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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A002057 Fourth convolution of Catalan numbers: a(n) = 4*binomial(2*n+3,n)/(n+4).

Original entry on oeis.org

1, 4, 14, 48, 165, 572, 2002, 7072, 25194, 90440, 326876, 1188640, 4345965, 15967980, 58929450, 218349120, 811985790, 3029594040, 11338026180, 42550029600, 160094486370, 603784920024, 2282138106804, 8643460269248, 32798844771700, 124680849918352
Offset: 0

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Author

Keywords

Comments

a(n) is sum of the (flattened) list obtained by the iteration of: replace each integer k with the list 0,...,k+1 on the starting value 0. Length of this list is Catalan(n) or A000108. - Wouter Meeussen, Nov 11 2001
a(n-2) is the number of n-th generation vertices in the tree of sequences with unit increase labeled by 3 (cf. Zoran Sunic reference). - Benoit Cloitre, Oct 07 2003
Number of standard tableaux of shape (n+2,n-1). - Emeric Deutsch, May 30 2004
a(n) = CatalanNumber(n+3) - 2*CatalanNumber(n+2). Proof. From its definition as a convolution of Catalan numbers, a(n) counts lists of 4 Dyck paths of total size (semilength) = n. Connect the 4 paths by 3 upsteps (U) and append 3 downsteps (D). This is a reversible procedure. So a(n) is also the number of Dyck (n+3)-paths that end DDD (D for downstep). Let C(n) denote CatalanNumber(n) (A000108). Since C(n+3) is the total number of Dyck (n+3)-paths and C(n+2) is the number that end UD, we have (*) C(n+3) - C(n+2) is the number of Dyck (n+3)-paths that end DD. Also, (**) C(n+2) is the number of Dyck (n+3)-paths that end UDD (change the last D in a Dyck (n+2)-path to UDD). Subtracting (**) from (*) yields a(n) = C(n+3) - 2C(n+2) as claimed. - David Callan, Nov 21 2006
Convolution square of the Catalan sequence without one of the initial "1"'s: (1 + 4x + 14x^2 + 48x^3 + ...) = (1/x^2) * square(x + 2x^2 + 5x^3 + 14x^4 + ...)
a(n) is the number of binary trees with n+3 internal nodes in which both subtrees of the root are nonempty. Cf. A068875 [Sedgewick and Flajolet]. - Geoffrey Critzer, Jan 05 2013
With offset 4, a(n) is the number of permutations on {1,2,...,n} that are 123-avoiding, i.e., do not contain a three-term monotone subsequence, for which the first ascent is at positions (4,5); for example, there are 48 123-avoiding permutations on n=7 for which the first ascent is at spots (4,5). See Connolly link. There it is shown in general that the k-th Catalan Convolution is the number of 123-avoiding permutations for which the first ascent is at (k, k+1). (For n=k, the first ascent is defined to be at positions (k,k+1) if the permutation is the decreasing permutation with no ascents.) - Anant Godbole, Jan 17 2014
With offset 4, a(n) is the number of permutations on {1,2,...,n} that are 123-avoiding and for which the integer n is in the 4th spot; see Connolly link. - Anant Godbole, Jan 17 2014
a(n) is the number of North-East lattice paths from (0,0) to (n+2,n+2) that have exactly one east step below the subdiagonal y = x-1. Details can be found in Section 3.1 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
a(n) is the number of North-East lattice paths from (0,0) to (n+2,n+2) that bounce off the diagonal y = x to the right exactly once but do not bounce off y = x to the left. Details can be found in Section 4.2 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
a(n) is the number of North-East lattice paths from (0,0) to (n+2,n+2) that horizontally cross the diagonal y = x exactly once but do not cross the diagonal vertically. Details can be found in Section 4.3 in Pan and Remmel's link. - Ran Pan, Feb 04 2016
Apparently also Young tableaux of (non-partition) shape [n+1, 1, 1, n+1], see example file. - Joerg Arndt, Dec 30 2023

Examples

			From _Peter Bala_, Apr 14 2017: (Start)
This sequence appears on the main diagonal of a generalized Catalan triangle. Construct a lower triangular array (T(n,k)), n,k >= 0 by placing the sequence [0,0,0,1,1,1,1,...] in the first column and then filling in the remaining entries in the array using the rule T(n,k) = T(n,k-1) + T(n-1,k). The resulting array begins
  n\k| 0 1  2  3  4   5   6   7  ...
  ---+-------------------------------
   0 | 0
   1 | 0 0
   2 | 0 0  0
   3 | 1 1  1  1
   4 | 1 2  3  4  4
   5 | 1 3  6 10 14  14
   6 | 1 4 10 20 34  48  48
   7 | 1 5 15 35 69 117 165 165
   ...
(see Tedford 2011; this is essentially the array C_4(n,k) in the notation of Lee and Oh). Compare with A279004. (End)
		

References

  • Pierre de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 11, coefficients of P_4(z).
  • C. Krishnamachary and M. Bheemasena Rao, Determinants whose elements are Eulerian, prepared Bernoullian and other numbers, J. Indian Math. Soc., Vol. 14 (1922), pp. 55-62, 122-138 and 143-146.
  • Robert Sedgewick and Phillipe Flajolet, Analysis of Algorithms, Addison Wesley, 1996, page 225.
  • 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

T(n, n+4) for n=0, 1, 2, ..., array T as in A047072. Also a diagonal of A059365 and of A009766.
Cf. A001003.
A diagonal of any of the essentially equivalent arrays A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Cf. A145596 (row sums), A279004.

Programs

  • GAP
    List([0..25],n->4*Binomial(2*n+3,n)/(n+4)); # Muniru A Asiru, Mar 05 2018
    
  • Magma
    [4*Binomial(2*n+3,n)/(n+4): n in [0..30]]; // Vincenzo Librandi, Feb 04 2016
    
  • Maple
    a := n -> 32*4^n*GAMMA(5/2+n)*(1+n)/(sqrt(Pi)*GAMMA(5+n)):
    seq(a(n),n=0..23); # Peter Luschny, Dec 14 2015
    A002057List := proc(m) local A, P, n; A := [1]; P := [1,1,1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-1]]);
    A := [op(A), P[-1]] od; A end: A002057List(27); # Peter Luschny, Mar 26 2022
  • Mathematica
    Table[Plus@@Flatten[Nest[ #/.a_Integer:> Range[0, a+1]&, {0}, n]], {n, 0, 10}]
    Table[4 Binomial[2n+3,n]/(n+4),{n,0,30}] (* or *) CoefficientList[ Series[ (1-Sqrt[1-4 x]+2 x (-2+Sqrt[1-4 x]+x))/(2 x^4),{x,0,30}],x] (* Harvey P. Dale, May 05 2011 *)
  • PARI
    {a(n) = if( n<0, 0, n+=2; 2*binomial(2*n, n-2) / n)}; /* Michael Somos, Jul 31 2005 */
    
  • PARI
    x='x+O('x^100); Vec((1-(1-4*x)^(1/2)+2*x*(-2+(1-4*x)^(1/2)+x))/(2*x^4)) \\ Altug Alkan, Dec 14 2015
    
  • SageMath
    [2*(n+1)*catalan_number(n+2)/(n+4) for n in (0..30)] # G. C. Greubel, May 27 2022

Formula

a(n) = A033184(n+4, 4) = 4*binomial(2*n+3, n)/(n+4) = 2*(n+1)*A000108(n+2)/(n+4).
G.f.: c(x)^4 with c(x) g.f. of A000108 (Catalan).
Row sums of A145596. Column 4 of A033184. By specializing the identities for the row polynomials given in A145596 we obtain the results a(n) = Sum_{k = 0..n} (-1)^k*binomial(n+1,k+1)*a(k)*4^(n-k) and a(n) = Sum_{k = 0..floor(n/2)} binomial(n+1,2*k+1) * Catalan(k+1) * 2^(n-2*k). From the latter identity we can derive the congruences a(2n+1) == 0 (mod 4) and a(2n) == Catalan(n+1) (mod 4). It follows that a(n) is odd if and only if n = (2^m - 4) for some m >= 2. - Peter Bala, Oct 14 2008
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n>=3, a(n-3) = (-1)^(n-3) * coeff(charpoly(A,x), x^3). - Milan Janjic, Jul 08 2010
G.f.: (1-sqrt(1-4*x) + 2*x*(-2+sqrt(1-4*x) + x))/(2*x^4). - Harvey P. Dale, May 05 2011
a(n+1) = A214292(2*n+4,n). - Reinhard Zumkeller, Jul 12 2012
D-finite with recurrence: (n+4)a(n) = 8*(2*n-1)*a(n-3) - 20*(n+1)*a(n-2) + 4*(2*n+5)*a(n-1). - Fung Lam, Jan 29 2014
D-finite with recurrence: (n+4)*a(n) - 2*(3*n+7)*a(n-1) + 4*(2*n+1)*a(n-2) = 0. - R. J. Mathar, Jun 03 2014
Asymptotics: a(n) ~ 4^(n+3)/sqrt(4*Pi*n^3). - Fung Lam, Mar 31 2014
a(n) = 32*4^n*Gamma(5/2+n)*(1+n)/(sqrt(Pi)*Gamma(5+n)). - Peter Luschny, Dec 14 2015
a(n) = C(n+1) - 2*C(n) where C is Catalan number A000108. Yuchun Ji, Oct 18 2017 [Note: Offset is off by 2]
E.g.f.: d/dx ( 2*exp(2*x)*BesselI(2,2*x)/x ). - Ilya Gutkovskiy, Nov 01 2017
From Bradley Klee, Mar 05 2018: (Start)
With F(x) = 16/(1+sqrt(1-4*x))^4 g.f. of A002057, xi(x) = F(x/4)*(x/4)^2, K(16*x) = 2F1(1/2,1/2;1;16*x) g.f. of A002894, q(x) g.f. of A005797, and q'(x) g.f. of A274344:
K(x) = (1+sqrt(xi(x)))*K(xi(x)).
2*K(1-x) = (1+sqrt(xi(x)))*K(1-xi(x)).
q(x) = sqrt(q(xi(16*x)/16)) = q'(xi(16*x)/16)/sqrt(xi(16*x)/16). (End)
From Amiram Eldar, Jan 02 2022: (Start)
Sum_{n>=0} 1/a(n) = 5/4 + Pi/(18*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 183*log(phi)/(25*sqrt(5)) - 77/100, where phi is the golden ratio (A001622). (End)
a(n) = Integral_{x=0..4} x^n*W(x) dx where W(x) = -x^(3/2)*(1 - x/2)*sqrt(4 - x)/Pi, defined on the open interval (0,4). - Karol A. Penson, Nov 13 2022

A130534 Triangle T(n,k), 0 <= k <= n, read by rows, giving coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in increasing powers of x. T(n,k) is also the unsigned Stirling number |s(n+1, k+1)|, denoting the number of permutations on n+1 elements that contain exactly k+1 cycles.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 11, 6, 1, 24, 50, 35, 10, 1, 120, 274, 225, 85, 15, 1, 720, 1764, 1624, 735, 175, 21, 1, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

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Author

Philippe Deléham, Aug 09 2007

Keywords

Comments

This triangle is an unsigned version of the triangle of Stirling numbers of the first kind, A008275, which is the main entry for these numbers. - N. J. A. Sloane, Jan 25 2011
Or, triangle T(n,k), 0 <= k <= n, read by rows given by [1,1,2,2,3,3,4,4,5,5,6,6,...] DELTA [1,0,1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938.
Reversal of A094638.
Equals A132393*A007318, as infinite lower triangular matrices. - Philippe Deléham, Nov 13 2007
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the exponential integrals E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + n*(n+1)/x^2 - n*(n+1)*(n+2)/x^3 + ...), see Abramowitz and Stegun. This formula follows from the general formula for the asymptotic expansion, see A163932. We rewrite E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) and observe that the T(n,m) are the polynomials coefficients in the denominators. Looking at the a(n,m) formula of A028421, A163932 and A163934, and shifting the offset given above to 1, we can write T(n-1,m-1) = a(n,m) = (-1)^(n+m)*Stirling1(n,m), see the Maple program.
The asymptotic expansion leads for values of n from one to eleven to known sequences, see the cross-references. With these sequences one can form the triangles A008279 (right-hand columns) and A094587 (left-hand columns).
See A163936 for information about the o.g.f.s. of the right-hand columns of this triangle.
(End)
The number of elements greater than i to the left of i in a permutation gives the i-th element of the inversion vector. (Skiena-Pemmaraju 2003, p. 69.) T(n,k) is the number of n-permutations that have exactly k 0's in their inversion vector. See evidence in Mathematica code below. - Geoffrey Critzer, May 07 2010
T(n,k) counts the rooted trees with k+1 trunks in forests of "naturally grown" rooted trees with n+2 nodes. This corresponds to sums of coefficients of iterated derivatives representing vectors, Lie derivatives, or infinitesimal generators for flow fields and formal group laws. Cf. links in A139605. - Tom Copeland, Mar 23 2014
A refinement is A036039. - Tom Copeland, Mar 30 2014
From Tom Copeland, Apr 05 2014: (Start)
With initial n=1 and row polynomials of T as p(n,x)=x(x+1)...(x+n-1), the powers of x correspond to the number of trunks of the rooted trees of the "naturally-grown" forest referred to above. With each trunk allowed m colors, p(n,m) gives the number of such non-plane colored trees for the forest with each tree having n+1 vertices.
p(2,m) = m + m^2 = A002378(m) = 2*A000217(m) = 2*(first subdiag of |A238363|).
p(3,m) = 2m + 3m^2 + m^3 = A007531(m+2) = 3*A007290(m+2) = 3*(second subdiag A238363).
p(4,m) = 6m + 11m^2 + 6m^3 + m^4 = A052762(m+3) = 4*A033487(m) = 4*(third subdiag).
From the Joni et al. link, p(n,m) also represents the disposition of n distinguishable flags on m distinguishable flagpoles.
The chromatic polynomial for the complete graph K_n is the falling factorial, which encodes the colorings of the n vertices of K_n and gives a shifted version of p(n,m).
E.g.f. for the row polynomials: (1-y)^(-x).
(End)
A relation to derivatives of the determinant |V(n)| of the n X n Vandermonde matrix V(n) in the indeterminates c(1) thru c(n):
|V(n)| = Product_{1<=jTom Copeland, Apr 10 2014
From Peter Bala, Jul 21 2014: (Start)
Let M denote the lower unit triangular array A094587 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well defined). See the Example section. (End)
For the relation of this rising factorial to the moments of Viennot's Laguerre stories, see the Hetyei link, p. 4. - Tom Copeland, Oct 01 2015
Can also be seen as the Bell transform of n! without column 0 (and shifted enumeration). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle  T(n,k) begins:
n\k         0        1        2       3       4      5      6     7    8  9 10
n=0:        1
n=1:        1        1
n=2:        2        3        1
n=3:        6       11        6       1
n=4:       24       50       35      10       1
n=5:      120      274      225      85      15      1
n=6:      720     1764     1624     735     175     21      1
n=7:     5040    13068    13132    6769    1960    322     28     1
n=8:    40320   109584   118124   67284   22449   4536    546    36    1
n=9:   362880  1026576  1172700  723680  269325  63273   9450   870   45  1
n=10: 3628800 10628640 12753576 8409500 3416930 902055 157773 18150 1320 55  1
[Reformatted and extended by _Wolfdieter Lang_, Feb 05 2013]
T(3,2) = 6 because there are 6 permutations of {1,2,3,4} that have exactly 2 0's in their inversion vector: {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {2, 1, 3, 4},{2, 3, 1, 4}, {2, 3, 4, 1}. The respective inversion vectors are {0, 0, 1}, {0, 1, 0}, {0, 2, 0}, {1, 0, 0}, {2, 0, 0}, {3, 0, 0}. - _Geoffrey Critzer_, May 07 2010
T(3,1)=11 since there are exactly 11 permutations of {1,2,3,4} with exactly 2 cycles, namely, (1)(234), (1)(243), (2)(134), (2)(143), (3)(124), (3)(142), (4)(123), (4)(143), (12)(34), (13)(24), and (14)(23). - _Dennis P. Walsh_, Jan 25 2011
From _Peter Bala_, Jul 21 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
  / 1          \/1        \/1        \      / 1           \
  | 1  1       ||0 1      ||0 1      |      | 1  1        |
  | 2  2  1    ||0 1 1    ||0 0 1    |... = | 2  3  1     |
  | 6  6  3 1  ||0 2 2 1  ||0 0 1 1  |      | 6 11  6  1  |
  |24 24 12 4 1||0 6 6 3 1||0 0 2 2 1|      |24 50 35 10 1|
  |...         ||...      ||...      |      |...          |
(End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 93-94.
  • Sriram Pemmaraju and Steven Skiena, Computational Discrete Mathematics, Cambridge University Press, 2003, pp. 69-71. [Geoffrey Critzer, May 07 2010]

Crossrefs

See A008275, which is the main entry for these numbers; A094638 (reversed rows).
From Johannes W. Meijer, Oct 07 2009: (Start)
Row sums equal A000142.
The asymptotic expansions lead to A000142 (n=1), A000142(n=2; minus a(0)), A001710 (n=3), A001715 (n=4), A001720 (n=5), A001725 (n=6), A001730 (n=7), A049388 (n=8), A049389 (n=9), A049398 (n=10), A051431 (n=11), A008279 and A094587.
Cf. A163931 (E(x,m,n)), A028421 (m=2), A163932 (m=3), A163934 (m=4), A163936.
(End)
Cf. A136662.

Programs

  • Haskell
    a130534 n k = a130534_tabl !! n !! k
    a130534_row n = a130534_tabl !! n
    a130534_tabl = map (map abs) a008275_tabl
    -- Reinhard Zumkeller, Mar 18 2013
  • Maple
    with(combinat): A130534 := proc(n,m): (-1)^(n+m)*stirling1(n+1,m+1) end proc: seq(seq(A130534(n,m), m=0..n), n=0..10); # Johannes W. Meijer, Oct 07 2009, revised Sep 11 2012
    # The function BellMatrix is defined in A264428.
    # Adds (1,0,0,0, ..) as column 0 (and shifts the enumeration).
    BellMatrix(n -> n!, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    Table[Table[ Length[Select[Map[ToInversionVector, Permutations[m]], Count[ #, 0] == n &]], {n, 0, m - 1}], {m, 0, 8}] // Grid (* Geoffrey Critzer, May 07 2010 *)
    rows = 10;
    t = Range[0, rows]!;
    T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)

Formula

T(0,0) = 1, T(n,k) = 0 if k > n or if n < 0, T(n,k) = T(n-1,k-1) + n*T(n-1,k). T(n,0) = n! = A000142(n). T(2*n,n) = A129505(n+1). Sum_{k=0..n} T(n,k) = (n+1)! = A000142(n+1). Sum_{k=0..n} T(n,k)^2 = A047796(n+1). T(n,k) = |Stirling1(n+1,k+1)|, see A008275. (x+1)(x+2)...(x+n) = Sum_{k=0..n} T(n,k)*x^k. [Corrected by Arie Bos, Jul 11 2008]
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. - Philippe Deléham, Nov 13 2007
For k=1..n, let A={a_1,a_2,...,a_k} denote a size-k subset of {1,2,...,n}. Then T(n,n-k) = Sum(Product_{i=1..k} a_i) where the sum is over all subsets A. For example, T(4,1)=50 since 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 50. - Dennis P. Walsh, Jan 25 2011
The preceding formula means T(n,k) = sigma_{n-k}(1,2,3,..,n) with the (n-k)-th elementary symmetric function sigma with the indeterminates chosen as 1,2,...,n. See the Oct 24 2011 comment in A094638 with sigma called there a. - Wolfdieter Lang, Feb 06 2013
From Gary W. Adamson, Jul 08 2011: (Start)
n-th row of the triangle = top row of M^n, where M is the production matrix:
1, 1;
1, 2, 1;
1, 3, 3, 1;
1, 4, 6, 4, 1;
... (End)
Exponential Riordan array [1/(1 - x), log(1/(1 - x))]. Recurrence: T(n+1,k+1) = Sum_{i=0..n-k} (n + 1)!/(n + 1 - i)!*T(n-i,k). - Peter Bala, Jul 21 2014

A001705 Generalized Stirling numbers: a(n) = n! * Sum_{k=0..n-1} (k+1)/(n-k).

Original entry on oeis.org

0, 1, 5, 26, 154, 1044, 8028, 69264, 663696, 6999840, 80627040, 1007441280, 13575738240, 196287356160, 3031488633600, 49811492505600, 867718162483200, 15974614352793600, 309920046408806400, 6320046028584960000, 135153868608460800000, 3024476051557847040000
Offset: 0

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Author

Keywords

Comments

a(n) is also the sum of the positions of the right-to-left minima in all permutations of [n]. Example: a(3)=26 because the positions of the right-to-left minima in the permutations 123,132,213,231,312 and 321 are 123, 13, 23, 3, 23 and 3, respectively and 1 + 2 + 3 + 1 + 3 + 2 + 3 + 3 + 2 + 3 + 3 = 26. - Emeric Deutsch, Sep 22 2008
The asymptotic expansion of the higher order exponential integral E(x,m=2,n=2) ~ exp(-x)/x^2*(1 - 5/x + 26/x^2 - 154/x^3 + 1044/x^4 - 8028/x^5 + 69264/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009
a(n) is the total number of cycles (excluding fixed points) in all permutations of [n+1]. - Olivier Gérard, Oct 23 2012; Dec 31 2012
A length n sequence is formed by randomly selecting (one-by-one) n real numbers in (0,1). a(n)/(n+1)! is the expected value of the sum of the new maximums in such a sequence. For example for n=3: If we select (in this order): 0.591996, 0.646474, 0.163659 we would add 0.591996 + 0.646474 which would be a bit above the average of a(3)/4! = 26/24. - Geoffrey Critzer, Oct 17 2013

Examples

			(1-x)^-2 * (-log(1-x)) = x + 5/2*x^2 + 13/3*x^3 + 77/12*x^4 + ...
Examples: a(6) = 6!*(1/6 + 2/5 + 3/4 + 4/3 + 5/2 + 6/1) = 8028; a(20) = 20!*(1/20 + 2/19 + 3/18 + 4/17 + 5/16 + ... + 16/5 + 17/4 + 18/3 + 19/2 + 20/1) = 135153868608460800000. - _Alexander Adamchuk_, Oct 09 2004
From _Olivier Gérard_, Dec 31 2012: (Start)
The cycle decomposition of all permutations of 4 elements gives the following list: {{{1},{2},{3},{4}}, {{1},{2},{3,4}}, {{1},{2,3},{4}}, {{1},{2,4,3}}, {{1},{2,3,4}}, {{1},{2,4},{3}}, {{1,2},{3},{4}}, {{1,2},{3,4}}, {{1,3,2},{4}},{{1,4,3,2}}, {{1,3,4,2}}, {{1,4,2},{3}}, {{1,2,3},{4}}, {{1,2,4,3}},{{1,3},{2},{4}}, {{1,4,3},{2}}, {{1,3},{2,4}}, {{1,4,2,3}}, {{1,2,3,4}}, {{1,2,4},{3}}, {{1,3,4},{2}}, {{1,4},{2},{3}}, {{1,3,2,4}}, {{1,4},{2,3}}}.
Deleting the fixed points gives the following 26 items: {{3,4}, {2,3}, {2,4,3}, {2,3,4}, {2,4}, {1,2}, {1,2}, {3,4}, {1,3,2}, {1,4,3,2}, {1,3,4,2}, {1,4,2}, {1,2,3}, {1,2,4,3}, {1,3}, {1,4,3}, {1,3}, {2,4}, {1,4,2,3}, {1,2,3,4}, {1,2,4}, {1,3,4}, {1,4}, {1,3,2,4}, {1,4}, {2,3}}. (End)
		

References

  • 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

Cf. A000254 (total number of cycles in permutations, including fixed points).
Cf. A002104 (number of different cycles in permutations, without fixed points).
Cf. A006231 (number of different cycles in permutations, including fixed points).
Related to n!*the k-th successive summation of the harmonic numbers:
(k=0) A000254, (k=1) A001705, (k=2) A001711, (k=3) A001716,
(k=4) A001721, (k=5) A051524, (k=6) A051545, (k=7) A051560,
(k=8) A051562, (k=9) A051564.

Programs

  • Maple
    a := n-> add((n+1)!/k, k=2..n+1): seq(a(n), n=0..21); # Zerinvary Lajos, Jan 22 2008; edited Johannes W. Meijer, Nov 28 2012
    a := n -> ((n+1)!*(h(n+1)-1)): h := n-> harmonic(n): seq(a(n), n=0..21); # Gary Detlefs, Dec 18 2009; corrected by Johannes W. Meijer, Nov 28 2012
  • Mathematica
    Table[n!*Sum[Sum[1/k,{k,1,m}], {m,1,n}], {n,0,20}] (* Alexander Adamchuk, Apr 14 2006 *)
    a[n_] := (n + 1)! (EulerGamma - 1 + PolyGamma[n + 2]);
    Table[a[n], {n, 0, 21}] (* Peter Luschny, Feb 19 2022 *)
  • Maxima
    a(n):=n!*sum(((-1)^(k+1)*binomial(n+1,k+1))/k,k,1,n); /* Vladimir Kruchinin, Oct 10 2016 */
    
  • PARI
    for(n=0,25, print1(n!*sum(k=0,n-1,(k+1)/(n-k)), ", ")) \\ G. C. Greubel, Jan 20 2017
    
  • Python
    from math import factorial
    def A001705(n):
        f = factorial(n)
        return sum(f*(k+1)//(n-k) for k in range(n)) # Chai Wah Wu, Jun 23 2022

Formula

Partial sum of first n harmonic numbers multiplied by n!.
a(n) = n!*Sum_{m=1..n} Sum_{k=1..m} 1/k = n!*Sum_{m=1..n} H(m), where H(m) = Sum_{k=1..m} 1/k = A001008(m)/A002805(m) is m-th Harmonic number.
E.g.f.: - log (1 - x) / (1 - x)^2.
a(n) = (n+1)! * H(n) - n*n!, H(n) = Sum_{k=1..n} (1/k).
a(n) = A112486(n, 1).
a(n) = a(n-1)*(n+1) + n! = A000254(n+1) - A000142(n+1) = A067176(n+1, 1). - Henry Bottomley, Jan 09 2002
a(n) = Sum_{k=0..n-1} ((-1)^(n-1+k) * (k+1) * 2^k * Stirling1(n, k+1)). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
With alternating signs: Ramanujan polynomials psi_2(n, x) evaluated at 0. - Ralf Stephan, Apr 16 2004
a(n) = Sum_{k=1..n} (k*StirlingCycle(n+1,k+1)). - David Callan, Sep 25 2006
a(n) = Sum_{k=n..n*(n+1)/2} k*A143947(n,k). - Emeric Deutsch, Sep 22 2008
For n >= 1, a(n) = Sum_{j=0..n-1} ((-1)^(n-j-1) * 2^j * (j+1) * Stirling1(n,j+1)). - Milan Janjic, Dec 14 2008
a(n) = (2*n+1)*a(n-1) - n^2*a(n-2). - Gary Detlefs, Nov 27 2009
a(n) = (n+1)!*(H(n+1) - 1) where H(n) is the n-th harmonic number. - Gary Detlefs, Dec 18 2009
a(n) = n!*Sum_{k=1..n} (-1)^(k+1)*binomial(n+1,k+1)/k. - Vladimir Kruchinin, Oct 10 2016
a(n) = (n+1)!*Sum_{k = 1..n} (-1)^(k+1)*binomial(n+1,k+1)*k/(k+1). - Peter Bala, Feb 15 2022
a(n) = Gamma(n + 2) * (Digamma(n + 2) + EulerGamma - 1). - Peter Luschny, Feb 19 2022
From Mélika Tebni, Jun 22 2022: (Start)
a(n) = -Sum_{k=0..n} k!*A066667(n, k+1).
a(n) = Sum_{k=0..n} k!*A132159(n, k+1). (End)
a(n) = n*(n + 1)!*hypergeom([1, 1, 1 - n], [2, 3], 1)/2. - Peter Luschny, Jun 22 2022

Extensions

More terms from Sascha Kurz, Mar 22 2002

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

A000399 Unsigned Stirling numbers of first kind s(n,3).

Original entry on oeis.org

1, 6, 35, 225, 1624, 13132, 118124, 1172700, 12753576, 150917976, 1931559552, 26596717056, 392156797824, 6165817614720, 102992244837120, 1821602444624640, 34012249593822720, 668609730341153280, 13803759753640704000
Offset: 3

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Author

Keywords

Comments

Number of permutations of n elements with exactly 3 cycles.
The asymptotic expansion of the higher order exponential integral E(x,m=3,n=1) ~ exp(-x)/x^3*(1 - 6/x + 35/x^2 - 225/x^3 + 1624/x^4 - 13132/x^5 + ...) leads to the sequence given above. See A163931 and A163932 for more information. - Johannes W. Meijer, Oct 20 2009

Examples

			(-log(1-x))^3 = x^3 + 3/2*x^4 + 7/4*x^5 + 15/8*x^6 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 217.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
  • Shanzhen Gao, Permutations with Restricted Structure (in preparation). - Shanzhen Gao, Sep 14 2010
  • 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

Programs

  • Magma
    A000399:=func< n | Abs(StirlingFirst(n, 3)) >; [ A000399(n): n in [3..25] ]; // Klaus Brockhaus, Jan 14 2011
  • Maple
    seq(abs(Stirling1(n,3)),n=3..30); # Robert Israel, Jul 05 2015
  • Mathematica
    a=Log[1/(1-x)];Range[0,20]! CoefficientList[Series[a^3/3!,{x,0,20}],x]
    f[n_] := Abs@ StirlingS1[n, 3]; Array[f, 19, 3]
    Abs[StirlingS1[Range[3,30],3]] (* Harvey P. Dale, Jun 23 2014 *)
    f[n_] := Gamma[n]*(HarmonicNumber[n - 1]^2 + Zeta[2, n] - Zeta[2])/2; Array[f, 19, 3] (* Robert G. Wilson v, Jul 05 2015 *)
  • MuPAD
    f := proc(n) option remember; begin n^3*f(n-3)-(3*n^2+3*n+1)*f(n-2)+3*(n+1)*f(n-1) end_proc: f(0) := 1: f(1) := 6: f(2) := 35:
    
  • PARI
    for(n=2,50,print1(polcoeff(prod(i=1,n,x+i),2,x),","))
    
  • Sage
    [stirling_number1(i+2,3) for i in range(1,22)] # Zerinvary Lajos, Jun 27 2008
    

Formula

Let P(n-1,X) = (X+1)(X+2)(X+3)...(X+n-1); then a(n) is the coefficient of X^2; or a(n) = P''(n-1,0)/2!. - Benoit Cloitre, May 09 2002 [Edited by Petros Hadjicostas, Jun 29 2020 to agree with the offset 3]
E.g.f.: -log(1-x)^3/3!.
a(n) is the coefficient of x^(n+3) in (-log(1-x))^3, multiplied by (n+3)!/6.
a(n) = ((Sum_{i=1..n-1} 1/i)^2 - Sum_{i=1..n-1} 1/i^2)*(n-1)!/2 for n >= 3. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 18 2000
a(n) = det(|S(i+3,j+2)|, 1 <= i,j <= n-3), where S(n,k) are Stirling numbers of the second kind. - Mircea Merca, Apr 06 2013
a(n) = Gamma(n)*(HarmonicNumber(n-1)^2 + Zeta(2,n) - Zeta(2))/2. - Gerry Martens, Jul 05 2015
From Petros Hadjicostas, Jun 28 2020: (Start)
a(n) = (n-3)! + (2*n-3)*a(n-1) - (n-2)^2*a(n-2) for n >= 5.
a(n) = 3*(n-2)*a(n-1) - (3*n^2-15*n+19)*a(n-2) + (n-3)^3*a(n-3) for n >= 6. (End)

A028421 Triangle read by rows: T(n, k) = (k+1)*A132393(n+1, k+1), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 2, 2, 6, 3, 6, 22, 18, 4, 24, 100, 105, 40, 5, 120, 548, 675, 340, 75, 6, 720, 3528, 4872, 2940, 875, 126, 7, 5040, 26136, 39396, 27076, 9800, 1932, 196, 8, 40320, 219168, 354372, 269136, 112245, 27216, 3822, 288, 9
Offset: 0

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Author

Peter Wiggen (wiggen(AT)math.psu.edu)

Keywords

Comments

Previous name was: Number triangle f(n, k) from n-th differences of the sequence {1/m^2}{m >= 1}, for n >= 0; the n-th difference sequence is {(-1)^n*n!*P(n, m)/D(n, m)^2}{m >= 1} where P(n, x) is the row polynomial P(n, x) = Sum_{k=0..n} f(n,k)*x^k and D(n, x) = x*(x+1)*...*(x+n).
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher-order exponential integrals E(x,m,n) are defined in A163931 and the general formula of the asymptotic expansion of E(x,m,n) can be found in A163932.
We used the general formula and the asymptotic expansion of E(x,m=1,n), see A130534, to determine that E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2 + 6*n + 3*n^2)/x^2 - (6 + 22*n + 18*n^2 + 4*n^3)/x^3 + ...) which can be verified with the EA(x,2,n) formula, see A163932. The coefficients in the denominators of this expansion lead to the sequence given above.
The asymptotic expansion of E(x,m=2,n) leads for n from one to ten to known sequences, see the cross-references. With these sequences one can form the triangles A165674 (left hand columns) and A093905 (right hand columns).
(End)
For connections to an operator relation between log(x) and x^n(d/dx)^n, see A238363. - Tom Copeland, Feb 28 2014
From Wolfdieter Lang, Nov 25 2018: (Start)
The signed triangle t(n, k) := (-1)^{n-k}*f(n, k) gives (n+1)*N(-1;n,x) = Sum_{k=0..n} t(n, k)*x^k, where N(-1;n,x) are the Narumi polynomials with parameter a = -1 (see the Weisstein link).
The members of the n-th difference sequence of the sequence {1/m^2}_{m>=1} mentioned above satisfies the recurrence delta(n, m) = delta(n-1, m+1) - delta(n-1, m), for n >= 1, m >= 1, with input delta(0, m) = 1/m^2. The solution is delta(n, m) = (n+1)!*N(-1;n,-m)/risefac(m, n+1)^2, with Narumi polynomials N(-1;n,x) and the rising factorials risefac(x, n+1) = D(n, x) = x*(x+1)*...*(x+n).
The above mentioned row polynomials P satisfy P(n, x) = (-1)^n*(n + 1)*N(-1;n,-x), for n >= 0. The recurrence is P(n, x) = (-x^2*P(n-1, x+1) + (n+x)^2*P(n-1, x))/n, for n >= 1, and P(0, x) = 1. (End)
The triangle is the exponential Riordan square (cf. A321620) of -log(1-x) with an additional main diagonal of zeros. - Peter Luschny, Jan 03 2019

Examples

			The triangle T(n, k) begins:
n\k       0        1        2        3        4       5       6      7     8   9 10
------------------------------------------------------------------------------------
0:        1
1:        1        2
2:        2        6        3
3:        6       22       18        4
4:       24      100      105       40        5
5:      120      548      675      340       75       6
6:      720     3528     4872     2940      875     126       7
7:     5040    26136    39396    27076     9800    1932     196      8
8:    40320   219168   354372   269136   112245   27216    3822    288     9
9:   362880  2053152  3518100  2894720  1346625  379638   66150   6960   405  10
10: 3628800 21257280 38260728 33638000 17084650 5412330 1104411 145200 11880 550 11
... - _Wolfdieter Lang_, Nov 23 2018
		

Crossrefs

Row sums give A000254(n+1), n >= 0.
Cf. A132393 (unsigned Stirling1), A061356, A139526, A321620.
From Johannes W. Meijer, Oct 07 2009: (Start)
A000142, A052517, 3*A000399, 5*A000482 are the first four left hand columns; A000027, A002411 are the first two right hand columns.
The asymptotic expansion of E(x,m=2,n) leads to A000254 (n=1), A001705 (n=2), A001711 (n=3), A001716 (n=4), A001721 (n=5), A051524 (n=6), A051545 (n=7), A051560 (n=8), A051562 (n=9), A051564 (n=10), A093905 (triangle) and A165674 (triangle).
Cf. A163931 (E(x,m,n)), A130534 (m=1), A163932 (m=3), A163934 (m=4), A074246 (E(x,m=2,n+1)). (End)

Programs

  • Maple
    A028421 := proc(n,k) (-1)^(n+k)*(k+1)*Stirling1(n+1,k+1) end:
    seq(seq(A028421(n,k), k=0..n), n=0..8);
    # Johannes W. Meijer, Oct 07 2009, Revised Sep 09 2012
    egf := (1 - t)^(-x - 1)*(1 - x*log(1 - t)):
    ser := series(egf, t, 16): coefft := n -> expand(coeff(ser,t,n)):
    seq(seq(n!*coeff(coefft(n), x, k), k = 0..n), n = 0..8); # Peter Luschny, Jun 12 2022
  • Mathematica
    f[n_, k_] = (k + 1) StirlingS1[n + 1, k + 1] // Abs; Flatten[Table[f[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, Jun 01 2011, after formula *)
  • Sage
    # uses[riordan_square from A321620]
    riordan_square(-ln(1 - x), 10, True) # Peter Luschny, Jan 03 2019

Formula

E.g.f.: d/dt(-log(1-t)/(1-t)^x). - Vladeta Jovovic, Oct 12 2003
The e.g.f. with offset 1: y = x + (1 + 2*t)*x^2/2! + (2 + 6*t + 3*t^2)*x^3/3! + ... has series reversion with respect to x equal to y - (1 + 2*t)*y^2/2! + (1 + 3*t)^2*y^3/3! - (1 + 4*t)^3*y^4/4! + .... This is an e.g.f. for a signed version of A139526. - Peter Bala, Jul 18 2013
Recurrence: T(n, k) = 0 if n < k; if k = 0 then T(0, 0) = 1 and T(n, 0) = n * T(n-1, 0) for n >= 1, otherwise T(n, k) = n*T(n-1, k) + ((k+1)/k)*T(n-1, k-1). From the unsigned Stirling1 recurrence. - Wolfdieter Lang, Nov 25 2018

Extensions

Edited by Wolfdieter Lang, Nov 23 2018

A001711 Generalized Stirling numbers.

Original entry on oeis.org

1, 7, 47, 342, 2754, 24552, 241128, 2592720, 30334320, 383970240, 5231113920, 76349105280, 1188825724800, 19675048780800, 344937224217600, 6386713749964800, 124548748102195200, 2551797512248320000, 54804198761303040000, 1231237843834521600000
Offset: 0

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Author

Keywords

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=3) ~ exp(-x)/x^2*(1 - 7/x + 47/x^2 - 342/x^3 + 2754/x^4 - 24552/x^5 + 241128/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009
For n > 4, a(n) mod n = 0 for n composite, = n-3 for n prime. - Gary Detlefs, Jul 18 2011
From Petros Hadjicostas, Jun 11 2020: (Start)
For nonnegative integers n, m and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) using slightly different notation. They were further examined by Mitrinovic and Mitrinovic (1962).
These numbers are defined via the g.f. Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0.
As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m. (Because an empty product is by definition 1, we may let R_0^0(a,b) = 1.)
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m). (Array A008275 is the same as array A048994 but with no zero row and no zero column.)
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current sequence, a(n) = R_{n+1}^1(a=-3, b=-1) for n >= 0. (End)

References

  • 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

Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k=2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564.

Programs

  • Maple
    a := n-> add(1/2*((n+3)!/(k+3)), k=0..n): seq(a(n), n=0..19); # Zerinvary Lajos, Jan 22 2008
    a := n -> (n+1)!*hs2(n+1): hs2 := n-> add(hs(k), k=0..n): hs := n-> add(h(k), k=0..n): h := n-> add(1/k, k=1..n): seq(a(n), n=0..19); # Gary Detlefs, Jan 01 2011
  • Mathematica
    f[k_] := k + 2; t[n_] := Table[f[k], {k, 1, n}]; a[n_] := SymmetricPolynomial[n - 1, t[n]]; Table[a[n], {n, 1, 16}]; (* Clark Kimberling, Dec 29 2011 *)
    Table[(n + 3)!*Sum[1/(2*k + 4), {k, 1, n + 1}], {n,0,100}] (* G. C. Greubel, Jan 15 2017 *)
  • PARI
    for(n=0, 19, print1((n+1)! * sum(k=0, n, binomial(k + 2, 2) / (n + 1 - k)),", ")) \\ Indranil Ghosh, Mar 13 2017
    
  • PARI
    R(n,m,a,b) =  sum(k=0, n-m, (-1)^k*a^k*b^(n-m-k)*binomial(m+k,k)*stirling(n, m+k,1));
    aa(n) = R(n+1,1,-3,-1);
    for(n=0, 19, print1(aa(n), ",")) \\ Petros Hadjicostas, Jun 11 2020

Formula

E.g.f.: -log(1 - x)/(1 - x)^3 if offset 1. With offset 0: (d/dx)(-log(1 - x)/(1 - x)^3) = (1 - 3*log(1 - x))/(1 - x)^4.
a(n) = Sum_{k=0..n} ((-1)^(n+k)*(k+1)*3^k*Stirling1(n+1, k+1)). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
a(n) = n!*Sum_{k=0..n-1} ((-1)^k*binomial(-3,k)/(n-k)). - Milan Janjic, Dec 14 2008
a(n) = ( A000254(n+3) - 3*A001710(n+3) )/2. - Gary Detlefs, May 24 2010
a(n) = ((n+3)!/4) * (2*h(n+3) - 3), where h(n) = Sum_{k=1..n} (1/k) is the n-th harmonic number. - Gary Detlefs, Aug 15 2010
a(n) = n!*[2]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n. With offset 1. - Gary Detlefs, Jan 04 2011
a(n) = (n+3)! * Sum_{k=1..n+1} (1/(2*k+4)). - Gary Detlefs, Sep 14 2011
a(n) = (n+1)! * Sum_{k=0..n} (binomial(k+2,2)/(n+1-k)). - Gary Detlefs, Dec 01 2011
a(n) = A001705(n+2) - A182541(n+4). - Anton Zakharov, Jul 02 2016
a(n) ~ n^(n+7/2) * exp(-n) * sqrt(Pi/2) * log(n) * (1 + (gamma - 3/2)/log(n)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Jul 12 2016
Conjectural D-finite with recurrence: a(n) + (-2*n-5)*a(n-1) + (n+2)^2*a(n-2)=0. - R. J. Mathar, Feb 16 2020
From Petros Hadjicostas, Jun 11 2020: (Start)
Since a(n) = R_{n+1}^1(a=-3, b=-1), it follows from Mitrinovic (1961) and Mitrinovic and Mitrinovic (1962) that:
a(n) = [x] Product_{r=0}^n (x + 3 + r) = (Product_{r=0}^n (3 + r)) * Sum_{s=0}^n 1/(3 + s).
a(n) = (n + 2)!/2 + (n + 3)*a(n-1) for n >= 1. [This can be used to prove R. J. Mathar's recurrence above.] (End)

Extensions

More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
Maple programs corrected and edited by Johannes W. Meijer, Nov 28 2012

A163932 Triangle related to the asymptotic expansion of E(x,m=3,n).

Original entry on oeis.org

1, 3, 3, 11, 18, 6, 50, 105, 60, 10, 274, 675, 510, 150, 15, 1764, 4872, 4410, 1750, 315, 21, 13068, 39396, 40614, 19600, 4830, 588, 28, 109584, 354372, 403704, 224490, 68040, 11466, 1008, 36, 1026576, 3518100, 4342080, 2693250, 949095, 198450
Offset: 1

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Author

Johannes W. Meijer & Nico Baken (n.h.g.baken(AT)tudelft.nl), Aug 13 2009, Oct 22 2009

Keywords

Comments

The higher order exponential integrals E(x,m,n) are defined in A163931. The general formula for the asymptotic expansion E(x,m,n) ~ E(x,m-1,n+1)/x - n*E(x,m-1,n+2)/x^2 + n*(n+1) * E(x,m-1,n+3)/x^3 - n*(n+1)*(n+2)*E(x,m-1,n+4)/x^4 + ...., m >= 1 and n >= 1.
We used this formula and the asymptotic expansion of E(x,m=2,n), see A028421, to determine that E (x,m=3,n) ~ (exp(-x)/x^3)*(1 - (3+3*n)/x + (11+18*n+6*n^2)/x^2 - (50+105*n+ 60*n^2+ 10*n^3)/x^3 + .. ). This formula leads to the triangle coefficients given above.
The asymptotic expansion leads for the values of n from one to ten to known sequences, see the cross-references.
The numerators of the o.g.f.s. of the right hand columns of this triangle lead for z=1 to A001879, see A163938 for more information.
The first Maple program generates the sequence given above and the second program generates the asymptotic expansion of E(x,m=3,n).

Examples

			The first few rows of the triangle are:
[1]
[3, 3]
[11, 18, 6]
[50, 105, 60, 10]
		

Crossrefs

Cf. A163931 (E(x,m,n)) and A163938.
Cf. A048994 (Stirling1), A000399 (row sums).
A000254, 3*A000399, 6*A000454, 10*A000482, 15*A001233, 21*A001234 equal the first six left hand columns.
A000217, A006011 and A163933 equal the first three right hand columns.
The asymptotic expansion leads to A000399 (n=1), A001706 (n=2), A001712 (n=3), A001717 (n=4), A001722 (n=5), A051525 (n=6), A051546 (n=7), A051561 (n=8), A051563 (n=9) and A051565 (n=10).
Cf. A130534 (m=1), A028421 (m=2) and A163934 (m=4).

Programs

  • Maple
    nmax:=8; with(combinat): for n1 from 1 to nmax do for m from 1 to n1 do a(n1, m) := (-1)^(n1+m)*binomial(m+1, 2)*stirling1(n1+1, m+1) od: od: seq(seq(a(n1,m), m=1..n1), n1=1..nmax);
    # End program 1
    with(combinat): imax:=6; EA:=proc(x, m, n) local E, i; E := 0: for i from m-1 to imax+1 do E := E + sum((-1)^(m+k1+1)*binomial(k1, m-1)*n^(k1-m+1)* stirling1(i, k1), k1=m-1..i)/x^(i-m+1) od: E := exp(-x)/x^(m)*E: return(E); end: EA(x, 3, n);
    # End program 2
  • Mathematica
    a[n_, m_] /; n >= 1 && 1 <= m <= n = (-1)^(n+m)*Binomial[m+1, 2] * StirlingS1[n+1, m+1]; Flatten[Table[a[n, m], {n, 1, 9}, {m, 1, n}]][[1 ;; 42]] (* Jean-François Alcover, Jun 01 2011, after formula *)
  • PARI
    for(n=1,10, for(m=1,n, print1((-1)^(n+m)*binomial(m+1,2) *stirling(n+1,m+1,1), ", "))) \\ G. C. Greubel, Aug 08 2017

Formula

a(n,m) = (-1)^(n+m)*binomial(m+1,2)*stirling1(n+1,m+1) for n >= 1 and 1 <= m <= n.

Extensions

Edited by Johannes W. Meijer, Sep 22 2012

A001716 Generalized Stirling numbers.

Original entry on oeis.org

1, 9, 74, 638, 5944, 60216, 662640, 7893840, 101378880, 1397759040, 20606463360, 323626665600, 5395972377600, 95218662067200, 1773217155225600, 34758188233574400, 715437948072960000, 15429680577561600000, 347968129734973440000, 8190600438533990400000
Offset: 0

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Author

Keywords

Comments

The asymptotic expansion of the higher order exponential integral E(x,m=2,n=4) ~ exp(-x)/x^2*(1 - 9/x + 74/x^2 - 638/x^3 + 5944/x^4 - 60216/x^5 + 662640/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009
From Petros Hadjicostas, Jun 23 2020: (Start)
For nonnegative integers n, m and complex numbers a, b (with b <> 0), the numbers R_n^m(a,b) were introduced by Mitrinovic (1961) and Mitrinovic and Mitrinovic (1962) using slightly different notation.
These numbers are defined via the g.f. Product_{r=0..n-1} (x - (a + b*r)) = Sum_{m=0..n} R_n^m(a,b)*x^m for n >= 0.
As a result, R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b) for n >= m >= 1 with R_0^0(a,b) = 1, R_1^0(a,b) = a, R_1^1(a,b) = 1, and R_n^m(a,b) = 0 for n < m.
With a = 0 and b = 1, we get the Stirling numbers of the first kind S1(n,m) = R_n^m(a=0, b=1) = A048994(n,m) for n, m >= 0.
We have R_n^m(a,b) = Sum_{k=0}^{n-m} (-1)^k * a^k * b^(n-m-k) * binomial(m+k, k) * S1(n, m+k) for n >= m >= 0.
For the current sequence, a(n) = R_{n+1}^1(a=-4, b=-1) for n >= 0. (End)

References

  • 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

Related to n!*the k-th successive summation of the harmonic numbers: k=0..A000254, k=1..A001705, k= 2..A001711, k=3..A001716, k=4..A001721, k=5..A051524, k=6..A051545, k=7..A051560, k=8..A051562, k=9..A051564.

Programs

  • Mathematica
    f[k_] := k + 3; t[n_] := Table[f[k], {k, 1, n}]; a[n_] := SymmetricPolynomial[n - 1, t[n]]; Table[a[n], {n, 1, 16}] (* Clark Kimberling, Dec 29 2011 *)
    Rest[CoefficientList[Series[(1-x)^(-4)*Log[1/(1-x)],{x,0,20}],x]*Range[0,20]!] (* Vaclav Kotesovec, Jan 19 2014 *)
  • PARI
    R(n, m, a, b) =  sum(k=0, n-m, (-1)^k*a^k*b^(n-m-k)*binomial(m+k, k)*stirling(n, m+k, 1));
    aa(n) = R(n+1, 1, -4, -1);
    for(n=0, 19, print1(aa(n), ", ")) \\ Petros Hadjicostas, Jun 23 2020

Formula

a(n) = Sum_{k=0..n} (-1)^(n+k) * (k+1) * 4^k * stirling1(n+1, k+1). - Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004
a(n-1) = n!*Sum_{k=0..n-1} (-1)^k*binomial(-4,k)/(n-k) for n >= 1. [Milan Janjic, Dec 14 2008] [Edited by Petros Hadjicostas, Jun 23 2020]
a(n)= n! * [3]h(n), where [k]h(n) denotes the k-th successive summation of the harmonic numbers from 0 to n (with offset 1). [Gary Detlefs, Jan 04 2011]
a(n) = (n+1)! * Sum_{k=0..n} (-1)^k*binomial(-4,k)/(n+1-k). [Gary Detlefs, Jul 16 2011]
a(n) = (n+4)! * Sum_{k=1..n+1} 1/(k+3)/6. [Gary Detlefs, Sep 14 2011]
E.g.f. (for offset 1): 1/(1-x)^4 * log(1/(1-x)). - Vaclav Kotesovec, Jan 19 2014
E.g.f.: (1 + 4*log(1/(1 - x)))/(1 - x)^5. - Ilya Gutkovskiy, Jan 23 2017
From Petros Hadjicostas, Jun 23 2020: (Start)
a(n) = [x] Product_{r=0..n} (x + 4 + r) = (Product_{r=0..n} (4 + r)) * Sum_{i=0..n} 1/(4 + i).
Since a(n) = R_{n+1}^1(a=-4, b=-1) and R_n^m(a,b) = R_{n-1}^{m-1}(a,b) - (a + b*(n-1))*R_{n-1}^m(a,b), we conclude that:
(i) a(n) = (n+3)!/6 + (n+4)*a(n-1) for n >= 1;
(ii) a(n) = (2*n+7)*a(n-1) - (n+3)^2*a(n-2) for n >= 2. (End)

Extensions

More terms from Borislav Crstici (bcrstici(AT)etv.utt.ro), Jan 26 2004

A001819 Central factorial numbers: second right-hand column of triangle A008955.

Original entry on oeis.org

0, 1, 5, 49, 820, 21076, 773136, 38402064, 2483133696, 202759531776, 20407635072000, 2482492033152000, 359072203696128000, 60912644957448192000, 11977654199703478272000, 2702572249389834608640000
Offset: 0

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Author

Keywords

Comments

Coefficient of x^2 in Product_{k=0..n}(x + k^2). - Ralf Stephan, Aug 22 2004
p divides a(p-1) for prime p > 3. p divides a((p-1)/2) for prime p > 3. For prime p, p^2 divides a(n) for n > 2*p+1. - Alexander Adamchuk, Jul 11 2006; last comment corrected by Michel Marcus, May 20 2020
The ratio a(n)/A001044(n) is the partial sum of the reciprocals of squares. E.g., a(4)/A001044(4) = 820/576 = 1/1 + 1/4 + 1/9 + 1/16. - Pierre CAMI, Oct 30 2006
a(n) is the (n-1)-st elementary symmetric function of the squares of the first n numbers. - Anton Zakharov, Nov 06 2016
Primes p such that p^2 | a(p-1) are the Wolstenholme primes A088164. - Amiram Eldar and Thomas Ordowski, Aug 08 2019

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
  • 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

Second right-hand column of triangle A008955.
Equals row sums of A162990(n)/(n+1)^2 for n >= 1.

Programs

Formula

a_n = (n!)^2 * Sum_{k=1..n} 1/k^2. - Joe Keane (jgk(AT)jgk.org)
a(n) ~ (1/3)*Pi^3*n*e^(-2*n)*n^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
Sum_{n>=0} a(n)*x^n/n!^2 = polylog(2, x)/(1-x). - Vladeta Jovovic, Jan 23 2003
a(n) = Sum_{i=1..n} 1/i^2 / Product_{i=1..n} 1/i^2. - Alexander Adamchuk, Jul 11 2006
a(0) = 0, a(n) = a(n-1)*n^2 + A001044(n-1). E.g., a(1) = 0*1 + 1 = 1 since A001044(0) = 1; a(2) = 1*2^2 + 1 = 5 since A001044(1) = 1; a(3) = 5*3^2 + 4 = 49 since A001044(2) = 4; and so on. - Pierre CAMI, Oct 30 2006
Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 1)*a(n) - n^4*a(n-1). The sequence b(n) = n!^2 satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(1 - 1^4/(5 - 2^4/(13 - 3^4/(25 - ... -(n-1)^4/((2*n^2 - 2*n + 1)))))), leading to the infinite continued fraction expansion zeta(2) = 1/(1-1^4/(5 - 2^4/(13 - 3^4/(25 - ... - n^4/((2*n^2 + 2*n + 1) - ...))))). Compare with A142995. Compare also with A024167 and A066989. - Peter Bala, Jul 18 2008
a(n)/(n!)^2 -> zeta(2) = A013661 as n -> infinity, rewriting the Keane formula. - Najam Haq (njmalhq(AT)yahoo.com), Jan 13 2010
a(n) = s(n+1,2)^2 - 2*s(n+1,1)*s(n+1,3), where s(n,k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 03 2012

Extensions

Minor edits by Vaclav Kotesovec, Jan 28 2015
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