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.

Showing 1-7 of 7 results.

A052852 Expansion of e.g.f.: (x/(1-x))*exp(x/(1-x)).

Original entry on oeis.org

0, 1, 4, 21, 136, 1045, 9276, 93289, 1047376, 12975561, 175721140, 2581284541, 40864292184, 693347907421, 12548540320876, 241253367679185, 4909234733857696, 105394372192969489, 2380337795595885156, 56410454014314490981, 1399496554158060983080
Offset: 0

Views

Author

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

Keywords

Comments

A simple grammar.
Number of {121,212}-avoiding n-ary words of length n. - Ralf Stephan, Apr 20 2004
The infinite continued fraction (1+n)/(1+(2+n)/(2+(3+n)/(3+...))) converges to the rational number A052852(n)/A000262(n) when n is a positive integer. - David Angell (angell(AT)maths.unsw.edu.au), Dec 18 2008
a(n) is the total number of components summed over all nilpotent partial permutations of [n]. - Geoffrey Critzer, Feb 19 2022

Crossrefs

Row sums of unsigned triangle A062139 (generalized a=2 Laguerre).

Programs

  • Magma
    [n eq 0 select 0 else Factorial(n)*Evaluate(LaguerrePolynomial(n-1, 0), -1): n in [0..30]]; // G. C. Greubel, Feb 23 2021
  • Maple
    spec := [S,{B=Set(C),C=Sequence(Z,1 <= card),S=Prod(B,C)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
    a := n -> ifelse(n = 0, 0, n!*hypergeom([-n+1], [1], -1)): seq(simplify(a(n)), n = 0..18);  # Peter Luschny, Dec 30 2024
  • Mathematica
    Table[n!*SeriesCoefficient[(x/(1-x))*E^(x/(1-x)),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 09 2012 *)
    Table[If[n==0, 0, n!*LaguerreL[n-1, 0, -1]], {n, 0, 30}] (* G. C. Greubel, Feb 23 2021 *)
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(serlaplace((x/(1-x))*exp(x/(1-x))))) \\ G. C. Greubel, May 15 2018
    
  • Sage
    [0 if n==0 else factorial(n)*gen_laguerre(n-1, 0, -1) for n in (0..30)] # G. C. Greubel, Feb 23 2021
    

Formula

D-finite with recurrence: a(1)=1, a(0)=0, (n^2+2*n)*a(n)+(-4-2*n)*a(n+1)+ a(n+2)=0.
a(n) = Sum_{m=0..n} n!*binomial(n+2, n-m)/m!. - Wolfdieter Lang, Jun 19 2001
a(n) = n*A002720(n-1). [Riordan] - Vladeta Jovovic, Mar 18 2005
Related to an n-dimensional series: for n>=1, a(n) = (n!/e)*Sum_{k_n>=k_{n-1}>=...>=k_1>=0} 1/(k_n)!. - Benoit Cloitre, Sep 30 2006
E.g.f.: (x/(1-x))*exp((x/(1-x))) = (x/(1-x))*G(0); G(k)=1+x/((2*k+1)*(1-x)-x*(1-x)*(2*k+1)/(x+(1-x)*(2*k+2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 24 2011
a(n) = D^n(x*exp(x)) evaluated at x = 0, where D is the operator (1+x)^2*d/dx. Cf. A000262 and A005493. - Peter Bala, Nov 25 2011
a(n) ~ exp(2*sqrt(n)-n-1/2)*n^(n+1/4)/sqrt(2). - Vaclav Kotesovec, Oct 09 2012
a(n) = n!*hypergeom([-n+1], [1], -1) for n>=1. - Peter Luschny, Oct 18 2014 [Simplified by Natalia L. Skirrow, 30 December 2024]
a(n) = Sum_{k=0..n} L(n,k)*k; L(n,k) the unsigned Lah numbers. - Peter Luschny, Oct 18 2014
a(n) = n!*LaguerreL(n-1, 0, -1) for n>=1. - Peter Luschny, Apr 08 2015, simplified Dec 30 2024
The series reversion of the e.g.f. equals W(x)/(1 + W(x)) = x - 2^2*x^2/2! + 3^3*x^3/3! - 4^4*x^4/4! + ..., essentially the e.g.f. for a signed version of A000312, where W(x) is Lambert's W-function (see A000169). - Peter Bala, Jun 14 2016
Equals column A059114(n, 2) for n >= 1. - G. C. Greubel, Feb 23 2021
a(n) = Sum_{k=1..n} k * A271703(n,k). - Geoffrey Critzer, Feb 19 2022

A059114 Triangle T(n,m)= Sum_{i=0..n} L'(n,i)*Product_{j=1..m} (i-j+1), read by rows.

Original entry on oeis.org

1, 1, 1, 3, 4, 2, 13, 21, 18, 6, 73, 136, 156, 96, 24, 501, 1045, 1460, 1260, 600, 120, 4051, 9276, 15030, 16320, 11160, 4320, 720, 37633, 93289, 170142, 219450, 192360, 108360, 35280, 5040, 394353, 1047376, 2107448, 3116736, 3294480, 2405760, 1149120, 322560, 40320
Offset: 0

Views

Author

Vladeta Jovovic, Jan 04 2001

Keywords

Comments

L'(n,i) are unsigned Lah numbers (Cf. A008297): L'(n,i) = (n!/i!)*binomial(n-1,i-1) for i >= 1, L'(0,0) = 1, L'(n,0) = 0 for n > 0.

Examples

			Triangle begins as:
    1;
    1,    1;
    3,    4,    2;
   13,   21,   18,    6;
   73,  136,  156,   96,  24;
  501, 1045, 1460, 1260, 600, 120;
  ...;
E.g.f. for T(n, 2) = (x/(1-x))^2*e^(x/(x-1)) = x^2 + 3*x^3 + 13/2*x^4 + 73/6*x^5 + 167/8*x^6 + 4051/120*x^7 + ...
		

Crossrefs

Row sums give A059115. Alternating row sums give A288268.

Programs

  • Magma
    [Factorial(n)*Evaluate(LaguerrePolynomial(n-k, k-1), -1): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 23 2021
    
  • Mathematica
    Table[n!*LaguerreL[n-k, k-1, -1], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 23 2021 *)
  • PARI
    T(n, k) = n! * pollaguerre(n-k, k-1, -1); \\ Michel Marcus, Feb 23 2021
  • Sage
    flatten([[factorial(n)*gen_laguerre(n-k, k-1, -1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 23 2021
    

Formula

E.g.f. for T(n, k) = (x/(1-x))^k * exp(x/(x-1)).
T(n, k)= Sum_{i=0..n} L'(n,i) * ( Product_{j=1..k} (i-j+1) ).
T(n, 0) = A000262(n).
T(n, 1) = A052852(n).
From G. C. Greubel, Feb 23 2021: (Start)
T(n, k) = n! * k! * Sum_{j=0..n} binomial(j, k)*binomial(n-1, j-1)/j!.
T(n, k) = n! * Laguerre(n-k, k-1, -1).
T(n, k) = n!*binomial(n-1, k-1)*Hypergeometric1F1([k-n], [k], -1) with T(n, 0) = Hypergeometric2F0([1-n, -n], [], 1). (End)

A361649 a(n) = (1+n)*(2*a(n-1) - (n-2)*a(n-2)) with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 6, 44, 380, 3768, 42112, 523072, 7141248, 106209920, 1708188416, 29525850624, 545607622144, 10730032423936, 223691600732160, 4926284479250432, 114255071320260608, 2783085758131765248, 71023717127647854592, 1894699527341113999360, 52730415074075898937344
Offset: 0

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Author

Keywords

Crossrefs

For m=1 the formula gives the sequence A052852.

Programs

  • Maple
    # For recursion:
    N:=20;a[0]:=1;a[1]:=1;for n from 1 to N do
    a[n+1]:=(n+2)*(2*a[n]-(n-1)*a[n-1]);od;
    # For closed form:
    C := binomial:
    a := n -> `if`(n=0, 1, add(C(n-1, i)*C(n+1, n-i)*(n-i)!*2^(i-1), i = 0..n-1)):
    seq(a(n), n = 0..20);
  • PARI
    memo=Map([0, 1; 1, 1]);
    a(n)=if(mapisdefined(memo, n),mapget(memo, n), mapput(memo, n, (n+1)* (2*a(n-1) - (n-2)*a(n-2))); a(n)); \\ Winston de Greef, Mar 20 2023

Formula

a(n) = (m+n-1)*(2*a(n-1) - (n-2)*a(n-2)) where m = 2.
a(n) = Sum_{i=0..n-1} binomial(n-1,i) * binomial(n+m-1,n-i)*(n-i)!*m^(i-1) where m = 2 for n >= 1.
E.g.f.: (1+x^2)*exp(2/(1-x))/(4*(1-x)^2*exp(2))+3/4. - Alois P. Heinz, Mar 19 2023
a(n) ~ 2^(-9/4) * exp(2*sqrt(2*n) - n - 1) * n^(n + 3/4). - Vaclav Kotesovec, Mar 20 2023

A288269 Expansion of e.g.f.: exp(Sum_{k>=1} (k-1)*k*x^k).

Original entry on oeis.org

1, 0, 4, 36, 336, 3840, 52800, 836640, 14864640, 291755520, 6264276480, 145962432000, 3665362821120, 98604459233280, 2827182573895680, 86016204578304000, 2766450467708928000, 93741871082943283200, 3336807307530977280000, 124443669133537276723200
Offset: 0

Views

Author

Seiichi Manyama, Oct 20 2017

Keywords

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30);
    Coefficients(R!(Laplace( Exp(2*x^2/(1-x)^3) ))); // G. C. Greubel, Mar 10 2021
  • Mathematica
    With[{m = 30}, CoefficientList[Series[Exp[2*x^2/(1-x)^3], {x, 0, m}], x]*Range[0, m]!] (* G. C. Greubel, Mar 10 2021 *)
  • PARI
    {a(n) = n!*polcoeff(exp(sum(k=1, n, (k-1)*k*x^k)+x*O(x^n)), n)}
    
  • Sage
    [factorial(n)*( exp(2*x^2/(1-x)^3) ).series(x,n+1).list()[n] for n in (0..30)] # G. C. Greubel, Mar 10 2021
    

Formula

a(0) = 1 and a(n) = (n-1)! * Sum_{k=1..n} (k-1)*k^2*a(n-k)/(n-k)! for n > 0.
E.g.f.: exp(2*x^2/(1 - x)^3). - Ilya Gutkovskiy, Jul 27 2020
a(n) ~ 3^(1/8) * exp(2/27 - (n/6)^(1/4)/12 - (n/6)^(1/2) + 8*(n/6)^(3/4) - n) * n^(n - 1/8) / 2^(7/8) * (1 - 3203/34560 * (6/n)^(1/4)). - Vaclav Kotesovec, Mar 10 2021
a(n) = 4*(n-1)*a(n-1) - 2*(n-1)*(3*n-8)*a(n-2) + 2*(n-2)*(n-1)*(2*n-5)*a(n-3) - (n-4)*(n-3)*(n-2)*(n-1)*a(n-4). - Vaclav Kotesovec, Dec 01 2021

A361528 a(n) = (2+n)*(2*a(n-1) - (n-2)*a(n-2)) with a(0)=a(1)=1.

Original entry on oeis.org

1, 1, 8, 75, 804, 9681, 129168, 1889379, 30037500, 515342817, 9484627608, 186305208219, 3888697965012, 85920579594225, 2002828537732896, 49107722192594739, 1263165207424720812, 34004577057249890241, 955970215914084949800, 28011115058953357075563, 853924857091970071203972
Offset: 0

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Author

Keywords

Crossrefs

For m=1 the formula gives the sequence A052852.
Cf. A288268. For m=2 the formula gives the sequence A361649.

Programs

  • Maple
    # For recursion:
    N:=10;a[0]:=1;a[1]:=1;for n from 1 to N do
    a[n+1]:=(n+3)*(2*a[n]-(n-1)*a[n-1]);od;
    # For closed form:
    C := binomial:
    a := n -> `if`(n=0, 1, add(C(n-1, i)*C(n+2, n-i)*(n-i)!*3^(i-1), i = 0..n-1)):
    seq(a(n), n = 0..20);
    # Alternative:
    a := n -> `if`(n=0, 1, (n + 2)!*hypergeom([1 - n], [3], -3) / 6):
    seq(simplify(a(n)), n = 0..20); # Peter Luschny, Mar 23 2023
  • Mathematica
    nmax = 20; CoefficientList[Series[23/27 + (4 + 3*x + 2*x^3)*E^(3*x/(1 - x))/(27*(1 - x)^3), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Mar 23 2023 *)
  • PARI
    a(n) = if(n==0, 1, my(m=3); sum(i=0, n-1, binomial(n-1, i)*binomial(n+m-1, n-i)*(n-i)!*m^(i-1))) \\ Andrew Howroyd, Mar 23 2023

Formula

a(n) = (m+n-1)*(2*a(n-1) - (n-2)*a(n-2)) where m=3, a(0)=a(1)=1.
a(n) = Sum_{i=0..n-1} binomial(n-1,i) * binomial(n+m-1,n-i)*(n-i)!*m^(i-1) where m = 3 for n >= 1.
a(n) = (n + 2)!*hypergeom([1 - n], [3], -3) / 6 for n >= 1. - Peter Luschny, Mar 23 2023
From Vaclav Kotesovec, Mar 23 2023: (Start)
E.g.f.: 23/27 + (4 + 3*x + 2*x^3) * exp(3*x/(1-x)) / (27*(1-x)^3).
a(n) ~ exp(2*sqrt(3*n) - n - 3/2) * n^(n + 5/4) / (sqrt(2) * 3^(9/4)). (End)

Extensions

Terms a(12) and beyond from Andrew Howroyd, Mar 23 2023

A347340 E.g.f.: exp( exp(exp(x) - 1) - exp(x) ).

Original entry on oeis.org

1, 0, 1, 4, 17, 91, 587, 4327, 35604, 323316, 3210600, 34574453, 400893066, 4975247460, 65755573847, 921535225267, 13643496840808, 212688569520955, 3480978391442106, 59657975022473437, 1068151956803180295, 19937983367649562025, 387243759600707804811, 7812456801157894913964
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 27 2021

Keywords

Comments

Exponential transform of A058692.
Stirling transform of A000296.

Crossrefs

Programs

  • Maple
    g:= proc(n) option remember; `if`(n=0, 1,
          add(g(n-j)*binomial(n-1, j-1), j=2..n))
        end:
    b:= proc(n, m) option remember; `if`(n=0,
          g(m), m*b(n-1, m)+b(n-1, m+1))
        end:
    a:= n-> b(n, 0):
    seq(a(n), n=0..23);  # Alois P. Heinz, Aug 27 2021
    # second Maple program:
    b:= proc(n, t) option remember; `if`(n=0, 1, add(b(n-j, t)*
          `if`(t=0, 1, b(j, 0)-1)*binomial(n-1, j-1), j=1..n))
        end:
    a:= n-> b(n, 1):
    seq(a(n), n=0..23);  # Alois P. Heinz, Sep 02 2021
  • Mathematica
    nmax = 23; CoefficientList[Series[Exp[Exp[Exp[x] - 1] - Exp[x]], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k - 1] (BellB[k] - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 23}]
  • PARI
    my(x='x+O('x^25)); Vec(serlaplace(exp(exp(exp(x)-1)-exp(x)))) \\ Michel Marcus, Aug 27 2021

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * (Bell(k) - 1) * a(n-k).
a(n) = Sum_{k=0..n} Stirling2(n,k) * A000296(k).
a(n) = Sum_{k=0..n} binomial(n,k) * A000258(k) * A000587(n-k).

A372205 a(n) = (-1)^n*a((n - 2^A007814(n))/2) + a(floor((2*n - 2^A007814(n))/2)) for n > 0 and a(0) = 1.

Original entry on oeis.org

1, 0, 1, 1, 2, 1, 2, 1, 3, 1, 3, 2, 5, 3, 5, 4, 4, 1, 4, 3, 7, 4, 7, 5, 10, 5, 10, 7, 15, 10, 15, 11, 5, 1, 5, 4, 9, 5, 9, 6, 13, 6, 13, 9, 20, 13, 20, 15, 17, 7, 17, 12, 27, 17, 27, 20, 37, 22, 37, 27, 52, 37, 52, 41, 6, 1, 6, 5, 11, 6, 11, 7, 16, 7, 16, 11, 25, 16, 25, 19
Offset: 0

Views

Author

Peter Luschny, Apr 22 2024

Keywords

Comments

This sequence was originally introduced by Mikhail Kurkov in A217924 where he conjectured that A217924(n) = Sum_{k=0..2^n-1} a(k).

Crossrefs

Programs

  • Maple
    f := n -> padic[ordp](n, 2):
    a := proc(n) option remember; if n = 0 then return 1 fi;
    (-1)^n*a((n - 2^f(n))/2) + a(floor((2*n - 2^f(n))/2)) end:
    seq(a(n), n = 0..79);

Formula

Conjecture (by Mikhail Kurkov): a(2^n - 1) = A000296(n).
Conjecture (by Mikhail Kurkov): a((4^n - 1)/3) = A288268(n).
Showing 1-7 of 7 results.