, 5, 25, 149, 1081, 9365, 94585, 1091669, 14174521, 204495125, 3245265145, 56183135189, 1053716696761, 21282685940885, 460566381955705, 10631309363962709, 260741534058271801, 6771069326513690645
Offset: 0

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Author

Keywords

Comments

Stirling transform of A052849(n)=[1,4,12,48,240,...] is a(n)=[1,5,25,149,1081,...]. - Michael Somos, Mar 04 2004
Stirling transform of A000142(n-1)=[0,1,2,6,24,...] is a(n-1)=[0,1,5,25,149,...]. - Michael Somos, Mar 04 2004
Stirling transform of 2*A005359(n-1)=[1,0,4,0,48,0,...] is a(n-1)=[1,1,5,25,149,...]. - Michael Somos, Mar 04 2004
"Stirling-Bernoulli transform" of A000225. - Paul Barry, Apr 20 2005
a(n) is the number of nonempty words that can be formed from an alphabet of nonempty subsets of [n] so that the letters in each word are pairwise disjoint. - Geoffrey Critzer, Apr 12 2009
Row sums of A053440. - Peter Bala, Jul 12 2014
Conjecture: Let k be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 9 we obtain the sequence [0, 1, 5, 7, 5, 1, 5, 4, 5, 7, 5, 1, 5, 4, 5, 7, 5, 1, 5, 4, 5, 7, 5, ...], with an apparent period of 6 = phi(9) beginning at a(5). - Peter Bala, Aug 03 2023

References

  • R. Austin, R. K. Guy, and R. Nowakowski, unpublished notes, circa 1987.
  • Ulrike Sattler, Decidable classes of formal power series with nice closure properties, Diplomarbeit im Fach Informatik, Univ. Erlangen - Nuernberg, Jul 27 1994
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A000629, A000670, A002050, A052856, A076726 are all more-or-less the same sequence. - N. J. A. Sloane, Jul 04 2012
A diagonal of the triangle in A241168. Row sums of A053440.

Programs

  • Mathematica
    Table[Sum[Binomial[n, i]*Sum[StirlingS2[i, k]*k!, {k, 1, i}], {i, 1, n}], {n, 0, 20}] (* Geoffrey Critzer, Apr 12 2009 *)
    With[{nn=20},CoefficientList[Series[(Exp[2x]-Exp[x])/(2-Exp[x]),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, May 28 2013 *)
    a[0] = 0; a[n_] := 2*Sum[k!*StirlingS2[n, k], {k, 2, n}] + 1; Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Sep 27 2013, after Vladimir Kruchinin *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(subst((y+y^2)/(1-y),y,exp(x+x*O(x^n))-1),n));

Formula

E.g.f.: (exp(2x)-exp(x))/(2-exp(x)).
a(n) = A000629(n) - 1.
a(n) = Sum_{k=0..n} (-1)^(n-k)k!*S2(n, k)(2^k-1). - Paul Barry, Apr 20 2005
a(n) = Sum_{k=1...n} binomial(n,k)*A000670(k). - Geoffrey Critzer, Apr 12 2009
a(n) ~ n!/log(2)^(n+1). - Vaclav Kotesovec, Jul 29 2013
a(n) = 1 + 2*Sum_{k=2..n} k!*Stirling2(n,k), n > 0, a(0)=1. - Vladimir Kruchinin, Sep 27 2013
G.f.: T(0)/(1-2*x) - 1/(1-x), where T(k) = 1 - 2*x^2*(k+1)^2/(2*x^2*(k+1)^2 - (1 - 2*x - 3*x*k)*(1 - 5*x - 3*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 29 2013
G.f.: Sum_{j>=1} j!*x^j / Product_{k=0..j} (1 - (k + 1)*x). - Ilya Gutkovskiy, Apr 04 2019

Extensions

More terms from James Sellers, Aug 22 2000

A326703 BII-numbers of chains of nonempty sets.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 16, 17, 24, 32, 34, 40, 64, 65, 66, 68, 69, 70, 72, 80, 81, 88, 96, 98, 104, 128, 256, 257, 384, 512, 514, 640, 1024, 1025, 1026, 1028, 1029, 1030, 1152, 1280, 1281, 1408, 1536, 1538, 1664, 2048, 2056, 2176, 4096, 4097, 4104, 4112, 4113, 4120
Offset: 1

Views

Author

Gus Wiseman, Jul 21 2019

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. We define the set-system with BII-number n to be obtained by taking the binary indices of each binary index of n. Every finite set of finite nonempty sets has a different BII-number. For example, 18 has reversed binary expansion (0,1,0,0,1), and since the binary indices of 2 and 5 are {2} and {1,3} respectively, it follows that the BII-number of {{2},{1,3}} is 18.
Elements of a set-system are sometimes called edges. In a chain of sets, every edge is a subset or superset of every other edge.

Examples

			The sequence of all chains of nonempty sets together with their BII-numbers begins:
    0: {}
    1: {{1}}
    2: {{2}}
    4: {{1,2}}
    5: {{1},{1,2}}
    6: {{2},{1,2}}
    8: {{3}}
   16: {{1,3}}
   17: {{1},{1,3}}
   24: {{3},{1,3}}
   32: {{2,3}}
   34: {{2},{2,3}}
   40: {{3},{2,3}}
   64: {{1,2,3}}
   65: {{1},{1,2,3}}
   66: {{2},{1,2,3}}
   68: {{1,2},{1,2,3}}
   69: {{1},{1,2},{1,2,3}}
   70: {{2},{1,2},{1,2,3}}
   72: {{3},{1,2,3}}
   80: {{1,3},{1,2,3}}
   81: {{1},{1,3},{1,2,3}}
   88: {{3},{1,3},{1,2,3}}
   96: {{2,3},{1,2,3}}
   98: {{2},{2,3},{1,2,3}}
		

Crossrefs

Chains of nonempty sets are counted by A000629.
MM-numbers of chains of multisets are A318991.
BII-numbers of antichains of nonempty sets are A326704.

Programs

  • Mathematica
    bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    stableQ[u_,Q_]:=!Apply[Or,Outer[#1=!=#2&&Q[#1,#2]&,u,u,1],{0,1}];
    Select[Range[0,100],stableQ[bpe/@bpe[#],!SubsetQ[#1,#2]&&!SubsetQ[#2,#1]&]&]
  • Python
    from itertools import chain, count, combinations, islice
    from sympy.combinatorics.subsets import ksubsets
    def subsets(x):
        for i in range(1,len(x)):
            for j in ksubsets(x,i):
                yield(list(j))
    def a_gen(): #generator of terms
        yield 0
        for n in count(1):
            t,v,j = [[]],[],0
            for i in chain.from_iterable(combinations(range(1, n+1), r) for r in range(n+1)):
                if n in i:
                    t[j].append([list(i)])
            while n:
                t.append([])
                for i in t[j]:
                    if len(i[-1]) > 1:
                        for k in list(subsets(i[-1])):
                            t[j+1].append(i.copy()+[k])
                if len(t[j+1]) < 1:
                    break
                j += 1
            for j in chain.from_iterable(t):
                v.append(sum(2**(sum(2**(m-1) for m in k)-1) for k in j))
            yield from sorted(v)
    A326703_list = list(islice(a_gen(), 55)) # John Tyler Rascoe, Jun 07 2024

A052882 A simple grammar: rooted ordered set partitions.

Original entry on oeis.org

0, 1, 2, 9, 52, 375, 3246, 32781, 378344, 4912515, 70872610, 1124723193, 19471590876, 365190378735, 7376016877334, 159620144556645, 3684531055645648, 90366129593683035, 2346673806524446218, 64325158601880061137, 1856031746386568222660, 56231443721132068265415
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Recurrence (see Mathematica line) is similar to that for Genocchi numbers A001469. - Wouter Meeussen, Jan 09 2001
Stirling transform of A024167(n) = [ 1, 1, 5, 14, 94, ...] is a(n) = [ 1, 2, 9, 52, 375, ...]. Stirling transform of a(n) = [ 0, 2, 9, 52, 375, ...] is A087301(n+1) = [ 0, 2, 3, 20, ...]. - Michael Somos, Mar 04 2004
Starting with offset 1 = the right border of triangle A208744. - Gary W. Adamson, Mar 05 2012
a(n) is the number of ordered set partitions of {1,2,...,n} such that the first block is a singleton. - Geoffrey Critzer, Jul 22 2013
Ramanujan gives a method of finding a continued fraction of the solution x of an equation 1 = x + a2*x^2 + ... and uses log(2) as the solution of 1 = x + x^2/2 + x^3/6 + ... as an example giving the sequence of simplified convergents as 0/1, 1/1, 2/3, 9/13, 52/75, 375/541, ... of which the sequence of numerators is this sequence while A000670 is the denominators. - Michael Somos, Jun 19 2015

Examples

			G.f. = x + 2*x^2 + 9*x^3 + 52*x^4 + 375*x^5 + 3246*x^6 + 32781*x^7 + ...
		

References

  • S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay 1957 Vol. 1, see page 19.

Crossrefs

Programs

  • Maple
    spec := [S,{C=Sequence(B),B=Set(Z,1 <= card),S=Prod(Z,C)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
    with(combinat): a:=n-> add(add(add((-1)^(k-i)*binomial(k, i)*i^(n-1), i=0..n-1), k=0..n-1), m=0..n-1): seq(a(n), n=0..20); # Zerinvary Lajos, Jun 03 2007
    # next Maple program:
    b:= proc(n, k) option remember;
         `if`(n<1, k!, k*b(n-1, k)+b(n-1, k+1))
        end:
    a:= n-> b(n-1, 0)*n:
    seq(a(n), n=0..25);  # Alois P. Heinz, Apr 15 2023
  • Mathematica
    a[1] := 1; a[n_] := a[n]=Sum[ Binomial[n, m] a[ n-m], {m, 1, n-1}]
    Range[0, 30]!* CoefficientList[Series[x/(2 - Exp[x]),{x, 0, 30}], x] (* Vincenzo Librandi, Dec 06 2012 *)
    a[ n_] := If[ n < 2, Boole[n == 1], n PolyLog[ 1 - n, 1/2] / 2]; (* Michael Somos, Jun 19 2015 *)
    a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ x / (2 - Exp@x), {x, 0, n}]]; (* Michael Somos, Jun 19 2015 *)
    Fubini[n_, r_] := Sum[k!*Sum[(-1)^(i+k+r)*(i+r)^(n-r)/(i!*(k-i-r)!), {i, 0, k-r}], {k, r, n}]; Fubini[0, 1] = 1; a[n_] := n*Fubini[n-1, 1]; Table[ a[n], {n, 0, 18}] (* Jean-François Alcover, Mar 30 2016 *)
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( subst( x / (1 - y), y, exp(x + x*O(x^n)) - 1), n))};
    
  • Python
    from math import factorial
    from sympy.functions.combinatorial.numbers import stirling
    def A052882(n): return n*sum(factorial(k)*stirling(n-1,k) for k in range(n)) # Chai Wah Wu, Apr 15 2023

Formula

E.g.f.: x / (2 - exp(x)).
a(n) = n * A000670(n-1) if n>0.
a(n) = (1/2)*sum(k=0, n-1, B_k*A000629(k)*binomial(n, k)) where B_k is the k-th Bernoulli number. - Benoit Cloitre, Oct 19 2005
a(n) ~ n!/(2*(log(2))^n). - Vaclav Kotesovec, Aug 09 2013
a(0) = 0, a(1) = 1; a(n) = n! * [x^n] exp(x)*Sum_{k=1..n-1} a(k)*x^k/k!. - Ilya Gutkovskiy, Oct 17 2017

A069321 Stirling transform of A001563: a(0) = 1 and a(n) = Sum_{k=1..n} Stirling2(n,k)*k*k! for n >= 1.

Original entry on oeis.org

1, 1, 5, 31, 233, 2071, 21305, 249271, 3270713, 47580151, 760192505, 13234467511, 249383390393, 5057242311031, 109820924003705, 2542685745501751, 62527556173577273, 1627581948113854711, 44708026328035782905, 1292443104462527895991, 39223568601129844839353
Offset: 0

Views

Author

Karol A. Penson, Mar 14 2002

Keywords

Comments

The number of compatible bipartitions of a set of cardinality n for which at least one subset is not underlined. E.g., for n=2 there are 5 such bipartitions: {1 2}, {1}{2}, {2}{1}, {1}{2}, {2}{1}. A005649 is the number of bipartitions of a set of cardinality n. A000670 is the number of bipartitions of a set of cardinality n with none of the subsets underlined. - Kyle Petersen, Mar 31 2005
a(n) is the cardinality of the image set summed over "all surjections". All surjections means: onto functions f:{1, 2, ..., n} -> {1, 2, ..., k} for every k, 1 <= k <= n. a(n) = Sum_{k=1..n} A019538(n, k)*k. - Geoffrey Critzer, Nov 12 2012
From Gus Wiseman, Jan 15 2022: (Start)
For n > 1, also the number of finite sequences of length n + 1 covering an initial interval of positive integers with at least two adjacent equal parts, or non-anti-run patterns, ranked by the intersection of A348612 and A333217. The complement is counted by A005649. For example, the a(3) = 31 patterns, grouped by sum, are:
(1111) (1222) (1122) (1112) (1233) (1223)
(2122) (1221) (1121) (1332) (1322)
(2212) (2112) (1211) (2133) (2213)
(2221) (2211) (2111) (2331) (2231)
(1123) (3312) (3122)
(1132) (3321) (3221)
(2113)
(2311)
(3112)
(3211)
Also the number of ordered set partitions of {1,...,n + 1} with two successive vertices together in some block.
(End)

Crossrefs

The complement is counted by A005649.
A version for permutations of prime indices is A336107.
A version for factorizations is A348616.
Dominated (n > 1) by A350252, complement A345194, compositions A345192.
A000670 = patterns, ranked by A333217.
A001250 = alternating permutations, complement A348615.
A003242 = anti-run compositions, ranked by A333489.
A019536 = necklace patterns.
A226316 = patterns avoiding (1,2,3), weakly A052709, complement A335515.
A261983 = not-anti-run compositions, ranked by A348612.
A333381 = anti-runs of standard compositions.

Programs

  • Maple
    b:= proc(n) option remember; `if`(n=0, 1,
          add(b(n-j)*binomial(n, j), j=1..n))
        end:
    a:= n-> `if`(n=0, 2, b(n+1)-b(n))/2:
    seq(a(n), n=0..30);  # Alois P. Heinz, Feb 02 2018
  • Mathematica
    max = 20; t = Sum[n^(n - 1)x^n/n!, {n, 1, max}]; Range[0, max]!CoefficientList[Series[D[1/(1 - y(Exp[x] - 1)), y] /. y -> 1, {x, 0, max}], x] (* Geoffrey Critzer, Nov 12 2012 *)
    Prepend[Table[Sum[StirlingS2[n, k]*k*k!, {k, n}], {n, 18}], 1] (* Michael De Vlieger, Jan 03 2016 *)
    a[n_] := (PolyLog[-n-1, 1/2] - PolyLog[-n, 1/2])/4; a[0] = 1; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 30 2016 *)
    allnorm[n_]:=If[n<=0,{{}},Function[s,Array[Count[s,y_/;y<=#]+1&,n]]/@Subsets[Range[n-1]+1]];
    Table[Length[Select[Join@@Permutations/@allnorm[n],MemberQ[Differences[#],0]&]],{n,0,8}] (* Gus Wiseman, Jan 15 2022 *)
  • PARI
    {a(n)=polcoeff(1+sum(m=1, n, (2*m-1)!/(m-1)!*x^m/prod(k=1, m, 1+(m+k-1)*x+x*O(x^n))), n)} \\ Paul D. Hanna, Oct 28 2013

Formula

Representation as an infinite series: a(0) = 1 and a(n) = Sum_{k>=2} (k^n*(k-1)/(2^k))/4 for n >= 1. This is a Dobinski-type summation formula.
E.g.f.: (exp(x) - 1)/((2 - exp(x))^2).
a(n) = (1/2)*(A000670(n+1) - A000670(n)).
O.g.f.: 1 + Sum_{n >= 1} (2*n-1)!/(n-1)! * x^n / (Product_{k=1..n} (1 + (n + k - 1)*x)). - Paul D. Hanna, Oct 28 2013
a(n) = (A000629(n+1) - A000629(n))/4. - Benoit Cloitre, Oct 20 2002
a(n) = A232472(n-1)/2. - Vincenzo Librandi, Jan 03 2016
a(n) ~ n! * n / (4 * (log(2))^(n+2)). - Vaclav Kotesovec, Jul 01 2018
a(n > 0) = A000607(n + 1) - A005649(n). - Gus Wiseman, Jan 15 2022

A154921 Triangle read by rows, T(n, k) = binomial(n, k) * Sum_{j=0..n-k} E(n-k, j)*2^j, where E(n, k) are the Eulerian numbers A173018(n, k), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 13, 9, 3, 1, 75, 52, 18, 4, 1, 541, 375, 130, 30, 5, 1, 4683, 3246, 1125, 260, 45, 6, 1, 47293, 32781, 11361, 2625, 455, 63, 7, 1, 545835, 378344, 131124, 30296, 5250, 728, 84, 8, 1, 7087261, 4912515, 1702548, 393372, 68166, 9450, 1092, 108, 9, 1
Offset: 0

Views

Author

Mats Granvik, Jan 17 2009

Keywords

Comments

Previous name: Matrix inverse of A154926.
A000670 appears in the first column. A052882 appears in the second column. A000027 and A045943 appear as diagonals. An alternative to calculating the matrix inverse of A154926 is to move the term in the lower right corner to a position in the same column and calculate the determinant instead, which yields the same answer.
Matrix inverse of (2*I - P), where P is Pascal's triangle and I the identity matrix. See A162312 for the matrix inverse of (2*P - I) and some general remarks about arrays of the form M(a) := (I - a*P)^-1 and their connection with weighted sums of powers of integers. The present array equals (1/2)*M(1/2). - Peter Bala, Jul 01 2009
From Mats Granvik, Aug 11 2009: (Start)
The values in this triangle can be seen as permanents of the Pascal triangle analogous to the method in the Redheffer matrix. The elements satisfy (T(n,k)/T(n,k-1))*k = (T(n-1,k)/T(n,k))*n which converges to log(2) as n->oo and k->0. More generally to calculate log(x) multiply the negative values in A154926 by 1/(x-1) and calculate the matrix inverse. Then (T(n,k)/T(n,k-1))*k and (T(n-1,k)/T(n,k))*n in the resulting triangle converge to log(x).
This method for calculating log(x) converges faster than the Taylor series when x is greater than 5 or so. See chapter on Taylor series in Spiegel for comparison. (End)
Exponential Riordan array [1/(2-exp(x)),x]. - Paul Barry, Apr 06 2011
T(n,k) is the number of ordered set partitions of {1,2,...,n} such that the first block contains k elements. For k=0 the first block contains arbitrarily many elements. - Geoffrey Critzer, Jul 22 2013
A natural (signed) refinement of these polynomials is given by the Appell sequence e.g.f. e^(xt)/ f(t) = exp[tP.(x)] with the formal Taylor series f(x) = 1 + x[1] x + x[2] x^2/2! + ... and with raising operator R = x - d[log(f(D)]/dD (cf. A263634). - Tom Copeland, Nov 06 2015

Examples

			From _Peter Bala_, Jul 01 2009: (Start)
Triangle T(n, k) begins:
n\k|     0     1     2     3     4     5     6
==============================================
0  |     1
1  |     1     1
2  |     3     2     1
3  |    13     9     3     1
4  |    75    52    18     4     1
5  |   541   375   130    30     5     1
6  |  4683  3246  1125   260    45     6     1
...
(End)
From _Mats Granvik_, Aug 11 2009: (Start)
Row 4 equals 75,52,18,4,1 because permanents of:
  1,0,0,0,1  1,0,0,0,0  1,0,0,0,0  1,0,0,0,0  1,0,0,0,0
  1,1,0,0,0  1,1,0,0,1  1,1,0,0,0  1,1,0,0,0  1,1,0,0,0
  1,2,1,0,0  1,2,1,0,0  1,2,1,0,1  1,2,1,0,0  1,2,1,0,0
  1,3,3,1,0  1,3,3,1,0  1,3,3,1,0  1,3,3,1,1  1,3,3,1,0
  1,4,6,4,0  1,4,6,4,0  1,4,6,4,0  1,4,6,4,0  1,4,6,4,1
are:
     75         52         18          4          1
(End)
		

References

  • Murray R. Spiegel, Mathematical handbook, Schaum's Outlines, p. 111.

Crossrefs

Cf. A000629 (row sums), A000670, A007047, A052822 (column 1), A052841 (alt. row sums), A080253, A162312, A162313.
Cf. A263634, A099880 (T(2n,n)).

Programs

  • Maple
    A154921_row := proc(n) local i,p; p := proc(n,x) option remember; local k;
    if n = 0 then 1 else add(p(k,0)*binomial(n,k)*(1+x^(n-k)),k=0..n-1) fi end:
    seq(coeff(p(n,x),x,i),i=0..n) end: for n from 0 to 5 do A154921_row(n) od;
    # Peter Luschny, Jul 15 2012
    T := (n,k) -> binomial(n,k)*add(combinat:-eulerian1(n-k,j)*2^j, j=0..n-k):
    seq(print(seq(T(n,k), k=0..n)),n=0..6); # Peter Luschny, Feb 07 2015
    # third Maple program:
    b:= proc(n) b(n):= `if`(n=0, 1, add(b(n-j)/j!, j=1..n)) end:
    T:= (n, k)-> n!/k! *b(n-k):
    seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Feb 03 2019
    # fourth Maple program:
    p := proc(n, m) option remember; if n = 0 then 1 else
        (m + x)*p(n - 1, m) + (m + 1)*p(n - 1, m + 1) fi end:
    row := n -> local k; seq(coeff(p(n, 0), x, k), k = 0..n):
    for n from 0 to 6 do row(n) od;  # Peter Luschny, Jun 23 2023
  • Mathematica
    nn = 8; a = Exp[x] - 1;
    Map[Select[#, # > 0 &] &,
      Transpose[
       Table[Range[0, nn]! CoefficientList[
    Series[x^n/n!/(1 - a), {x, 0, nn}], x], {n, 0, nn}]]] // Grid (* Geoffrey Critzer, Jul 22 2013 *)
    E1[n_ /; n >= 0, 0] = 1; E1[n_, k_] /; k < 0 || k > n = 0; E1[n_, k_] := E1[n, k] = (n - k) E1[n - 1, k - 1] + (k + 1) E1[n - 1, k];
    T[n_, k_] := Binomial[n, k] Sum[E1[n - k, j] 2^j, {j, 0, n - k}];
    Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 30 2018, after Peter Luschny *)
  • Sage
    @CachedFunction
    def Poly(n, x):
        return 1 if n == 0 else add(Poly(k,0)*binomial(n,k)*(x^(n-k)+1) for k in range(n))
    R = PolynomialRing(ZZ, 'x')
    for n in (0..6): print(R(Poly(n,x)).list()) # Peter Luschny, Jul 15 2012

Formula

From Peter Bala, Jul 01 2009: (Start)
TABLE ENTRIES
(1) T(n,k) = binomial(n,k)*A000670(n-k).
GENERATING FUNCTION
(2) exp(x*t)/(2-exp(t)) = 1 + (1+x)*t + (3+2*x+x^2)*t^2/2! + ....
PROPERTIES OF THE ROW POLYNOMIALS
The row generating polynomials R_n(x) form an Appell sequence. They appear in the study of the poset of power sets [Nelsen and Schmidt].
The first few values are R_0(x) = 1, R_1(x) = 1+x, R_2(x) = 3+2*x+x^2 and R_3(x) = 13+9*x+3*x^2+x^3.
The row polynomials may be recursively computed by means of
(3) R_n(x) = x^n + Sum_{k = 0..n-1} binomial(n,k)*R_k(x).
Explicit formulas include
(4) R_n(x) = (1/2)*Sum_{k >= 0} (1/2)^k*(x+k)^n,
(5) R_n(x) = Sum_{j = 0..n} Sum_{k = 0..j} (-1)^(j-k)*binomial(j,k) *(x+k)^n,
and
(6) R_n(x) = Sum_{j = 0..n} Sum_{k = j..n} k!*Stirling2(n,k) *binomial(x,k-j).
SUMS OF POWERS OF INTEGERS
The row polynomials satisfy the difference equation
(7) 2*R_m(x) - R_m(x+1) = x^m,
which easily leads to the evaluation of the weighted sums of powers of integers
(8) Sum_{k = 1..n-1} (1/2)^k*k^m = 2*R_m(0) - (1/2)^(n-1)*R_m(n).
For example, m = 2 gives
(9) Sum_{k = 1..n-1} (1/2)^k*k^2 = 6 - (1/2)^(n-1)*(n^2+2*n+3).
More generally we have
(10) Sum_{k=0..n-1} (1/2)^k*(x+k)^m = 2*R_m(x) - (1/2)^(n-1)*R_m(x+n).
RELATIONS WITH OTHER SEQUENCES
Sequences in the database given by particular values of the row polynomials are
(11) A000670(n) = R_n(0)
(12) A052841(n) = R_n(-1)
(13) A000629(n) = R_n(1)
(14) A007047(n) = R_n(2)
(15) A080253(n) = 2^n*R_n(1/2).
This last result is the particular case (x = 0) of the result that the polynomials 2^n*R_n(1/2+x/2) are the row generating polynomials for A162313.
The above formulas should be compared with those for A162312. (End)
From Peter Luschny, Jul 15 2012: (Start)
(16) A151919(n) = R_n(1/3)*3^n*(-1)^n
(17) A052882(n) = [x^1] R_n(x)
(18) A045943(n) = [x^(n-1)] R_n+1(x)
(19) A099880(n) = [x^n] R_2n(x). (End)
The coefficients in ascending order of x^i of the polynomials p{0}(x) = 1 and p{n}(x) = Sum_{k=0..n-1} binomial(n,k)*p{k}(0)*(1+x^(n-k)). - Peter Luschny, Jul 15 2012

Extensions

New name by Peter Luschny, Feb 07 2015

A201354 Expansion of e.g.f. exp(x) / (4 - 3*exp(x)).

Original entry on oeis.org

1, 4, 28, 292, 4060, 70564, 1471708, 35810212, 995827420, 31153998244, 1082931514588, 41407678132132, 1727226633730780, 78051253062575524, 3798351192214837468, 198049421007186054052, 11014905131587945490140, 650903915009792820650404, 40726453234725158535472348
Offset: 0

Views

Author

Paul D. Hanna, Nov 30 2011

Keywords

Examples

			E.g.f.: E(x) = 1 + 4*x + 28*x^2/2! + 292*x^3/3! + 4060*x^4/4! + 70564*x^5/5! + ...
O.g.f.: A(x) = 1 + 4*x + 28*x^2 + 292*x^3 + 4060*x^4 + 70564*x^5 + ...
where A(x) = 1 + 4*x/(1+x) + 2!*4^2*x^2/((1+x)*(1+2*x)) + 3!*4^3*x^3/((1+x)*(1+2*x)*(1+3*x)) + 4!*4^4*x^4/((1+x)*(1+2*x)*(1+3*x)*(1+4*x)) + ...
		

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 20); Coefficients(R!(Laplace( 1/(4*Exp(-x) -3) ))); // G. C. Greubel, Jun 08 2020
    
  • Maple
    seq(coeff(series(1/(4*exp(-x) -3), x, n+1)*n!, x, n), n = 0..20); # G. C. Greubel, Jun 08 2020
  • Mathematica
    Table[Sum[(-1)^(n-k)*4^k*StirlingS2[n,k]*k!,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 13 2013 *)
  • PARI
    {a(n)=n!*polcoeff(exp(x+x*O(x^n))/(4 - 3*exp(x+x*O(x^n))), n)}
    
  • PARI
    {a(n)=polcoeff(sum(m=0, n, 4^m*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)}
    
  • PARI
    {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, (-1)^(n-k)*4^k*Stirling2(n, k)*k!)}
    
  • Sage
    [sum( (-1)^(n-j)*4^j*factorial(j)*stirling_number2(n,j) for j in (0..n)) for n in (0..20)] # G. C. Greubel, Jun 08 2020

Formula

O.g.f.: A(x) = Sum_{n>=0} n! * 4^n*x^n / Product_{k=0..n} (1+k*x).
O.g.f.: A(x) = 1/(1 - 4*x/(1-3*x/(1 - 8*x/(1-6*x/(1 - 12*x/(1-9*x/(1 - 16*x/(1-12*x/(1 - 20*x/(1-15*x/(1 - ...))))))))))), a continued fraction.
a(n) = Sum_{k=0..n} (-1)^(n-k) * 4^k * Stirling2(n,k) * k!.
a(n) = 4*A050352(n) for n>0.
a(n) = Sum_{k=0..n} A123125(n,k)*4^k*3^(n-k). - Philippe Deléham, Nov 30 2011
a(n) = log(4/3) * Integral_{x = 0..oo} (ceiling(x))^n * (4/3)^(-x) dx. - Peter Bala, Feb 06 2015
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 8*x*(k+1)/(8*x*(k+1) - 1 + 6*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
a(n) ~ n! / (3*(log(4/3))^(n+1)). - Vaclav Kotesovec, Jun 13 2013
a(n) = 1 + 3 * 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) = -4*Sum_{k=1..n} (-1)^k * binomial(n,k) * a(n-k).
a(0) = 1; a(n) = 4*a(n-1) + 3*Sum_{k=1..n-1} binomial(n-1,k) * a(n-k). (End)
a(n) = (4/3)*A032033(n) - (1/3)*0^n. - Seiichi Manyama, Dec 21 2023
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