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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A000262 Number of "sets of lists": number of partitions of {1,...,n} into any number of lists, where a list means an ordered subset.

Original entry on oeis.org

1, 1, 3, 13, 73, 501, 4051, 37633, 394353, 4596553, 58941091, 824073141, 12470162233, 202976401213, 3535017524403, 65573803186921, 1290434218669921, 26846616451246353, 588633468315403843, 13564373693588558173, 327697927886085654441, 8281153039765859726341
Offset: 0

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Author

Keywords

Comments

Determinant of n X n matrix M=[m(i,j)] where m(i,i)=i, m(i,j)=1 if i > j, m(i,j)=i-j if j > i. - Vladeta Jovovic, Jan 19 2003
With p(n) = the number of integer partitions of n, d(i) = the number of different parts of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, Sum_{i=1..p(n)} = sum over i and Product_{j=1..d(i)} = product over j, one has: a(n) = Sum_{i=1..p(n)} n!/(Product_{j=1..d(i)} m(i,j)!). - Thomas Wieder, May 18 2005
Consider all n! permutations of the integer sequence [n] = 1,2,3,...,n. The i-th permutation, i=1,2,...,n!, consists of Z(i) permutation cycles. Such a cycle has the length lc(i,j), j=1,...,Z(i). For a given permutation we form the product of all its cycle lengths Product_{j=1..Z(i)} lc(i,j). Furthermore, we sum up all such products for all permutations of [n] which gives Sum_{i=1..n!} Product_{j=1..Z(i)} lc(i,j) = A000262(n). For n=4 we have Sum_{i=1..n!} Product_{j=1..Z(i)} lc(i,j) = 1*1*1*1 + 2*1*1 + 3*1 + 2*1*1 + 3*1 + 2*1 + 3*1 + 4 + 3*1 + 4 + 2*2 + 2*1*1 + 3*1 + 4 + 3*1 + 2*1*1 + 2*2 + 4 + 2*2 + 4 + 3*1 + 2*1*1 + 3*1 + 4 = 73 = A000262(4). - Thomas Wieder, Oct 06 2006
For a finite set S of size n, a chain gang G of S is a partially ordered set (S,<=) consisting only of chains. The number of chain gangs of S is a(n). For example, with S={a, b}and n=2, there are a(2)=3 chain gangs of S, namely, {(a,a),(b,b)}, {(a,a),(a,b),(b,b)} and {(a,a),(b,a),(b,b)}. - Dennis P. Walsh, Feb 22 2007
(-1)*A000262 with the first term set to 1 and A084358 form a reciprocal pair under the list partition transform and associated operations described in A133314. Cf. A133289. - Tom Copeland, Oct 21 2007
Consider the distribution of n unlabeled elements "1" onto n levels where empty levels are allowed. In addition, the empty levels are labeled. Their names are 0_1, 0_2, 0_3, etc. This sequence gives the total number of such distributions. If the empty levels are unlabeled ("0"), then the answer is A001700. Let the colon ":" separate two levels. Then, for example, for n=3 we have a(3)=13 arrangements: 111:0_1:0_2, 0_1:111:0_2, 0_1:0_2:111, 111:0_2:0_1, 0_2:111:0_1, 0_2:0_1:111, 11:1:0, 11:0:1, 0:11:1, 1:11:0, 1:0:11, 0:1:11, 1:1:1. - Thomas Wieder, May 25 2008
Row sums of exponential Riordan array [1,x/(1-x)]. - Paul Barry, Jul 24 2008
a(n) is the number of partitions of [n] (A000110) into lists of noncrossing sets. For example, a(3)=3 counts 12, 1-2, 2-1 and a(4) = 73 counts the 75 partitions of [n] into lists of sets (A000670) except for 13-24, 24-13 which fail to be noncrossing. - David Callan, Jul 25 2008
a(i-j)/(i-j)! gives the value of the non-null element (i, j) of the lower triangular matrix exp(S)/exp(1), where S is the lower triangular matrix - of whatever dimension - having all its (non-null) elements equal to one. - Giuliano Cabrele, Aug 11 2008, Sep 07 2008
a(n) is also the number of nilpotent partial one-one bijections (of an n-element set). Equivalently, it is the number of nilpotents in the symmetric inverse semigroup (monoid). - Abdullahi Umar, Sep 14 2008
A000262 is the exp transform of the factorial numbers A000142. - Thomas Wieder, Sep 10 2008
If n is a positive integer then the infinite continued fraction (1+n)/(1+(2+n)/(2+(3+n)/(3+...))) converges to the rational number A052852(n)/A000262(n). - David Angell (angell(AT)maths.unsw.edu.au), Dec 18 2008
Vladeta Jovovic's formula dated Sep 20 2006 can be restated as follows: a(n) is the n-th ascending factorial moment of the Poisson distribution with parameter (mean) 1. - Shai Covo (green355(AT)netvision.net.il), Jan 25 2010
The umbral exponential generating function for a(n) is (1-x)^{-B}. In other words, write (1-x)^{-B} as a power series in x whose coefficients are polynomials in B, and then replace B^k with the Bell number B_k. We obtain a(0) + a(1)x + a(2)x^2/2! + ... . - Richard Stanley, Jun 07 2010
a(n) is the number of Dyck n-paths (A000108) with its peaks labeled 1,2,...,k in some order, where k is the number of peaks. For example a(2)=3 counts U(1)DU(2)D, U(2)DU(1)D, UU(1)DD where the label at each peak is in parentheses. This is easy to prove using generating functions. - David Callan, Aug 23 2011
a(n) = number of permutations of the multiset {1,1,2,2,...,n,n} such that for 1 <= i <= n, all entries between the two i's exceed i and if any such entries are present, they include n. There are (2n-1)!! permutations satisfying the first condition, and for example: a(3)=13 counts all 5!!=15 of them except 331221 and 122133 which fail the second condition. - David Callan, Aug 27 2014
a(n) is the number of acyclic, injective functions from subsets of [n] to [n]. Let subset D of [n] have size k. The number of acyclic, injective functions from D to [n] is (n-1)!/(n-k-1)! and hence a(n) = Sum_{k=0..n-1} binomial(n,k)*(n-1)!/(n-k-1)!. - Dennis P. Walsh, Nov 05 2015
a(n) is the number of acyclic, injective, labeled directed graphs on n vertices with each vertex having outdegree at most one. - Dennis P. Walsh, Nov 05 2015
For n > 0, a(n) is the number of labeled-rooted skinny-tree forests on n nodes. A skinny tree is a tree in which each vertex has at most one child. Let k denote the number of trees. There are binomial(n,k) ways to choose the roots, binomial(n-1,k-1) ways to choose the number of descendants for each root, and (n-k)! ways to permute those descendants. Summing over k, we obtain a(n) = Sum_{k=1..n} C(n,k)*C(n-1,k-1)*(n-k)!. - Dennis P. Walsh, Nov 10 2015
This is the Sheffer transform of the Bell numbers A000110 with the Sheffer matrix S = |Stirling1| = (1, -log(1-x)) = A132393. See the e.g.f. formula, a Feb 21 2017 comment on A048993, and R. Stanley's Jun 07 2010 comment above. - Wolfdieter Lang, Feb 21 2017
All terms = {1, 3} mod 10. - Muniru A Asiru, Oct 01 2017
We conjecture that for k = 2,3,4,..., the difference a(n+k) - a(n) is divisible by k: if true, then for each k the sequence a(n) taken modulo k is periodic with period dividing k. - Peter Bala, Nov 14 2017
The above conjecture is true - see the Bala link. - Peter Bala, Jan 20 2018
The terms of this sequence can be derived from the numerators of the fractions, produced by the recursion: b(0) = 1, b(n) = 1 + ((n-1) * b(n-1)) / (n-1 + b(n-1)) for n > 0. The denominators give A002720. - Dimitris Valianatos, Aug 01 2018
a(n) is the number of rooted labeled forests on n nodes that avoid the patterns 213, 312, and 123. It is also the number of rooted labeled forests that avoid 312, 213, and 132, as well as the number of rooted labeled forests that avoid 132, 213, and 321. - Kassie Archer, Aug 30 2018
For n > 0, a(n) is the number of partitions of [2n-1] whose nontrivial blocks are of type {a,b}, with a <= n and b > n. In fact, for n > 0, a(n) = A056953(2n-1). - Francesca Aicardi, Nov 03 2022
For n > 0, a(n) is the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select one person from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them). - Enrique Navarrete, Oct 01 2023

Examples

			Illustration of first terms as sets of ordered lists of the first n integers:
  a(1) = 1  : (1)
  a(2) = 3  : (12), (21), (1)(2).
  a(3) = 13 : (123) (6 ways), (12)(3) (2*3 ways) (1)(2)(3) (1 way)
  a(4) = 73 : (1234) (24 ways), (123)(4) (6*4 ways), (12)(34) (2*2*3 ways), (12)(3)(4) (2*6 ways), (1)(2)(3)(4) (1 way).
G.f. = 1 + x + 3*x^2 + 13*x^3 + 73*x^4 + 501*x^4 + 4051*x^5 + 37633*x^6 + 394353*x^7 + ...
		

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 194.
  • 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

Row sums of A271703 and for n >= 1 of A008297. Unsigned row sums of A111596.
A002868 is maximal element of the n-th row of A271703 and for n >= 1 of A008297.
Main diagonal of A257740 and of A319501.

Programs

  • GAP
    a:=[1,1];; for n in [3..10^2] do a[n]:=(2*n-3)*a[n-1]-(n-2)*(n-3)*a[n-2]; od; A000262:=a;  # Muniru A Asiru, Oct 01 2017
    
  • Haskell
    a000262 n = a000262_list !! n
    a000262_list = 1 : 1 : zipWith (-)
                   (tail $ zipWith (*) a005408_list a000262_list)
                          (zipWith (*) a002378_list a000262_list)
    -- Reinhard Zumkeller, Mar 06 2014
    
  • Magma
    I:=[1,3]; [1] cat [n le 2 select I[n]  else (2*n-1)*Self(n-1) - (n-1)*(n-2)*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 14 2019
    
  • Magma
    [Factorial(n)*Evaluate(LaguerrePolynomial(n, -1), -1): n in [0..30]]; // G. C. Greubel, Feb 23 2021
    
  • Maple
    A000262 := proc(n) option remember: if n=0 then RETURN(1) fi: if n=1 then RETURN(1) fi: (2*n-1)*procname(n-1) - (n-1)*(n-2)*procname(n-2) end proc:
    for n from 0 to 20 do printf(`%d,`,a(n)) od:
    spec := [S, {S = Set(Prod(Z,Sequence(Z)))}, labeled]; [seq(combstruct[count](spec, size=n), n=0..40)];
    with(combinat):seq(sum(abs(stirling1(n, k))*bell(k), k=0..n), n=0..18); # Zerinvary Lajos, Nov 26 2006
    B:=[S,{S = Set(Sequence(Z,1 <= card),card <=13)},labelled]: seq(combstruct[count](B, size=n), n=0..19);# Zerinvary Lajos, Mar 21 2009
    a := n -> `if`(n=0, 1, n!*hypergeom([1 - n], [2], -1)): seq(simplify(a(n)), n=0..19); # Peter Luschny, Jun 05 2014
  • Mathematica
    Range[0, 19]! CoefficientList[ Series[E^(x/(1-x)), {x, 0, 19}], x] (* Robert G. Wilson v, Apr 04 2005 *)
    a[ n_]:= If[ n<0, 0, n! SeriesCoefficient[ Exp[x/(1-x)], {x, 0, n}]]; (* Michael Somos, Jul 19 2005 *)
    a[n_]:=If[n==0, 1, n! Sum[Binomial[n-1, j]/(j+1)!, {j, 0, n-1}]]; Table[a[n], {n, 0, 30}] (* Wilf *)
    a[0] = 1; a[n_]:= n!*Hypergeometric1F1[n+1, 2, 1]/E; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jun 18 2012, after Shai Covo *)
    Table[Sum[BellY[n, k, Range[n]!], {k, 0, n}], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *)
    a[ n_]:= If[ n<0, 0, n! SeriesCoefficient[ Product[ QPochhammer[x^k]^(-MoebiusMu[k]/k), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 02 2019 *)
    Table[n!*LaguerreL[n, -1, -1], {n, 0, 30}] (* G. C. Greubel, Feb 23 2021 *)
    RecurrenceTable[{a[n] == (2*n - 1)*a[n-1] - (n-1)*(n-2)*a[n-2], a[0] == 1, a[1] == 1}, a, {n, 0, 30}] (* Vaclav Kotesovec, Jul 21 2022 *)
  • Maxima
    makelist(sum(abs(stirling1(n,k))*belln(k),k,0,n),n,0,24); /* Emanuele Munarini, Jul 04 2011 */
    
  • Maxima
    makelist(hypergeometric([-n+1,-n],[],1),n,0,12); /* Emanuele Munarini, Sep 27 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( x / (1 - x) + x * O(x^n)), n))}; /* Michael Somos, Feb 10 2005 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( prod( k=1, n, eta(x^k + x * O(x^n))^( -moebius(k) / k)), n))}; /* Michael Somos, Feb 10 2005 */
    
  • PARI
    {a(n) = s = 1; for(k = 1, n-1, s = 1 + k * s / (k + s)); return( numerator(s))}; /* Dimitris Valianatos, Aug 03 2018 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( prod( k=1, n, (1 - x^k + x * O(x^n))^(-eulerphi(k) / k)), n))}; /* Michael Somos, Jun 02 2019 */
    
  • PARI
    a(n) = if (n==0, 1, (n-1)!*pollaguerre(n-1,1,-1)); \\ Michel Marcus, Feb 23 2021; Jul 13 2024
    
  • Python
    from sympy import hyper, hyperexpand
    def A000262(n): return hyperexpand(hyper((-n+1, -n), [], 1)) # Chai Wah Wu, Jan 14 2022
  • Sage
    A000262 = lambda n: hypergeometric([-n+1, -n], [], 1)
    [simplify(A000262(n)) for n in (0..19)] # Peter Luschny, Apr 08 2015
    

Formula

D-finite with recurrence: a(n) = (2*n-1)*a(n-1) - (n-1)*(n-2)*a(n-2).
E.g.f.: exp( x/(1-x) ).
a(n) = Sum_{k=0..n} |A008275(n,k)| * A000110(k). - Vladeta Jovovic, Feb 01 2003
a(n) = (n-1)!*LaguerreL(n-1,1,-1) for n >= 1. - Vladeta Jovovic, May 10 2003
Representation as n-th moment of a positive function on positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x-1)*BesselI(1, 2*x^(1/2))/x^(1/2) dx, n >= 1. - Karol A. Penson, Dec 04 2003
a(n) = Sum_{k=0..n} A001263(n, k)*k!. - Philippe Deléham, Dec 10 2003
a(n) = n! Sum_{j=0..n-1} binomial(n-1, j)/(j+1)!, for n > 0. - Herbert S. Wilf, Jun 14 2005
Asymptotic expansion for large n: a(n) -> (0.4289*n^(-1/4) + 0.3574*n^(-3/4) - 0.2531*n^(-5/4) + O(n^(-7/4)))*(n^n)*exp(-n + 2*sqrt(n)). - Karol A. Penson, Aug 28 2002
Minor part of this asymptotic expansion is wrong! Right is (in closed form): a(n) ~ n^(n-1/4)*exp(-1/2+2*sqrt(n)-n)/sqrt(2) * (1 - 5/(48*sqrt(n)) - 95/(4608*n)), numerically a(n) ~ (0.42888194248*n^(-1/4) - 0.0446752023417*n^(-3/4) - 0.00884196713*n^(-5/4) + O(n^(-7/4))) *(n^n)*exp(-n+2*sqrt(n)). - Vaclav Kotesovec, Jun 02 2013
a(n) = exp(-1)*Sum_{m>=0} [m]^n/m!, where [m]^n = m*(m+1)*...*(m+n-1) is the rising factorial. - Vladeta Jovovic, Sep 20 2006
Recurrence: D(n,k) = D(n-1,k-1) + (n-1+k) * D(n-1,k) n >= k >= 0; D(n,0)=0. From this, D(n,1) = n! and D(n,n)=1; a(n) = Sum_{i=0..n} D(n,i). - Stephen Dalton (StephenMDalton(AT)gmail.com), Jan 05 2007
Proof: Notice two distinct subsets of the lists for [n]: 1) n is in its own list, then there are D(n-1,k-1); 2) n is in a list with other numbers. Denoting the separation of lists by |, it is not hard to see n has (n-1+k) possible positions, so (n-1+k) * D(n-1,k). - Stephen Dalton (StephenMDalton(AT)gmail.com), Jan 05 2007
Define f_1(x), f_2(x), ... such that f_1(x) = exp(x), f_{n+1}(x) = (d/dx)(x^2*f_n(x)), for n >= 2. Then a(n-1) = exp(-1)*f_n(1). - Milan Janjic, May 30 2008
a(n) = (n-1)! * Sum_{k=1..n} (a(n-k)*k!)/((n-k)!*(k-1)!), where a(0)=1. - Thomas Wieder, Sep 10 2008
a(n) = exp(-1)*n!*M(n+1,2,1), n >= 1, where M (=1F1) is the confluent hypergeometric function of the first kind. - Shai Covo (green355(AT)netvision.net.il), Jan 20 2010
a(n) = n!* A067764(n) / A067653(n). - Gary Detlefs, May 15 2010
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x)^2*d/dx. Cf. A000110, A049118, A049119 and A049120. - Peter Bala, Nov 25 2011
From Sergei N. Gladkovskii, Nov 17 2011, Aug 02 2012, Dec 11 2012, Jan 27 2013, Jul 31 2013, Dec 25 2013: (Start)
Continued fractions:
E.g.f.: Q(0) where Q(k) = 1+x/((1-x)*(2k+1)-x*(1-x)*(2k+1)/(x+(1-x)*(2k+2)/Q(k+1))).
E.g.f.: 1 + x/(G(0)-x) where G(k) = (1-x)*k + 1 - x*(1-x)*(k+1)/G(k+1).
E.g.f.: exp(x/(1-x)) = 4/(2-(x/(1-x))*G(0))-1 where G(k) = 1 - x^2/(x^2 + 4*(1-x)^2*(2*k+1)*(2*k+3)/G(k+1) ).
E.g.f.: 1 + x*(E(0)-1)/(x+1) where E(k) = 1 + 1/(k+1)/(1-x)/(1-x/(x+1/E(k+1) )).
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - x/(x + (k+1)*(1-x)/E(k+1) )).
E.g.f.: E(0)-1, where E(k) = 2 + x/( (2*k+1)*(1-x) - x/E(k+1) ).
(End)
E.g.f.: Product {n >= 1} ( (1 + x^n)/(1 - x^n) )^( phi(2*n)/(2*n) ), where phi(n) = A000010(n) is the Euler totient function. Cf. A088009. - Peter Bala, Jan 01 2014
a(n) = n!*hypergeom([1-n],[2],-1) for n >= 1. - Peter Luschny, Jun 05 2014
a(n) = (-1)^(n-1)*KummerU(1-n,2,-1). - Peter Luschny, Sep 17 2014
a(n) = hypergeom([-n+1, -n], [], 1) for n >= 0. - Peter Luschny, Apr 08 2015
E.g.f.: Product_{k>0} exp(x^k). - Franklin T. Adams-Watters, May 11 2016
0 = a(n)*(18*a(n+2) - 93*a(n+3) + 77*a(n+4) - 17*a(n+5) + a(n+6)) + a(n+1)*(9*a(n+2) - 80*a(n+3) + 51*a(n+4) - 6*a(n+5)) + a(n+2)*(3*a(n+2) - 34*a(n+3) + 15*a(n+4)) + a(n+3)*(-10*a(n+3)) if n >= 0. - Michael Somos, Feb 27 2017
G.f. G(x)=y satisfies a differential equation: (1-x)*y-2*(1-x)*x^2*y'+x^4*y''=1. - Bradley Klee, Aug 13 2018
a(n) = n! * LaguerreL(n, -1, -1) = c_{n}(n-1; -1) where c_{n}(x; a) are the Poisson - Charlier polynomials. - G. C. Greubel, Feb 23 2021
3 divides a(3*n-1); 9 divides a(9*n-1); 11 divides a(11*n-1). - Peter Bala, Mar 26 2022
For n > 0, a(n) = Sum_{k=0..n-1}*k!*C(n-1,k)*C(n,k). - Francesca Aicardi, Nov 03 2022
For n > 0, a(n) = (n-1)! * (Sum_{i=0..n-1} A002720(i) / i!). - Werner Schulte, Mar 29 2024
a(n+1) = numerator of (1 + n/(1 + n/(1 + (n-1)/(1 + (n-1)/(1 + ... + 1/(1 + 1/(1))))))). See A002720 for the denominators. - Peter Bala, Feb 11 2025

A251568 Expansion of e.g.f. exp(x*C(x)^2) where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers, A000108.

Original entry on oeis.org

1, 1, 5, 43, 529, 8501, 169021, 4010455, 110676833, 3484717129, 123320412181, 4847038223171, 209536628422705, 9882471447634813, 505033804901100749, 27802319803528367791, 1640388588050579832001, 103275015543414629215505, 6910877628962983581031333
Offset: 0

Views

Author

Paul D. Hanna, Dec 05 2014

Keywords

Examples

			E.g.f.: A(x) = 1 + x + 5*x^2/2! + 43*x^3/3! + 529*x^4/4! + 8501*x^5/5! + ...
where
log(A(x)) = x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + ... + A000108(n)*x^n + ...
		

Crossrefs

Programs

  • Maple
    CatalanNumber := n -> binomial(2*n,n)/(n+1):
    a := n -> `if`(n=0, 1, n!*CatalanNumber(n)*hypergeom([1-n], [2+n], -1)):
    seq(simplify(a(n)), n=0..9); # Peter Luschny, May 04 2017
  • Mathematica
    Flatten[{1,Table[Sum[n!/k!*Binomial[2*n-1,n-k]*2*k/(n+k),{k,1,n}],{n,1,20}]}] (* Vaclav Kotesovec, Feb 14 2015 *)
    a[0] = 1; a[n_] := (2n)!/(n+1)! Hypergeometric1F1[1-n, n+2, -1];
    Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 03 2017, after Vladimir Kruchinin *)
  • PARI
    {a(n)=my(C=1);for(i=1,n,C=1+x*C^2 +x*O(x^n));n!*polcoef(exp(x*C^2),n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n) = if(n==0, 1, sum(k=1, n, n!/k! * binomial(2*n-1, n-k) * 2*k/(n+k) ))}
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(exp(serreverse(x*(1-x))^2/x))) \\ Seiichi Manyama, Mar 15 2025

Formula

a(n) = Sum_{k=0..n} (n!/k!) * binomial(2*n-1, n-k) * 2*k/(n+k) for n > 0 with a(0)=1.
E.g.f. A(x) satisfies: A'(x)/A(x) = C'(x) = C(x)^2 / sqrt(1-4*x) where C(x) = (1-sqrt(1-4*x))/(2*x) is the Catalan function.
Recurrence equation: a(n) = -(n^2 - 5*n +1)*a(n-1) + n*(2*n - 3)*(2*n - 4)*a(n-2) with a(0) = 1, a(1) = 1. It appears that a(n) - 1 is divisible by n*(n - 1) for n >= 2. - Peter Bala, Feb 14 2015
a(n) ~ 2^(2*n+1/2) * n^(n-1) / exp(n-1). - Vaclav Kotesovec, Feb 14 2015
a(n) are special values of the hypergeometric function 1F1: a(n) = 4^n*Gamma(n+1/2)*exp(-1)*hypergeom([2*n+1], [n+2], 1)/(sqrt(Pi)*(n+1)), for n>=1. - Karol A. Penson, Jun 01 2015
a(n) = ((2*n)!/(n+1)!)*hypergeometric([1-n],[n+2],-1), a(0)=1. - Vladimir Kruchinin, May 03 2017
From Seiichi Manyama, Mar 15 2025: (Start)
E.g.f.: exp( (1/x) * Series_Reversion( x*(1-x) )^2 ).
E.g.f.: exp( Series_Reversion( x/(1+x)^2 ) ). (End)

A091695 Expansion of e.g.f. exp(x/(1-x)^3).

Original entry on oeis.org

1, 1, 7, 55, 529, 6121, 82711, 1273567, 21945505, 417540529, 8680953511, 195582295591, 4742407056817, 123045795823705, 3399348471640759, 99573135106176271, 3081061456572152641, 100382623544966098657, 3433727597233037475655, 123000248740384210119319, 4603377404407810366309201
Offset: 0

Views

Author

Karol A. Penson, Jan 29 2004

Keywords

Comments

Special values of the hypergeometric function 3F3: a(n) = n!*binomial(n+1,n-1) * hypergeom([ -n+1, 1/2*n+1, 1/2*n+3/2], [4/3, 5/3, 2], -4/27) for n>0.

Crossrefs

Column k=3 of A293012.
Cf. A082579.

Programs

  • Mathematica
    CoefficientList[Series[E^(x/(1-x)^3), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 27 2013 *)
  • PARI
    x='x+O('x^33);
    Vec(serlaplace(exp( x/(1-x)^3 )))
    /* Joerg Arndt, Sep 14 2012 */

Formula

E.g.f.: exp(x/(1-x)^3).
a(n) ~ 1/2*exp(-1/27-n^(1/4)*3^(3/4)/72+sqrt(3*n)/6+4/3*n^(3/4)*3^(1/4)-n)*3^(1/8)*n^(n-1/8). - Vaclav Kotesovec, Jun 27 2013
a(n) = n! * Sum_{k=0..n} binomial(n+2*k-1,n-k)/k!. - Seiichi Manyama, Mar 06 2023

Extensions

Prepended a(0)=1, Joerg Arndt, Sep 14 2012.

A361278 Expansion of e.g.f. exp(x * (1+x)^2).

Original entry on oeis.org

1, 1, 5, 19, 97, 581, 3661, 26335, 202049, 1659817, 14621941, 135567851, 1326672865, 13624218349, 146056961597, 1633376573431, 18980051829121, 228677164878545, 2852155973178469, 36740599423566787, 488127224550517601, 6678832987859315221
Offset: 0

Views

Author

Seiichi Manyama, Mar 06 2023

Keywords

Crossrefs

Column k=2 of A361277.
Cf. A082579.

Programs

  • Maple
    A361278 := proc(n)
        n!*add(binomial(2*k,n-k)/k!,k=0..n) ;
    end proc:
    seq(A361278(n),n=0..60) ; #R. J. Mathar, Mar 12 2023
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(x*(1+x)^2)))
    
  • PARI
    a(n) = n!*sum(k=0, n, binomial(2*k, n-k)/k!);
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=1, i, j*binomial(2, j-1)*v[i-j+1]/(i-j)!)); v;

Formula

a(n) = n! * Sum_{k=0..n} binomial(2*k,n-k)/k!.
a(0) = 1; a(n) = (n-1)! * Sum_{k=1..n} k * binomial(2,k-1) * a(n-k)/(n-k)!.
D-finite with recurrence a(n) -a(n-1) +4*(-n+1)*a(n-2) -3*(n-1)*(n-2)*a(n-3)=0. - R. J. Mathar, Mar 12 2023
a(n) ~ 3^(n/3 - 1/2) * n^(2*n/3) / exp(2*n/3 - 2*3^(-2/3)*n^(2/3) - 3^(-7/3)*n^(1/3) + 4/81) * (1 + 953*3^(1/3)/(4374*n^(1/3)) - 2051059*3^(2/3)/(191318760*n^(2/3))). - Vaclav Kotesovec, Nov 11 2023

A293012 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of e.g.f. exp(x/(1 - x)^k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 13, 1, 1, 1, 7, 31, 73, 1, 1, 1, 9, 55, 241, 501, 1, 1, 1, 11, 85, 529, 2261, 4051, 1, 1, 1, 13, 121, 961, 6121, 24781, 37633, 1, 1, 1, 15, 163, 1561, 13041, 82711, 309835, 394353, 1, 1, 1, 17, 211, 2353, 24101, 207001, 1273567, 4342241, 4596553, 1
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 28 2017

Keywords

Examples

			E.g.f. of column k: A_k(x) =  1 + x/1! + (2*k + 1)*x^2/2! + (3*k^2 + 9*k + 1)*x^3/3! + (4*k^3 + 36*k^2 + 32*k + 1)*x^4/4! + ...
Square array begins:
  1,   1,    1,    1,     1,     1,  ...
  1,   1,    1,    1,     1,     1,  ...
  1,   3,    5,    7,     9,    11,  ...
  1,  13,   31,   55,    85,   121,  ...
  1,  73,  241,  529,   961,  1561,  ...
  1, 501, 2261, 6121, 13041, 24101,  ...
		

Crossrefs

Columns k=0..4 give A000012, A000262, A082579, A091695, A361283.
Main diagonal gives A293013.

Programs

  • Mathematica
    Table[Function[k, n! SeriesCoefficient[Exp[x/(1 - x)^k], {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten
  • PARI
    T(n, k) = n!*sum(j=0, n, binomial(n+(k-1)*j-1, n-j)/j!); \\ Seiichi Manyama, Mar 06 2023

Formula

E.g.f. of column k: exp(x/(1 - x)^k).
From Seiichi Manyama, Oct 21 2017: (Start)
Let B(j,k) = (-1)^(j-1)*binomial(-k,j-1) for j>0 and k>=0.
A(0,k) = 1 and A(n,k) = (n-1)! * Sum_{j=1..n} j*B(j,k)*A(n-j,k)/(n-j)! for n > 0. (End)
A(n,k) = n! * Sum_{j=0..n} binomial(n+(k-1)*j-1,n-j)/j!. - Seiichi Manyama, Mar 06 2023

A255807 E.g.f.: exp(Sum_{k>=1} k^2 * x^k).

Original entry on oeis.org

1, 1, 9, 79, 841, 10821, 162601, 2777419, 52960209, 1112813641, 25509407401, 632772511911, 16870674740569, 480717000225229, 14568646143888201, 467640968478534691, 15841420612530533281, 564519727866573515409, 21102817266052772063689, 825435163723385398719871
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 07 2015

Keywords

Comments

In general, if e.g.f. = exp(Sum_{k>=1} k^m * x^k) and m>0, then a(n) ~ (m+2)^(-1/2) * Gamma(m+2)^(1/(2*m+4)) * exp((m+2)/(m+1) * Gamma(m+2)^(1/(m+2)) * n^((m+1)/(m+2)) + zeta(-m) - n) * n^(n - 1/(2*m+4)).
It appears that the sequence a(n) taken modulo 10 is periodic with period 5. More generally, we conjecture that for k = 2,3,4,... the difference a(n+k) - a(n) is divisible by k: if true, then the sequence a(n) taken modulo k would be periodic with period dividing k. - Peter Bala, Nov 14 2017
The above conjecture is true - see the Bala link. - Peter Bala, Jan 20 2018

Crossrefs

Programs

  • Mathematica
    nmax=20; CoefficientList[Series[Exp[Sum[k^2*x^k,{k,1,nmax}]],{x,0,nmax}],x] * Range[0,nmax]!
    nn = 20; Range[0, nn]! * CoefficientList[Series[Product[Exp[k^2*x^k], {k, 1, nn}], {x, 0, nn}], x] (* Vaclav Kotesovec, Mar 21 2016 *)

Formula

E.g.f.: exp(x*(1+x)/(1-x)^3).
a(n) ~ 2^(-7/8) * 3^(1/8) * n^(n-1/8) * exp(2^(9/4) * 3^(-3/4) * n^(3/4) - n).
a(n) = n!*y(n) where y(0)=1 and y(n)=(Sum_{k=0..n-1} (n-k)^3*y(k))/n for n>=1. - Benedict W. J. Irwin, Jun 02 2016
a(n) = (4*n-3)*a(n-1) - 2*(n-1)*(3*n-8)*a(n-2) + (n-1)*(n-2)*(4*n-11)*a(n-3) - (n-1)*(n-2)*(n-3)*(n-4)*a(n-4). - Peter Bala, Nov 12 2017
E.g.f.: Product_{k>=1} 1/(1 - x^k)^(J_3(k)/k), where J_3() is the Jordan function (A059376). - Ilya Gutkovskiy, May 25 2019

A335344 Expansion of e.g.f. exp(x^2/(2*(1 - x)^2)).

Original entry on oeis.org

1, 0, 1, 6, 39, 300, 2715, 28350, 335265, 4422600, 64298745, 1020477150, 17542820295, 324552648420, 6426708843555, 135573281994150, 3034400481137025, 71801374285040400, 1790523094644709425, 46923435009924823350, 1289032229351717425575
Offset: 0

Views

Author

Ilya Gutkovskiy, Jun 02 2020

Keywords

Comments

For n>0, a(n) is also the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select 2 people from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them). See example below. - Enrique Navarrete, Oct 01 2023

Examples

			For n = 5, using one table, there are 4! circular seatings and binomial(5,2) ways to select 2 persons, hence 240 ways. Using two tables, the only way we can select 2 persons from each one is seating 3 persons in one table and 2 in the other, which can be done in 20 ways; then choosing 2 persons from each table can be done in 3 ways, hence giving another 60 ways for a total of 300. - _Enrique Navarrete_, Oct 01 2023
		

Crossrefs

Programs

  • Maple
    f:= gfun:-rectoproc({a(n) = -(n-1)*(3*n-7)*a(n-2) + 3*(n-1)*a(n-1) + (n - 1)*(n - 2)*(n - 3)*a(n-3),a(0)=1,a(1)=0,a(2)=1},a(n),remember):
    map(f, [$0..30]); # Robert Israel, Jun 04 2020
  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[x^2/(2 (1 - x)^2)], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = (1/2) Sum[Binomial[n - 1, k - 1] (k - 1) k! a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 20}]
  • PARI
    seq(n)=Vec(serlaplace(exp(x^2/(2*(1 - x)^2) + O(x*x^n)))) \\ Andrew Howroyd, Jun 02 2020

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * A001286(k) * a(n-k).
D-finite with recurrence a(n) = -(n - 1)*(3*n - 7)*a(n - 2) + 3*(n - 1)*a(n - 1) + (n - 1)*(n - 2)*(n - 3)*a(n - 3). - Robert Israel, Jun 04 2020
a(n) ~ n^(n - 1/6) * exp(1/6 - n^(1/3)/2 + 3*n^(2/3)/2 - n) / sqrt(3). - Vaclav Kotesovec, Jun 11 2020
a(n) = n! * Sum_{k=0..floor(n/2)} binomial(n-1,n-2*k)/(2^k * k!). - Seiichi Manyama, Mar 16 2023

A318215 Expansion of e.g.f. exp(x/(1 + x)^2).

Original entry on oeis.org

1, 1, -3, 7, 1, -219, 2581, -22973, 162177, -554039, -10506419, 343049631, -6846400703, 113528248237, -1609627861659, 17371462450651, -36066494745599, -5681921495461743, 243263898097515037, -7398126521141652809, 193119003246643917441, -4476119490014676723659, 89171014860669488040757
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 21 2018

Keywords

Crossrefs

Programs

  • Maple
    A318215 := proc(n)
        add((-1)^(n-k)*binomial(n+k-1,2*k-1)*n!/k!,k=0..n) ;
    end proc:
    seq(A318215(n),n=0..42) ; # R. J. Mathar, Aug 20 2021
  • Mathematica
    nmax = 22; CoefficientList[Series[Exp[x/(1 + x)^2], {x, 0, nmax}], x] Range[0, nmax]!
    Table[Sum[(-1)^(n - k) Binomial[n + k - 1, 2 k - 1] n!/k!, {k, 0, n}], {n, 0, 22}]
    a[n_] := a[n] = Sum[(-1)^(k + 1) k k! Binomial[n - 1, k - 1] a[n - k], {k, 1, n}]; a[0] = 1; Table[a[n], {n, 0, 22}]
    Join[{1}, Table[(-1)^(n + 1) n n! HypergeometricPFQ[{1 - n, 1 + n}, {3/2, 2}, 1/4], {n, 22}]]

Formula

E.g.f.: Product_{k>=1} exp((-1)^(k+1)*k*x^k).
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n+k-1,2*k-1)*n!/k!.
a(0) = 1; a(n) = Sum_{k=1..n} (-1)^(k+1)*k*k!*binomial(n-1,k-1)*a(n-k).
D-finite with recurrence a(n) +(3*n-4)*a(n-1) +(n-1)*(3*n-5)*a(n-2) +(n-1)*(n-2)*(n-3)*a(n-3)=0. - R. J. Mathar, Aug 20 2021

A255819 E.g.f.: exp(Sum_{k>=1} k^3 * x^k).

Original entry on oeis.org

1, 1, 17, 211, 3049, 54221, 1131601, 26714647, 700868561, 20208794329, 634445325361, 21512122643771, 782497124407417, 30364699568650981, 1251108918727992689, 54512805637285532671, 2502891521610396838561, 120718449425308259052977, 6099522639316776103853521
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 07 2015

Keywords

Comments

In general, if e.g.f. = exp(Sum_{k>=1} k^m * x^k) and m>0, then a(n) ~ (m+2)^(-1/2) * Gamma(m+2)^(1/(2*m+4)) * exp((m+2)/(m+1) * Gamma(m+2)^(1/(m+2)) * n^((m+1)/(m+2)) + zeta(-m) - n) * n^(n - 1/(2*m+4)).
It appears that the sequence a(n) taken modulo 10 is periodic with period 5. More generally, we conjecture that for k = 2,3,4,... the difference a(n+k) - a(n) is divisible by k: if true, then the sequence a(n) taken modulo k would be periodic with period dividing k. - Peter Bala, Nov 14 2017
The above conjecture is true - see the Bala link. - Peter Bala, Jan 20 2018

Crossrefs

Programs

  • Mathematica
    nmax=20; CoefficientList[Series[Exp[Sum[k^3*x^k,{k,1,nmax}]],{x,0,nmax}],x] * Range[0,nmax]!
    nn = 20; Range[0, nn]! * CoefficientList[Series[Product[Exp[k^3*x^k], {k, 1, nn}], {x, 0, nn}], x] (* Vaclav Kotesovec, Mar 21 2016 *)

Formula

E.g.f.: exp(x*(1 + 4*x + x^2)/(1-x)^4).
a(n) ~ 2^(3/10) * 3^(1/10) * 5^(-1/2) * n^(n-1/10) * exp(1/120 + 5 * 2^(-7/5) * 3^(1/5) * n^(4/5) - n).
a(n) = y(n)*n! where y(0)=1 and y(n)=(Sum_{k=0..n-1} (n-k)^4*y(k))/n for n>=1. - Benedict W. J. Irwin, Jun 02 2016
E.g.f.: Product_{k>=1} 1/(1 - x^k)^(J_4(k)/k), where J_4(k) is the Jordan function (A059377). - Ilya Gutkovskiy, May 25 2019

A293718 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of e.g.f. exp(Sum_{j=1..k} j*x^j).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 5, 13, 1, 1, 1, 5, 31, 73, 1, 1, 1, 5, 31, 145, 281, 1, 1, 1, 5, 31, 241, 1181, 1741, 1, 1, 1, 5, 31, 241, 1661, 9661, 8485, 1, 1, 1, 5, 31, 241, 2261, 16861, 77155, 57233, 1, 1, 1, 5, 31, 241, 2261, 20461, 181315, 794081
Offset: 0

Views

Author

Seiichi Manyama, Oct 15 2017

Keywords

Examples

			Square array begins:
   1,   1,    1,    1,    1, ...
   1,   1,    1,    1,    1, ...
   1,   5,    5,    5,    5, ...
   1,  13,   31,   31,   31, ...
   1,  73,  145,  241,  241, ...
   1, 281, 1181, 1661, 2261, ...
		

Crossrefs

Columns k=1..4 give A000012, A115329, A293716, A293717.
Rows n=0-1 give A000012.
Main diagonal gives A082579.

Formula

A(0,k) = 1 and A(n,k) = (n-1)! * Sum_{j=1..min(k,n)} j^2*A(n-j,k)/(n-j)!.
Showing 1-10 of 26 results. Next