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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A317855 Decimal expansion of a constant related to the asymptotics of A122400.

Original entry on oeis.org

3, 1, 6, 1, 0, 8, 8, 6, 5, 3, 8, 6, 5, 4, 2, 8, 8, 1, 3, 8, 3, 0, 1, 7, 2, 2, 0, 2, 5, 8, 8, 1, 3, 2, 4, 9, 1, 7, 2, 6, 3, 8, 2, 7, 7, 4, 1, 8, 8, 5, 5, 6, 3, 4, 1, 6, 2, 7, 2, 7, 8, 2, 0, 7, 5, 3, 7, 6, 9, 7, 0, 5, 9, 2, 1, 9, 3, 0, 4, 6, 1, 1, 2, 1, 9, 7, 5, 7, 4, 6, 8, 5, 4, 9, 7, 8, 4, 5, 9, 3, 2, 4, 2, 2, 7
Offset: 1

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Author

Vaclav Kotesovec, Aug 09 2018

Keywords

Examples

			3.161088653865428813830172202588132491726382774188556341627278...
		

Crossrefs

Programs

  • Mathematica
    r = r /. FindRoot[E^(1/r)/r + (1 + E^(1/r)) * ProductLog[-E^(-1/r)/r] == 0, {r, 3/4}, WorkingPrecision -> 120]; RealDigits[(1 + Exp[1/r])*r^2][[1]]
  • PARI
    r=solve(r=.8,1,exp(1/r)/r + (1+exp(1/r))*lambertw(-exp(-1/r)/r))
    (1+exp(1/r))*r^2 \\ Charles R Greathouse IV, Jun 15 2021

Formula

Equals (1+exp(1/r))*r^2, where r = 0.873702433239668330496568304720719298213992... is the root of the equation exp(1/r)/r + (1+exp(1/r))*LambertW(-exp(-1/r)/r) = 0.

A122399 a(n) = Sum_{k=0..n} k^n * k! * Stirling2(n,k).

Original entry on oeis.org

1, 1, 9, 211, 9285, 658171, 68504709, 9837380491, 1863598406805, 450247033371451, 135111441590583909, 49300373690091496171, 21495577955682021043125, 11037123350952586270549531, 6591700149366720366704735109
Offset: 0

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Author

Vladeta Jovovic, Aug 31 2006

Keywords

Comments

Conjecture: Let p be prime. The sequence obtained by reducing a(n) modulo p for n >= 1 is purely periodic with period p - 1. For example, modulo 7 the sequence becomes [1, 2, 1, 3, 3, 0, 1, 2, 1, 3, 3, 0, ...], with an apparent period of 6. Cf. A338040. - Peter Bala, May 31 2022

Examples

			E.g.f.: A(x) = 1 + x + 9*x^2/2! + 211*x^3/3! + 9285*x^4/4! + 658171*x^5/5! + ...
such that
A(x) = 1 + (exp(x)-1) + (exp(2*x)-1)^2 + (exp(3*x)-1)^3 + (exp(4*x)-1)^4 + ...
The e.g.f. is also given by the series:
A(x) = 1/2 + exp(x)/(1+exp(x))^2 + exp(4*x)/(1+exp(2*x))^3 + exp(9*x)/(1+exp(3*x))^4 + exp(16*x)/(1+exp(4*x))^5 + exp(25*x)/(1+exp(5*x))^6 + ...
or, equivalently,
A(x) = 1/2 + exp(-x)/(1+exp(-x))^2 + exp(-2*x)/(1+exp(-2*x))^3 + exp(-3*x)/(1+exp(-3*x))^4 + exp(-4*x)/(1+exp(-4*x))^5 + exp(-5*x)/(1+exp(-5*x))^6 + ...
		

Crossrefs

Programs

  • Maple
    a := n -> add(k^n*k!*combinat[stirling2](n,k),k=0..n); # Max Alekseyev, Feb 01 2007
  • Mathematica
    Flatten[{1,Table[Sum[k^n*k!*StirlingS2[n,k],{k,0,n}],{n,1,20}]}] (* Vaclav Kotesovec, Jun 21 2013 *)
  • PARI
    {a(n)=polcoeff(sum(m=0, n, m^m*m!*x^m/prod(k=1, m, 1-m*k*x+x*O(x^n))), n)}
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Jan 05 2013
    
  • PARI
    {a(n)=n!*polcoeff(sum(k=0, n, (exp(k*x +x*O(x^n)) - 1)^k), n)}
    for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Oct 26 2014
    
  • PARI
    /* From e.g.f. infinite series: */
    \p100 \\ set precision
    {A=Vec(serlaplace(sum(n=0, 500, 1.*exp(n^2*x +O(x^26))/(1 + exp(n*x +O(x^26)))^(n+1)) ))}
    for(n=0, #A-1, print1(round(A[n+1]), ", ")) \\ Paul D. Hanna, Oct 30 2014

Formula

E.g.f.: Sum_{n >= 0} (exp(n*x) - 1)^n. - Vladeta Jovovic, Sep 03 2006
E.g.f.: Sum_{n>=0} exp(n^2*x) / (1 + exp(n*x))^(n+1). - Paul D. Hanna, Oct 26 2014
E.g.f.: Sum_{n>=0} exp(-n*x) / (1 + exp(-n*x))^(n+1). - Paul D. Hanna, Oct 30 2014
O.g.f.: Sum_{n>=0} n^n * n! * x^n / Product_{k=1..n} (1 - n*k*x). - Paul D. Hanna, Jan 05 2013
Limit n->infinity (a(n)/n!)^(1/n)/n = ((1+exp(1/r))*r^2)/exp(1) = A317855/exp(1) = 1.162899527477400818845..., where r = 0.87370243323966833... is the root of the equation 1/(1+exp(-1/r)) = -r*LambertW(-exp(-1/r)/r). - Vaclav Kotesovec, Jun 21 2013
a(n) ~ c * A317855^n * (n!)^2 / sqrt(n), where c = 0.327628285569869481442286492410507030710253054522608... - Vaclav Kotesovec, Aug 09 2018
Let A(x) = 1 + x + 9*x^2/2! + 211*x^3/3! + ... denote the e.g.f. of the sequence. Let F(x) denote the series reversion of A(x) - 1 = x - 9*x^2/2 + 16*x^3/3 - 205*x^4/4 - 2714*x^5/5 - .... Then both dF/dx = 1 - 9*x + 16*x^2 - 205*x^3 - 2714*x^4 - ... and exp(F(x)) = 1 + x - 4*x^2 + x^3 - 38*x^4 - 606*x^5 - ... have integer coefficients. Note that 1 + series reversion(exp(F(x)) - 1) is the o.g.f. for A122400. - Peter Bala, Aug 09 2022

Extensions

More terms from Max Alekseyev, Feb 01 2007

A179525 G.f.: A(x) = Sum_{n>=0} Product_{k=1..n} ((1+x)^k - 1).

Original entry on oeis.org

1, 1, 2, 7, 33, 197, 1419, 11966, 115575, 1257718, 15223822, 202860828, 2950665011, 46516215168, 790009447590, 14379745626739, 279256447482090, 5763290215111558, 125960271446527241, 2906289188751628643, 70594767279197608011, 1800695322331687800336, 48122711251655255426539, 1344617808976210991187090, 39206731897407002624384182, 1190905492485213830900901986
Offset: 0

Views

Author

Paul D. Hanna, Jul 17 2010

Keywords

Comments

From Vít Jelínek, Feb 12 2012: (Start)
a(n) has the following combinatorial interpretations:
(1) the number of upper-triangular matrices over {0,1} having at least one '1'-entry in each row and having n '1'-entries in total. E.g., for n=2, this corresponds to these two matrices (with zeros represented as dots):
1. .1
.1 .1
(2) the number of upper-triangular matrices over {0,1} that are symmetric with respect to the northeast diagonal, have at least one '1'-entry in each row and column, have no '1'-entry on the northeast diagonal, and have 2n '1'-entries in total. For n=2, those are the two matrices
11. 1...
..1 .1..
..1 ..1.
...1
(3) the number of upper-triangular matrices over {0,1} that are symmetric with respect to the northeast diagonal, have at least one '1'-entry in each row and column, have at least one '1'-entry on the northeast diagonal, and have n '1'-entries on or above the northeast diagonal. For n=2, this corresponds to
11 1..
.1 .1.
..1
(End)
This is an example of Peter Bala's identity (cf. A158690):
Sum_{n>=0} Product_{k=1..n} (q^k - 1) = Sum_{n>=0} q^(-n^2) * Product_{k = 1..n} (q^(2*k-1) - 1) at q = 1 + x. See cross-references for other examples.

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 33*x^4 + 197*x^5 + 1419*x^6 +...
A(x) = 1 + ((1+x)-1) + ((1+x)-1)*((1+x)^2-1) + ((1+x)-1)*((1+x)^2-1)*((1+x)^3-1) +...
Let q = 1+x, then g.f. also equals:
A(x) = 1 + (q-1)/q + (q-1)*(q^3-1)/q^4 + (q-1)*(q^3-1)*(q^5-1)/q^9 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)/q^16 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)*(q^9-1)/q^25 +...
		

Crossrefs

Cf. A207434 (log).

Programs

  • Mathematica
    a[ n_] := SeriesCoefficient[ Sum[ Product[ (1 + x)^j - 1, {j, k}], {k, 0, n}], {x, 0, n}]; (* Michael Somos, Jun 27 2017 *)
  • PARI
    {a(n) = polcoeff(sum(i=0, n, prod(j=1, i, (1+x)^j-1, 1+x*O(x^n))), n)};
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    /* G.f. as a continued fraction: */
    {a(n) = local(CF=1+x*O(x)); for(k=0, n, CF=1/((1+x)^(n-k+1)-((1+x)^(n-k+2)-1)*CF)); polcoeff(1/(1-x*CF), n, x)}
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    {a(n) = local(A=1+x, q=(1+x +x*O(x^n))); A = sum(m=0, n, q^(-m^2)*prod(k=1, m, (q^(2*k-1)-1))); polcoeff(A, n)}
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    /* Sum_{n>=0} n!*Product_{k=0..n-1} [Integral (1+x)^k dx] */
    {a(n) = my(A=1); A = sum(m=0,n, m! * prod(k=0,m-1, intformal((1+x)^k) +x*O(x^n)) );polcoeff(A,n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 16 2016

Formula

G.f.: 1/(1 - ((1+x)-1)/((1+x) - ((1+x)^2-1)/((1+x)^2 - ((1+x)^3-1)/((1+x)^3 - ((1+x)^4-1)/((1+x)^4 - ((1+x)^5-1)/((1+x)^5 -...)))))), (continued fraction) [Paul D. Hanna, Jan 29 2012]
G.f.: Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1) where q = 1+x. [Based on Peter Bala's identity in comments]
Conjecturally, a(n) is asymptotically c*n!*(12/Pi^2)^n, where c=6*sqrt(2)*exp(-Pi^2/24)/Pi^2. - Vít Jelínek, Feb 12 2012 [This is correct: see Hwang and Jin, Table 3, p. 26. - Peter Bala, Jan 31 2021]
G.f.: Q(0), where Q(k)= 1 - (1-(1+x)^(2*k+1))/(1 - (1-(1+x)^(2*k+2))/(1 - (1+x)^(2*k+2) - 1/Q(k+1))); (continued fraction). Conjecture. - Sergei N. Gladkovskii, May 13 2013
From Peter Bala, May 16 2017: (Start)
G.f.: A(x) = 1/2*( 1 + Sum_{n >= 0} (1 + x)^(n+1)*Product_{k = 1..n} ((1 + x)^k - 1) ).
Conjectural g.f.: Sum_{n >= 0} 1/(1 + x)^(n+1)*Product_{k = 1..n} (1 - 1/(1 + x)^(2*k)).
Conjectural g.f.: Sum_{n >= 0} (1 + x)^(2*n+1)*Product_{k = 1..2*n} (1 - (1 + x)^k). Cf. A158690, which has e.g.f. A(exp(x) - 1). (End)

A338040 E.g.f.: Sum_{j>=0} 4^j * (exp(j*x) - 1)^j.

Original entry on oeis.org

1, 4, 132, 11140, 1763076, 449262724, 168055179012, 86720706877060, 59029852191779076, 51241585497612147844, 55245853646893977682692, 72423868722672448652558980, 113447698393867318106045295876, 209271794145089904620369489016964
Offset: 0

Views

Author

Vaclav Kotesovec, Oct 08 2020

Keywords

Comments

In general, if k > 0 and e.g.f.: Sum_{j>=0} k^j * (exp(j*x) - 1)^j, then a(n) ~ c * (1 + k*exp(1/r))^n * r^(2*n) * n!^2 / sqrt(n), where r is the root of the equation exp(1/r) * (1 + 1/(r*LambertW(-exp(-1/r)/r))) = -1/k and c is a constant (dependent only on k).
Conjecture: Let p be prime. The sequence obtained by reducing a(n) modulo p for n >= 1 is purely periodic with period p - 1. For example, modulo 7 the sequence becomes [4, 6, 3, 0, 1, 0, 4, 6, 3, 0, 1, 0, 4, 6, 3, 0, 1, 0, ...], with an apparent period of 6. - Peter Bala, May 31 2022

Crossrefs

Programs

  • Mathematica
    Flatten[{1, Table[Sum[4^j * j^n * j! * StirlingS2[n, j], {j, 0, n}], {n, 1, 20}]}]
    nmax = 20; CoefficientList[Series[1 + Sum[4^j*(Exp[j*x] - 1)^j, {j, 1, nmax}], {x, 0, nmax}], x] * Range[0, nmax]!
  • PARI
    {a(n)=local(X=x+x*O(x^n)); n!*polcoeff(sum(m=0, n, 4^m*(exp(m*X)-1)^m), n)}

Formula

a(n) = Sum_{j=0..n} 4^j * j^n * j! * Stirling2(n,j).
a(n) ~ c * (1 + 4*exp(1/r))^n * r^(2*n) * n!^2 / sqrt(n), where r = 0.95894043087329419322124137165060249611787608513866855417024... is the root of the equation exp(1/r) * (1 + 1/(r*LambertW(-exp(-1/r)/r))) = -1/4 and c = 0.37483929689722634406486945426531890297038414869116425498643733178324...

A301581 G.f.: Sum_{n>=0} 2^n * ((1+x)^n - 1)^n.

Original entry on oeis.org

1, 2, 16, 232, 4748, 125440, 4058312, 155336672, 6864980968, 343995674280, 19270975801600, 1193481831243584, 80966964261458368, 5971270693661978816, 475655179279901897536, 40699219246551726635840, 3722813577249648564213392, 362519587815189751405383520, 37442485808471509306691295808, 4088344078912544484116541775616, 470550859964811044524886252649760
Offset: 0

Views

Author

Paul D. Hanna, Mar 24 2018

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 16*x^2 + 232*x^3 + 4748*x^4 + 125440*x^5 + 4058312*x^6 + 155336672*x^7 + 6864980968*x^8 + 343995674280*x^9 + ...
such that
A(x) = 1 + 2*((1+x)-1) + 4*((1+x)^2-1)^2 + 8*((1+x)^3-1)^3 + 16*((1+x)^4-1)^4 + 32*((1+x)^5-1)^5 + 64*((1+x)^6-1)^6 + 128*((1+x)^7-1)^7 + ...
Also,
A(x) = 1/3 + 2*(1+x)/(1 + 2*(1+x))^2 + 4*(1+x)^4/(1 + 2*(1+x)^2)^3 + 8*(1+x)^9/(1 + 2*(1+x)^3)^4 + 16*(1+x)^16/(1 + 2*(1+x)^4)^5 + 32*(1+x)^25/(1 + 2*(1+x)^5)^6 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[1 + Sum[2^j*((1 + x)^j - 1)^j, {j, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 08 2020 *)
  • PARI
    {a(n) = my(A,o=x*O(x^n)); A = sum(m=0,n, 2^m * ((1+x +o)^m - 1)^m ); polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

G.f.: Sum_{n>=0} 2^n * (1+x)^(n^2) /(1 + 2*(1+x)^n)^(n+1).
a(n) ~ c * d^n * n! / sqrt(n), where d = (1 + 2*exp(1/r)) * r^2 = 5.90397654567002180048849478643459748789951071189960380570421209622277299181878..., where r = 0.925556278640887084941460444526398190071550948416... is the root of the equation exp(1/r) * (1 + 1/(r*LambertW(-exp(-1/r)/r))) = -1/2 and c = 0.3270084259351451234078681822585067892123772641092056... - Vaclav Kotesovec, Aug 09 2018, updated Oct 08 2020

A301582 G.f.: Sum_{n>=0} 3^n * ((1+x)^n - 1)^n.

Original entry on oeis.org

1, 3, 36, 765, 22932, 886707, 41971041, 2349915543, 151893243711, 11131097539221, 911906584505874, 82586031357156975, 8192750710914222984, 883506535094875209327, 102907862475072248379060, 12875067336646598300376165, 1722014444866824121524712497, 245185575019136812676809863351, 37027348593726417935247243009495, 5911490521308027393188499233189367, 994821814352463817234026392636083551
Offset: 0

Views

Author

Paul D. Hanna, Mar 24 2018

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 36*x^2 + 765*x^3 + 22932*x^4 + 886707*x^5 + 41971041*x^6 + 2349915543*x^7 + 151893243711*x^8 + ...
such that
A(x) = 1 + 3*((1+x)-1) + 9*((1+x)^2-1)^2 + 27*((1+x)^3-1)^3 + 81*((1+x)^4-1)^4 + 243*((1+x)^5-1)^5 + 729*((1+x)^6-1)^6 + 2187*((1+x)^7-1)^7 + ...
Also,
A(x) = 1/4 + 3*(1+x)/(1 + 3*(1+x))^2 + 9*(1+x)^4/(1 + 3*(1+x)^2)^3 + 27*(1+x)^9/(1 + 3*(1+x)^3)^4 + 81*(1+x)^16/(1 + 3*(1+x)^4)^5 + 243*(1+x)^25/(1 + 3*(1+x)^5)^6 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[1 + Sum[3^j*((1 + x)^j - 1)^j, {j, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 08 2020 *)
  • PARI
    {a(n) = my(A,o=x*O(x^n)); A = sum(m=0,n, 3^m * ((1+x +o)^m - 1)^m ); polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

G.f.: Sum_{n>=0} 3^n * (1+x)^(n^2) /(1 + 3*(1+x)^n)^(n+1).
a(n) ~ c * d^n * n! / sqrt(n), where d = (1 + 3*exp(1/r)) * r^2 = 8.632012704198046828204904686098781240870113556702123911346365466059061495897353..., where r = 0.947093169766093813913446822751643203941993193936... is the root of the equation exp(1/r) * (1 + 1/(r*LambertW(-exp(-1/r)/r))) = -1/3 and c = 0.34734097623709084937300542950550592394946492732014... - Vaclav Kotesovec, Aug 09 2018

A302598 G.f.: Sum_{n>=0} (1 + (1+x)^n)^n / (2 + (1+x)^n)^(n+1).

Original entry on oeis.org

1, 1, 6, 58, 798, 14150, 307076, 7881756, 233536532, 7844786248, 294582696686, 12228351266210, 556017625969246, 27482790417322218, 1467194712330407238, 84134395928742550138, 5157545958316518485420, 336574587493456290969620, 23296320082405927961459550, 1704662916739625989249415610, 131480805016085834305348796128
Offset: 0

Views

Author

Paul D. Hanna, Apr 10 2018

Keywords

Comments

The following identity holds for |y| <= 1 and fixed real k > 0:
Sum_{n>=0} (k + y^n)^n/(1+k + y^n)^(n+1) = Sum_{n>=0} (y^n - 1)^n/(1+k - k*y^n)^(n+1).

Examples

			G.f.: A(x) = 1 + x + 6*x^2 + 58*x^3 + 798*x^4 + 14150*x^5 + 307076*x^6 + 7881756*x^7 + 233536532*x^8 + 7844786248*x^9 + 294582696686*x^10 + ...
such that
A(x) = 1/3  +  (1 + (1+x))/(2 + (1+x))^2  +  (1 + (1+x)^2)^2/(2 + (1+x)^2)^3  +  (1 + (1+x)^3)^3/(2 + (1+x)^3)^4  +  (1 + (1+x)^4)^4/(2 + (1+x)^4)^5  +  (1 + (1+x)^5)^5/(2 + (1+x)^5)^6  +  (1 + (1+x)^6)^6/(2 + (1+x)^6)^7  + ...
Also,
A(x) = 1  +  ((1+x) - 1)/(2 - (1+x))^2  +  ((1+x)^2 - 1)^2/(2 - (1+x)^2)^3  +  ((1+x)^3 - 1)^3/(2 - (1+x)^3)^4  +  ((1+x)^4 - 1)^4/(2 - (1+x)^4)^5  +  ((1+x)^5 - 1)^5/(2 - (1+x)^5)^6  +  ((1+x)^6 - 1)^6/(2 - (1+x)^6)^7 + ...
RELATED INFINITE SERIES.
(1) At x = -1/2: the following sums are equal
S1 = Sum_{n>=0} 2^n * (2^n + 1)^n/(2^(n+1) + 1)^(n+1)
S1 = Sum_{n>=0} (-2)^n * (2^n - 1)^n/(2^(n+1) - 1)^(n+1).
Explicitly,
S1 = 1/3 + 2*3/5^2 + 4*5^2/9^3 + 8*9^3/17^4 + 16*17^4/33^5 + 32*33^5/65^6 + 64*65^6/129^7 + 128*129^7/257^8 + 256*257^8/513^9 + 512*513^9/1025^10 + ...
S1 = 1 - 2*1/3^2 + 4*3^2/7^3 - 8*7^3/15^4 + 16*15^4/31^5 - 32*31^5/63^6 + 64*63^6/127^7 - 128*127^7/255^8 + 256*255^8/511^9 - 512*511^9/1023^10 + ...
where S1 = 0.84714730053329880291591114748812485885366310294051236295420...
(2) At x = -2/3: the following sums are equal
S2 = Sum_{n>=0} 3^n * (1 + 3^n)^n / (2*3^n + 1)^(n+1)
S2 = Sum_{n>=0} (-3)^n * (3^n - 1)^n / (2*3^n - 1)^(n+1).
Explicitly,
S2 = 1/3 + 3*4/7^2 + 9*10^2/19^3 + 27*28^3/55^4 + 81*82^4/163^5 + 243*244^5/487^6 + 729*730^6/1459^7 + 2187*2188^7/4375^8 + 6561*6562^8/13123^9 + 19683*19684^9/39367^10 + ...
S2 = 1 - 3*2^1/5^2 + 9*8^2/17^3 - 27*26^3/53^4 + 81*80^4/161^5 - 243*242^5/485^6 + 729*728^6/1457^7 - 2187*2186^7/4373^8 + 6561*6560^8/13121^9 - 19683*19682^9/39365^10 + ...
where S2 = 0.837457334418049175936255584889342515316005199043439291643371...
(3) At x = -1/3: the following sums are equal
S3 = Sum_{n>=0} 3^n * (2^n + 3^n)^n/(2*3^n + 2^n)^(n+1)
S3 = Sum_{n>=0} (-3)^n * (3^n - 2^n)^n/(2*3^n - 2^n)^(n+1).
Explicitly,
S3 = 1/3 + 3*5/8^2 + 9*13^2/22^3 + 27*35^3/62^4 + 81*97^4/178^5 + 243*275^5/518^6 + 729*793^6/1522^7 + 2187*2315^7/4502^8 + 6561*6817^8/13378^9 + 19683*20195^9/39878^10 + ...
S3 = 1 - 3*1/4^2 + 9*5^2/14^3 - 27*19^3/46^4 + 81*65^4/146^5 - 243*211^5/454^6 + 729*665^6/1394^7 - 2187*2059^7/4246^8 + 6561*6305^8/12866^9 - 19683*19171^9/38854^10 + ...
where S3 = 0.867357695200699139470956415922046910279987551651352471994920...
		

Crossrefs

Programs

  • PARI
    {a(n) = my(A=1,o=x*O(x^n)); A = sum(m=0,n,((1+x +o)^m - 1)^m / (2 - (1+x +o)^m)^(m+1)); polcoeff(A,n)}
    for(n=0,30, print1(a(n),", "))

Formula

G.f.: Sum_{n>=0} ((1+x)^n - 1)^n / (2 - (1+x)^n)^(n+1).
G.f.: Sum_{n>=0} ((1+x)^n + 1)^n / (2 + (1+x)^n)^(n+1).
a(n) ~ c * d^n * n! / sqrt(n), where d = A317904 = 3.9561842030261697545408021818783008332999988095... and c = 0.274656497660429769528095546948772676444158... - Vaclav Kotesovec, Aug 09 2018

A301583 G.f.: Sum_{n>=0} 4^n * ((1+x)^n - 1)^n.

Original entry on oeis.org

1, 4, 64, 1792, 70736, 3600128, 224255040, 16521605376, 1405131880000, 135480346104896, 14602769310474240, 1739917222954854400, 227081534040721917952, 32217108743091290851328, 4936803887495636263284736, 812576030237749532251019264, 142976863303365903802301729024, 26781577193841845859144244087808, 5320767287406003709062843236972544, 1117525692987087894816123931091214336
Offset: 0

Views

Author

Paul D. Hanna, Mar 24 2018

Keywords

Comments

In general, if k > 0 and g.f.: Sum_{j>=0} k^j * ((1+x)^j - 1)^j, then a(n) ~ c * (1 + k*exp(1/r))^n * r^(2*n) * n! / sqrt(n), where r is the root of the equation exp(1/r) * (1 + 1/(r*LambertW(-exp(-1/r)/r))) = -1/k and c is a constant (dependent only on k). - Vaclav Kotesovec, Oct 08 2020

Examples

			G.f.: A(x) = 1 + 4*x + 64*x^2 + 1792*x^3 + 70736*x^4 + 3600128*x^5 + 224255040*x^6 + 16521605376*x^7 + 1405131880000*x^8 + ...
such that
A(x) = 1 + 4*((1+x)-1) + 16*((1+x)^2-1)^2 + 64*((1+x)^3-1)^3 + 256*((1+x)^4-1)^4 + 1024*((1+x)^5-1)^5 + 4096*((1+x)^6-1)^6 + ...
Also,
A(x) = 1/5 + 4*(1+x)/(1 + 4*(1+x))^2 + 16*(1+x)^4/(1 + 4*(1+x)^2)^3 + 64*(1+x)^9/(1 + 4*(1+x)^3)^4 + 256*(1+x)^16/(1 + 4*(1+x)^4)^5 + 1024*(1+x)^25/(1 + 4*(1+x)^5)^6 + ...
		

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[1 + Sum[4^j*((1 + x)^j - 1)^j, {j, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 08 2020 *)
  • PARI
    {a(n) = my(A,o=x*O(x^n)); A = sum(m=0,n, 4^m * ((1+x +o)^m - 1)^m ); polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

G.f.: Sum_{n>=0} 4^n * (1+x)^(n^2) /(1 + 4*(1+x)^n)^(n+1).
a(n) ~ c * d^n * n! / sqrt(n), where d = (1 + 4*exp(1/r)) * r^2 = 11.35554580636894436474777793373210745006910386794268638744346793426715754570218..., where r = 0.95894043087329419322124137165060249611787608513866855417024... is the root of the equation exp(1/r) * (1 + 1/(r*LambertW(-exp(-1/r)/r))) = -1/4 and c = 0.358692703763731594549618907599728117285634153... - Vaclav Kotesovec, Aug 09 2018, updated Oct 08 2020

A301463 G.f.: Sum_{n>=0} (2*(1+x)^n - 1)^n / 2^(n+1).

Original entry on oeis.org

1, 6, 134, 5102, 272694, 18758134, 1577807110, 156883546142, 18001728695894, 2341268080847014, 340346951612008454, 54686371000455538574, 9624103747115691611318, 1841049154379441320293142, 380367456989975381891133446, 84407842226680664984458744126, 20023121531700221583865582432854, 5056357801144690975957652265658438, 1354259474931265421064754160458035078, 383444904170987865090156939638756172846
Offset: 0

Views

Author

Paul D. Hanna, Mar 24 2018

Keywords

Examples

			G.f.: A(x) = 1 + 6*x + 134*x^2 + 5102*x^3 + 272694*x^4 + 18758134*x^5 + 1577807110*x^6 + 156883546142*x^7 + 18001728695894*x^8 + ...
such that
A(x) = 1/2  +  (2*(1+x) - 1)/2^2  +  (2*(1+x)^2 - 1)^2/2^3  +  (2*(1+x)^3 - 1)^3/2^4  +  (2*(1+x)^4 - 1)^4/2^5  +  (2*(1+x)^5 - 1)^5/2^6  + ...
Also,
A(x) = 1/3  +  2*(1+x)/(2 + (1+x))^2  +  2^2*(1+x)^4/(2 + (1+x)^2)^3  +  2^3*(1+x)^9/(2 + (1+x)^3)^4  +  2^4*(1+x)^16/(2 + (1+x)^4)^5  +  2^5*(1+x)^25/(2 + (1+x)^5)^6  +  2^6*(1+x)^36/(2 + (1+x)^6)^7  + ...
		

Crossrefs

Programs

  • Mathematica
    nmax = 20; Round[CoefficientList[Series[Sum[(2*(1 + x)^j - 1)^j/2^(j + 1), {j, 0, nmax^2}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Oct 08 2020 *)

Formula

G.f.: Sum_{n>=0} 2^n * (1+x)^(n^2) / (2 + (1+x)^n)^(n+1).
G.f.: Sum_{n>=0} ((1+x)^n - 1/2)^n / 2.
a(n) ~ c * d^n * n! / sqrt(n), where d = 15.305828173910545025228605110120647795... and c = 0.4246982835243422293505427496472772728... - Vaclav Kotesovec, Aug 09 2018

Extensions

b-file confirmed by Vaclav Kotesovec, Oct 08 2020

A192935 E.g.f.: Sum_{n>=0} ((1+x)^n - 1)^n / n!.

Original entry on oeis.org

1, 1, 4, 39, 592, 12965, 378276, 14062363, 643946920, 35426253465, 2295988778440, 172565368741931, 14847924324645996, 1446814927797156541, 158201328106874927980, 19258822568210913998955, 2592339296719295037808336, 383513887126740027040942577
Offset: 0

Views

Author

Paul D. Hanna, Jul 13 2011

Keywords

Comments

More generally, the following sums are equal:
(1) Sum_{n>=0} (q^n + p)^n * r^n/n!,
(2) Sum_{n>=0} q^(n^2) * exp(p*q^n*r) * r^n/n!;
here, q = (1+x) and p = -1, r = 1. - Paul D. Hanna, Jun 21 2019

Examples

			E.g.f: A(x) = 1 + x + 4*x^2/2! + 39*x^3/3! + 592*x^4/4! + 12965*x^5/5! + 378276*x^6/6! + 14062363*x^7/7! + 643946920*x^8/8! + 35426253465*x^9/9! + 2295988778440*x^10/10! +...
such that
A(x) = 1 + ((1+x) - 1) + ((1+x)^2 - 1)^2/2! + ((1+x)^3 - 1)^3/3! + ((1+x)^4 - 1)^4/4! + ((1+x)^5 - 1)^5/5! + ((1+x)^6 - 1)^6/6! + ((1+x)^7 - 1)^7/7! + ...
also
A(x) = 1 + (1+x)*exp(-(1+x)) + (1+x)^4*exp(-(1+x)^2)/2! + (1+x)^9*exp(-(1+x)^3)/3! + (1+x)^16*exp(-(1+x)^4)/4! + (1+x)^25*exp(-(1+x)^5)/5! + (1+x)^36*exp(-(1+x)^6)/6! + (1+x)^49*exp(-(1+x)^7)/7! + ...
RELATED SERIES.
Expansion of ((1+x)^n-1)^n suggests that the e.g.f. is related to LambertW(x):
((1+x)^2-1)^2 = 4*x^2 + 4*x^3 + x^4;
((1+x)^3-1)^3 = 27*x^3 + 81*x^4 + 108*x^5 + 81*x^6 + 36*x^7 + 9*x^8 + x^9;
((1+x)^4-1)^4 = 256*x^4 + 1536*x^5 + 4480*x^6 + 8320*x^7 + 10896*x^8 +...
		

Crossrefs

Programs

  • Magma
    m:=20; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( (&+[((1+x)^n -1)^n/Factorial(n): n in [0..m+2]]) )); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Feb 06 2019
    
  • Mathematica
    With[{m = 20}, CoefficientList[Series[Sum[If[n==0, 1, ((1+x)^n -1)^n/n!], {n,0,m+2}], {x,0,m}], x]*Range[0, m]!] (* G. C. Greubel, Feb 06 2019 *)
  • PARI
    {a(n)=n!*polcoeff(sum(m=0,n,((1+x+x*O(x^n))^m-1)^m/m!),n)}
    for(n=0, 30, print1(a(n)*n!, ", "))
    
  • PARI
    {Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)}
    {a(n)=sum(k=0, n, Stirling1(n, k)*k!*polcoeff(sum(m=0,k,(exp(m*x+x*O(x^n))-1)^m/m!),k))}
    for(n=0, 30, print1(a(n)*n!, ", "))
    
  • Sage
    m = 20; T = taylor(sum(((1+x)^k-1)^k/factorial(k) for k in range(m+2)), x, 0, m); [factorial(n)*T.coefficient(x, n) for n in (0..m)] # G. C. Greubel, Feb 06 2019

Formula

E.g.f.: Sum_{n>=0} (1+x)^(n^2) * exp(-(1+x)^n) / n!. - Paul D. Hanna, Jun 21 2019
a(n) = (1/n!)* Sum_{k=0..n} Stirling1(n,k)*A108459(k), where the e.g.f. of A108459 = Sum_{n>=0} (exp(n*x)-1)^n/n! (see Vladeta Jovovic's formula in A122400).
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