A001147 -id:A001147 - cp's OEIS frontend

A079484 a(n) = (2n-1)!! * (2n+1)!!, where the double factorial is A006882.

1, 3, 45, 1575, 99225, 9823275, 1404728325, 273922023375, 69850115960625, 22561587455281875, 9002073394657468125, 4348001449619557104375, 2500100833531245335015625, 1687568062633590601135546875, 1321365793042101440689133203125

Offset: 0

Benoit Cloitre, Jan 17 2003

nonn

a(n) is the determinant of M(2n+1) where M(k) is the k X k matrix with m(i,j)=j if i+j=k m(i,j)=i otherwise. - Adapted to offset 0, Rainer Rosenthal, Jun 19 2024
In the following two comments on the calculation of the terms using permanents, offset 1 is assumed. In the corresponding PARI code, this is implemented with offset 0. - Hugo Pfoertner, Jun 23 2024
(-1)^n*a(n)/2^(2n-1) is the permanent of the (m X m) matrix {1/(x_i-y_j), 1<=i<=m, 1<=j<=m}, where x_1,x_2,...,x_m are the zeros of x^m-1 and y_1,y_2,...,y_m the zeros of y^m+1 and m=2n-1.
In 1881, R. F. Scott posed a conjecture that the absolute value of permanent of square matrix with elements a(i,j)= (x_i - y_j)^(-1), where x_1,...,x_n are roots of x^n=1, while y_1,...,y_n are roots of y^n=-1, equals a((n-1)/2)/2^n, if n>=1 is odd, and 0, if n>=2 is even. After a century (in 1979), the conjecture was proved by H. Minc. - Vladimir Shevelev, Dec 01 2013
a(n) is the number of permutations in S_{2n+1} in which all cycles have odd length. - José H. Nieto S., Jan 09 2012
Number of 3-bundled increasing bilabeled trees with 2n labels. - Markus Kuba, Nov 18 2014
a(n) is the number of rooted, binary, leaf-labeled topologies with 2n+2 leaves that have n+1 cherry nodes. - Noah A Rosenberg, Feb 12 2019

			G.f. = 1 + 3*x + 45*x^2 + 1575*x^3 + 99225*x^4 + 9823275*x^5 + ...
M(5) =
  [1, 2, 3, 1, 5]
  [1, 2, 2, 4, 5]
  [1, 3, 3, 4, 5]
  [4, 2, 3, 4, 5]
  [1, 2, 3, 4, 5].
Integral_{x=0..oo} x^3*BesselK(1, sqrt(x)) = 1575*Pi. - _Olivier Gérard_, May 20 2009

Miklós Bóna, A walk through combinatorics, World Scientific, 2006.

Alois P. Heinz, Table of n, a(n) for n = 0..224
Cyril Banderier, Markus Kuba, and Michael Wallner, Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions, arXiv:2103.03751 [math.PR], 2021.
Guo-Niu Han and Christian Krattenthaler, Rectangular Scott-type permanents, arXiv:math/0003072 [math.RA], 2000.
Markus Kuba and Alois Panholzer, Combinatorial families of multilabelled increasing trees and hook-length formulas, arXiv:1411.4587 [math.CO], 17 Nov 2014.
MathOverflow, Geometric / physical / probabilistic interpretations of Riemann zeta(n>1)?, answer by Tom Copeland posted in Aug 2021.
Henryk Minc, On a conjecture of R. F. Scott (1881), Linear Algebra Appl., Vol. 28 (1979), pp. 141-153.
Theodoros Theodoulidis, On the Closed-Form Expression of Carson’s Integral, Period. Polytech. Elec. Eng. Comp. Sci., Vol. 59, No. 1 (2015), pp. 26-29.
Eric Weisstein's World of Mathematics, Struve function.

Cf. A001818, A000165.
Bisection of A000246, A053195, |A013069|, |A046126|. Cf. A000909.
Cf. A001044, A010791, |A129464|, A114779, are also values of similar moments.
Equals the row sums of A162005.
Cf. A316728.
Diagonal elements of A306364 in even-numbered rows.

I:=[1, 3]; [n le 2 select I[n] else (4*n^2-8*n+3)*Self(n-1): n in [1..20]]; // Vincenzo Librandi, Nov 18 2014

a:= n-> (d-> d(2*n-1)*d(2*n+1))(doublefactorial):
seq(a(n), n=0..15);  # Alois P. Heinz, Jan 30 2013
# second Maple program:
A079484 := n-> LinearAlgebra[Determinant](Matrix(2*n+1, (i, j)-> `if`(i+j=2*n+1, j, i))): seq(A079484(n), n=0..14); # Rainer Rosenthal, Jun 18 2024

a[n_] := (2n - 1)!!*(2n + 1)!!; Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Jan 30 2013 *)

/* Formula using the zeta function and a log integral:*/
L(n)= intnum(t=0, 1, log(1-1/t)^n);
Zetai(n)= -I*I^n*(2*Pi)^(n-1)/(n-1)*L(1-n);
a(m)={my(n=m+1);round(real(-I*2^(2*n-1)*Zetai(1/2-n)*L(-1/2+n)/(Zetai(-1/2+n)*L(1/2-n))))};
/* Gerry Martens, Mar 07 2011, adapted to offset 0 by Hugo Pfoertner, Jun 19 2024 */

{a(n) = if( n<0, -1 / self()(-1-n), (2*n + 1)! * (2*n)! / (n! * 2^n)^2 )}; /* Michael Somos, May 04 2017 */

{a(n) = if( n<0, -1 / self()(-1-n), my(m = 2*n + 1); m! * polcoeff( x / sqrt( 1 - x^2 + x * O(x^m) ), m))}; /* Michael Somos, May 04 2017 */

\\ using the Pochhammer symbol
a(n) = {my(P(x,k)=gamma(x+k)/gamma(x)); 4^n*round(P(1/2,n)*P(3/2,n))} \\ Hugo Pfoertner, Jun 20 2024

\\ Scott's (1881) method
a(n) = {my(m=2*n+1, X = polroots(x^m-1), Y = polroots(x^m+1), M = matrix(m, m, i, j, 1/(X[i]-Y[j]))); (-1)^n * round(2^m * real(matpermanent(M)))}; \\ Hugo Pfoertner, Jun 23 2024

D-finite with recurrence a(n) = (4*n^2 - 1) * a(n-1) for all n in Z.
a(n) = A001147(n)*A001147(n+1).
E.g.f.: 1/(1-x^2)^(3/2) (with interpolated zeros). - Paul Barry, May 26 2003
a(n) = (2n+1)! * C(2n, n) / 2^(2n). - Ralf Stephan, Mar 22 2004.
Alternatingly signed values have e.g.f. sqrt(1+x^2).
a(n) is the value of the n-th moment of (1/Pi)*BesselK(1, sqrt(x)) on the positive part of the real line. - Olivier Gérard, May 20 2009
a(n) = -2^(2*n-1)*exp(i*n*Pi)*gamma(1/2+n)/gamma(3/2-n). - Gerry Martens, Mar 07 2011
E.g.f. (odd powers) tan(arcsin(x)) = Sum_{n>=0} (2n-1)!!*(2n+1)!!*x^(2*n+1)/(2*n+1)!. - Vladimir Kruchinin, Apr 22 2011
G.f.: 1 + x*(G(0) - 1)/(x-1) where G(k) = 1 - ((2*k+2)^2-1)/(1-x/(x - 1/G(k+1))); ( continued fraction ). - Sergei N. Gladkovskii, Jan 15 2013
a(n) = (2^(2*n+3)*Gamma(n+3/2)*Gamma(n+5/2))/Pi. - Jean-François Alcover, Jul 20 2015
Limit_{n->oo} 4^n*(n!)^2/a(n) = Pi/2. - Daniel Suteu, Feb 05 2017
From Michael Somos, May 04 2017: (Start)
a(n) = (2*n + 1) * A001818(n).
E.g.f.: Sum_{n>=0} a(n) * x^(2*n+1) / (2*n+1)! = x / sqrt(1 - x^2) = tan(arcsin(x)).
Given e.g.f. A(x) = y, then x * y' = y + y^3.
a(n) = -1 / a(-1-n) for all n in Z.
0 = +a(n)*(+288*a(n+2) -60*a(n+3) +a(n+4)) +a(n+1)*(-36*a(n+2) -4*a(n+3)) +a(n+2)*(+3*a(n+2)) for all n in Z. (End)
a(n) = Sum_{k=0..2n} (k+1) * A316728(n,k). - Alois P. Heinz, Jul 12 2018
From Amiram Eldar, Mar 18 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + L_1(1)*Pi/2, where L is the modified Struve function.
Sum_{n>=0} (-1)^n/a(n) = 1 - H_1(1)*Pi/2, where H is the Struve function. (End)

Simpler description from Daniel Flath (deflath(AT)yahoo.com), Mar 05 2004

A156919 Table of coefficients of polynomials related to the Dirichlet eta function.

Original entry on oeis.org

1, 2, 1, 4, 10, 1, 8, 60, 36, 1, 16, 296, 516, 116, 1, 32, 1328, 5168, 3508, 358, 1, 64, 5664, 42960, 64240, 21120, 1086, 1, 128, 23488, 320064, 900560, 660880, 118632, 3272, 1, 256, 95872, 2225728

Offset: 0

Johannes W. Meijer, Feb 20 2009, Jun 24 2009

Essentially the same as A185411. Row reverse of A185410. - Peter Bala, Jul 24 2012
The SF(z; n) formulas, see below, were discovered while studying certain properties of the Dirichlet eta function.
From Peter Bala, Apr 03 2011: (Start)
Let D be the differential operator 2*x*d/dx. The row polynomials of this table come from repeated application of the operator D to the function g(x) = 1/sqrt(1 - x). For example,
  D(g) = x*g^3
  D^2(g) = x*(2 + x)*g^5
  D^3(g) = x*(4 + 10*x + x^2)*g^7
  D^4(g) = x*(8 + 60*x + 36*x^2 + x^3)*g^9.
Thus this triangle is analogous to the triangle of Eulerian numbers A008292, whose row polynomials come from the  repeated application of the operator x*d/dx to the function 1/(1 - x). (End)

			The first few rows of the triangle are:
  [1]
  [2, 1]
  [4, 10, 1]
  [8, 60, 36, 1]
  [16, 296, 516, 116, 1]
The first few P(z;n) are:
  P(z; n=0) = 1
  P(z; n=1) = 2 + z
  P(z; n=2) = 4 + 10*z + z^2
  P(z; n=3) = 8 + 60*z + 36*z^2 + z^3
The first few SF(z;n) are:
  SF(z; n=0) = (1/2)*(1)/(1-z)^(3/2);
  SF(z; n=1) = (1/4)*(2+z)/(1-z)^(5/2);
  SF(z; n=2) = (1/8)*(4+10*z+z^2)/(1-z)^(7/2);
  SF(z; n=3) = (1/16)*(8+60*z+36*z^2+z^3)/(1-z)^(9/2);
In the Savage-Viswanathan paper, the coefficients appear as
  1;
  1,    2;
  1,   10,     4;
  1,   36,    60,     8;
  1,  116,   516,   296,    16;
  1,  358,  3508,  5168,  1328,   32;
  1, 1086, 21120, 64240, 42960, 5664, 64;
  ...

D. H. Lehmer, Interesting Series Involving the Central Binomial Coefficient, Am. Math. Monthly 92 (1985) 449-457, Polynomial V in eq (17). [R. J. Mathar, Feb 24 2009]
Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, arXiv: 1204.4963v3 [math.CO], 2012.
Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, El J. Combinat. 20 (1) (2013) P11
S.-M. Ma and T. Mansour, The 1/k-Eulerian polynomials and k-Stirling permutations, arXiv preprint arXiv:1409.6525 [math.CO], 2014.
S.-M. Ma and Y.-N. Yeh, Stirling permutations, cycle structures of permutations and perfect matchings, arXiv preprint arXiv:1503.06601 [math.CO], 2015.
Carla D. Savage and Gopal Viswanathan, The 1/k-Eulerian polynomials, Elec. J. of Comb., Vol. 19, Issue 1, #P9 (2012).
Eric Weisstein's World of Mathematics, Dirichlet Eta Function

A142963 and this sequence can be mapped onto the A156920 triangle.
FP1 sequences A000340, A156922, A156923, A156924.
FP2 sequences A050488, A142965, A142966, A142968.
Appears in A162005, A000182, A162006 and A162007.
Cf. A186695, A187075.
Cf. A185410 (row reverse), A185411.

A156919 := proc(n,m) if n=m then 1; elif m=0 then 2^n ; elif m<0 or m>n then 0; else 2*(m+1)*procname(n-1,m)+(2*n-2*m+1)*procname(n-1,m-1) ; end if; end proc: seq(seq(A156919(n,m), m=0..n), n=0..7); # R. J. Mathar, Feb 03 2011

g[0] = 1/Sqrt[1-x]; g[n_] := g[n] = 2x*D[g[n-1], x]; p[n_] := g[n] / g[0]^(2n+1) // Cancel; row[n_] := CoefficientList[p[n], x] // Rest; Table[row[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, Aug 09 2012, after Peter Bala *)
Flatten[Table[Rest[CoefficientList[Nest[2 x D[#, x] &, (1 - x)^(-1/2), k] (1 - x)^(k + 1/2), x]], {k, 10}]] (* Jan Mangaldan, Mar 15 2013 *)

SF(z; n) = Sum_{m >= 1} m^(n-1)*4^(-m)*z^(m-1)*Gamma(2*m+1)/(Gamma(m)^2) = P(z;n) / (2^(n+1)*(1-z)^((2*n+3)/2)) for n >= 0. The polynomials P(z;n) = Sum_{k = 0..n} a(k)*z^k generate the a(n) sequence.
If we write the sequence as a triangle the following relation holds: T(n,m) = (2*m+2)*T(n-1,m) + (2*n-2*m+1)*T(n-1,m-1) with T(n,m=0) = 2^n and T(n,n) = 1, n >= 0 and 0 <= m <= n.
G.f.: 1/(1-xy-2x/(1-3xy/(1-4x/(1-5xy/(1-6x/(1-7xy/(1-8x/(1-... (continued fraction). - Paul Barry, Jan 26 2011
From Peter Bala, Apr 03 2011 (Start)
E.g.f.: exp(z*(x + 2)) * (1 - x)/(exp(2*x*z) - x*exp(2*z))^(3/2) = Sum_{n >= 0} P(x,n)*z^n/n! = 1 + (2 + x)*z + (4 + 10*x + x^2)*z^2/2! + (8 + 60*x + 36*x^2 + x^3)*z^3/3! + ... .
Explicit formula for the row polynomials:
P(x,n-1) = Sum_{k = 1..n} 2^(n-2*k)*binomial(2k,k)*k!*Stirling2(n,k)*x^(k-1)*(1 - x)^(n-k).
The polynomials x*(1 + x)^n * P(x/(x + 1),n) are the row polynomials of A187075.
The polynomials x^(n+1) * P((x + 1)/x,n) are the row polynomials of A186695.
Row sums are A001147(n+1). (End)
Sum_{k = 0..n} (-1)^k*T(n,k) = (-1)^binomial(n,2)*A012259(n+1). - Johannes W. Meijer, Sep 27 2011

Minor edits from Johannes W. Meijer, Sep 27 2011

A000140 Kendall-Mann numbers: the most common number of inversions in a permutation on n letters is floor(n*(n-1)/4); a(n) is the number of permutations with this many inversions.

Original entry on oeis.org

1, 1, 2, 6, 22, 101, 573, 3836, 29228, 250749, 2409581, 25598186, 296643390, 3727542188, 50626553988, 738680521142, 11501573822788, 190418421447330, 3344822488498265, 62119523114983224, 1214967840930909302, 24965661442811799655, 538134522243713149122

Offset: 1

N. J. A. Sloane

Row maxima of A008302, see example.
The term a(0) would be 1: the empty product is one and there is just one coefficient 1=x^0, corresponding to the 1 empty permutation (which has 0 inversions).
From Ryen Lapham and Anant Godbole, Dec 12 2006: (Start)
Also, the number of permutations on {1,2,...,n} for which the number A of monotone increasing subsequences of length 2 and the number D of monotone decreasing 2-subsequences are as close to each other as possible, i.e., 0 or 1. We call such permutations 2-balanced.
If 4|n(n-1) then (with A and D as above) the feasible values of A-D are C(n,2), C(n,2)-2,...,2,0,-2,...,-C(n,2), whereas if 4 does not divide n(n-1), A-D may equal C(n,2), C(n,2)-2,...,1,-1,...,-C(n,2). Let a_n(i) equal the number of permutations with A-D the i-th highest feasible value.
The sequence in question gives the number of permutations for which A-D=0 or A-D=1, i.e., it equals A_n(j) where j = floor((binomial(n,2)+2)/2). Here is the recursion: a_n(i) = a_n(i-1) + a_{n-1}(i) for 1 <= i <= n and a_n(n+k) = a_n(n+k-1) + a_{n-1}(n+k) - a_n(k) for k >= 1. (End)
The only two primes found < 301 are for n = 3 and 6.
Define an ordered list to have n terms with terms t(k) for k=1..n. Specify that t(k) ranges from 1 to k, hence the third term t(3) can be 1, 2, or 3. Find all sums of the terms for all n! allowable arrangements to obtain a maximum sum for the greatest number of arrangements. This number is a(n). For n=4, the maximum sum 7 appears in 6 arrangements: 1114, 1123, 1213, 1222, 1231, and 1132. - J. M. Bergot, May 14 2015
Named after the British statistician Maurice George Kendall (1907-1983) and the Austrian-American mathematician Henry Berthold Mann (1905-2000). - Amiram Eldar, Apr 07 2023

			From _Joerg Arndt_, Jan 16 2011: (Start)
a(4) = 6 because the among the permutations of 4 elements those with 3 inversions are the most frequent and appear 6 times:
       [inv. table]  [permutation]  number of inversions
   0:    [ 0 0 0 ]    [ 0 1 2 3 ]    0
   1:    [ 1 0 0 ]    [ 1 0 2 3 ]    1
   2:    [ 0 1 0 ]    [ 0 2 1 3 ]    1
   3:    [ 1 1 0 ]    [ 2 0 1 3 ]    2
   4:    [ 0 2 0 ]    [ 1 2 0 3 ]    2
   5:    [ 1 2 0 ]    [ 2 1 0 3 ]    3  (*)
   6:    [ 0 0 1 ]    [ 0 1 3 2 ]    1
   7:    [ 1 0 1 ]    [ 1 0 3 2 ]    2
   8:    [ 0 1 1 ]    [ 0 3 1 2 ]    2
   9:    [ 1 1 1 ]    [ 3 0 1 2 ]    3  (*)
  10:    [ 0 2 1 ]    [ 1 3 0 2 ]    3  (*)
  11:    [ 1 2 1 ]    [ 3 1 0 2 ]    4
  12:    [ 0 0 2 ]    [ 0 2 3 1 ]    2
  13:    [ 1 0 2 ]    [ 2 0 3 1 ]    3  (*)
  14:    [ 0 1 2 ]    [ 0 3 2 1 ]    3  (*)
  15:    [ 1 1 2 ]    [ 3 0 2 1 ]    4
  16:    [ 0 2 2 ]    [ 2 3 0 1 ]    4
  17:    [ 1 2 2 ]    [ 3 2 0 1 ]    5
  18:    [ 0 0 3 ]    [ 1 2 3 0 ]    3  (*)
  19:    [ 1 0 3 ]    [ 2 1 3 0 ]    4
  20:    [ 0 1 3 ]    [ 1 3 2 0 ]    4
  21:    [ 1 1 3 ]    [ 3 1 2 0 ]    5
  22:    [ 0 2 3 ]    [ 2 3 1 0 ]    5
  23:    [ 1 2 3 ]    [ 3 2 1 0 ]    6
The statistics are reflected by the coefficients of the polynomial
(1+x)*(1+x+x^2)*(1+x+x^2+x^3) ==
x^6 + 3*x^5 + 5*x^4 + 6*x^3 + 5*x^2 + 3*x^1 + 1*x^0
There is 1 permutation (the identity) with 0 inversions,
3 permutations with 1 inversion, 5 with 2 inversions,
6 with 3 inversions (the most frequent, marked with (*) ), 5 with 4 inversions, 3 with 5 inversions, and one with 6 inversions. (End)
G.f. = x + x^2 + 2*x^3 + 6*x^4 + 22*x^5 + 101*x^6 + 573*x^7 + 3836*x^8 + ...

F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 241.
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).

Robert Israel, Robert G. Wilson v and N. J. A. Sloane Table of n, a(n) for n = 1..400 (a(1) to a(61) from Sloane, a(62) to a(350) from Wilson).
Dorin Andrica and Ovidiu Bagdasar, On some results concerning the polygonal polynomials, Carpathian Journal of Mathematics (2019) Vol. 35, No. 1, 1-11.
Dominique Foata, Distributions eulériennes et mahoniennes sur le groupe des permutations, pp. 27-49 of M. Aigner, editor, Higher Combinatorics, Reidel, Dordrecht, Holland, 1977.
Mikhail Gaichenkov, The property of Kendall-Mann numbers, MathOverflow, 2010.
Richard K. Guy, Letter to N. J. A. Sloane with attachment, Mar 1988.
A. Waksman, On the complexity of inversions, IEEE Trans. Computers, 19 (1970), 1225-1226.
Wikipedia, Inversion.
Wikipedia, q-Pochhammer symbol. - _Paul Muljadi_, Jan 18 2011
Index entries for "core" sequences.

Row maxima of A008302.
Odd terms are A186888.
Cf. A000124, A001147, A128566, A181609.

/* based on David W. Wilson's formula */ PS:=PowerSeriesRing(Integers()); [ Max(Coefficients(&*[&+[ x^i: i in [0..j] ]: j in [0..n-1] ])): n in [1..21] ]; // Klaus Brockhaus, Jan 18 2011

f := 1: for n from 0 to 40 do f := f*add(x^i, i=0..n): s := series(f, x, n*(n+1)/2+1): m := max(coeff(s, x, j) $ j=0..n*(n+1)/2): printf(`%d,`,m) od: # James Sellers, Dec 07 2000 [offset is off by 1 - N. J. A. Sloane, May 23 2006]
P:= [1]: a[1]:= 1:
for n from 2 to 100 do
P:= expand(P * add(x^j,j=0..n-1));
a[n]:= max(eval(convert(P,list),x=1));
od:
seq(a[i],i=1..100); # Robert Israel, Dec 14 2014

f[n_] := Max@ CoefficientList[ Expand@ Product[ Sum[x^i, {i, 0, j}], {j, n-1}], x]; Array[f, 20]
Flatten[{1, 1, Table[Coefficient[Expand[Product[Sum[x^k, {k, 0, m-1}], {m, 1, n}]], x^Floor[n*(n-1)/4]], {n, 3, 20}]}] (* Vaclav Kotesovec, May 13 2016 *)
Table[SeriesCoefficient[QPochhammer[x, x, n]/(1-x)^n, {x, 0, Floor[n*(n-1)/4]}], {n, 1, 20}] (* Vaclav Kotesovec, May 13 2016 *)

{a(n) = if( n<0, 0, vecmax( Vec( prod(k=1, n, 1 - x^k) / (1 - x)^n)))}; /* Michael Somos, Apr 21 2014 */

from math import prod
from sympy import Poly
from sympy.abc import x
def A000140(n): return 1 if n == 1 else max(Poly(prod(sum(x**i for i in range(j+1)) for j in range(n))).all_coeffs()) # Chai Wah Wu, Feb 02 2022

Largest coefficient of (1)(x+1)(x^2+x+1)...(x^(n-1) + ... + x + 1). - David W. Wilson
The number of terms is given in A000124.
a(n+1)/a(n) = n - 1/2 + O(1/n^{1-epsilon}) as n --> infinity (compare with A008302, A181609, A001147). - Mikhail Gaichenkov, Apr 11 2014
Asymptotics (Mikhail Gaichenkov, 2010): a(n) ~ 6 * n^(n-1) / exp(n). - Vaclav Kotesovec, May 17 2015

Edited by N. J. A. Sloane, Mar 05 2011

A039683 Signed double Pochhammer triangle: expansion of x(x-2)(x-4)..(x-2n+2).

Original entry on oeis.org

1, -2, 1, 8, -6, 1, -48, 44, -12, 1, 384, -400, 140, -20, 1, -3840, 4384, -1800, 340, -30, 1, 46080, -56448, 25984, -5880, 700, -42, 1, -645120, 836352, -420224, 108304, -15680, 1288, -56, 1, 10321920, -14026752, 7559936, -2153088, 359184, -36288, 2184, -72, 1

Offset: 1

Wouter Meeussen

T(n,m) = R_n^m(a=0,b=2) in the notation of the given reference.
Exponential Riordan array [1/(1+2x),log(1+2x)/2]. The unsigned triangle is [1/(1-2x),log(1/sqrt(1-2x))]. - Paul Barry_, Apr 29 2009
The n-th row is related to the expansion of z^(-2n)*(z^3 d/dz)^n in polynomials of the Euler operator D=(z d/dz). E.g., z^(-6)(z^3 d/dz)^3 = D^3 + 6 D^2 + 8 D. See Copeland link for relations to Bell / Exponential / Touchard polynomial operators. - Tom Copeland, Nov 14 2013
A refinement of this array is given by A231846. - Tom Copeland, Nov 15 2013
Also the Bell transform of the double factorial of even numbers A000165 except that the values are unsigned and in addition a first column (1,0,0 ...) is added on the left side of the triangle. For the Bell transform of the double factorial of odd numbers A001147 see A132062. For the definition of the Bell transform see A264428. - Peter Luschny, Dec 20 2015
The signed triangle is also the inverse Bell transform of A000079 (see Luschny link). - John Keith, Nov 24 2020

			Triangle starts:
  {1},
  {2,1},
  {8,6,1},
  {48,44,12,1},
  ...
From _Paul Barry_, Apr 29 2009: (Start)
The unsigned triangle [1/(1-2x),log(1/sqrt(1-2x))] has production matrix:
  2, 1,
  4, 4, 1,
  8, 12, 6, 1,
  16, 32, 24, 8, 1,
  32, 80, 80, 40, 10, 1,
  64, 192, 240, 160, 60, 12, 1
which is A007318^{2} beheaded. (End)

Richell O. Celeste, Roberto B. Corcino, and Ken Joffaniel M. Gonzales. Two Approaches to Normal Order Coefficients, Journal of Integer Sequences, Vol. 20 (2017), Article 17.3.5.
Tom Copeland, Addendum to Mathemagical Forests.
P. Feijão, F. V. Martinez, and A. Thévenin, On the distribution of cycles and paths in multichromosomal breakpoint graphs and the expected value of rearrangement distance, BMC Bioinformatics 16:Suppl19 (2015), S1. doi:10.1186/1471-2105-16-S19-S1
Lisa Glaser, Causal set actions in various dimensions, J. Phys.: Conf. Ser. 306 (2011), 012041.
Wolfdieter Lang, First 9 rows and comment.
Peter Luschny, The Bell transform
D. S. Mitrinovic and M. S. Mitrinovic, Tableaux d'une classe de nombres relies aux nombres de Stirling, Univ. Beograd. Pubi. Elektrotehn. Fak. Ser. Mat. Fiz. 77 (1962).

First column (unsigned triangle) is (2(n-1))!! = 1, 2, 8, 48, 384...= A000165(n-1) and the row sums (unsigned) are (2n-1)!! = 1, 3, 15, 105, 945... = A001147(n-1).
Cf. A051141, A051142.
Cf. A000165, A132062, A264428.
Cf. A038207.

Table[ Rest@ CoefficientList[ Product[ z-k, {k, 0, 2p-2, 2} ], z ], {p, 6} ]

# uses[bell_transform from A264428]
# Unsigned values and an additional first column (1,0,0,...).
def A039683_unsigned_row(n):
    a = sloane.A000165
    dblfact = a.list(n)
    return bell_transform(n, dblfact)
[A039683_unsigned_row(n) for n in (0..9)] # Peter Luschny, Dec 20 2015

T(n, m) = T(n-1, m-1) - 2*(n-1)*T(n-1, m), n >= m >= 1; T(n, m) := 0, n

E.g.f. for m-th column of signed triangle: (((log(1+2*x))/2)^m)/m!.

E.g.f.: (1+2*x)^(y/2). O.g.f. for n-th row of signed triangle: Sum_{m=0..n} Stirling1(n, m)*2^(n-m)*x^m. - Vladeta Jovovic, Feb 11 2003

T(n, m) = S1(n, m)*2^(n-m), with S1(n, m) := A008275(n, m) (signed Stirling1 triangle).

The production matrix below is A038207 with the first row removed. With the initial index n = 0, the associated differential raising operator is R = e^(2D)*x = (2+x)*e^(2D) with D = d/dx, i.e., R p_n(x) = p_(n+1)(x) where p_n(x) is the n-th unsigned row polynomial and p_0(x) = 1, so p_(n+1)(x) = (2+x) * p_n(2+x). - Tom Copeland, Oct 11 2016

Additional comments from Wolfdieter Lang

Title revised by Tom Copeland, Dec 21 2013

A094216 Triangle read by rows giving the coefficients of formulas generating each variety of S1(n,k) (unsigned Stirling numbers of first kind). The p-th row (p>=1) contains T(i,p) for i=1 to 2p, where T(i,p) satisfies Sum_{i=1..2p} T(i,p) * C(n,i).

Original entry on oeis.org

1, 1, 2, 7, 8, 3, 6, 38, 93, 111, 65, 15, 24, 226, 874, 1821, 2224, 1600, 630, 105, 120, 1524, 8200, 24860, 47185, 58465, 47474, 24430, 7245, 945, 720, 11628, 81080, 326712, 852690, 1522375, 1905168, 1676325, 1018682, 407925, 97020, 10395, 5040

Offset: 1

André F. Labossière, May 27 2004, Feb 21 2007

The formulas S1(n+p,n) obtained are those of S1(n+2,n) { A000914 }, S1(n+3,n) { A001303 }, S1(n+4,n) { A000915 }, S1(n+5,n) { A053567 } and so on.

			Row 5 contains 120,1524,8200,24860,47185,58465,47474,24430,7245,945, so the formula generating S1(n+5,n) numbers { A053567 } will be the following : 120*n +1524*C(n,2) +8200*C(n,3) +24860*C(n,4) +47185*C(n,5) +58465*C(n,6) +47474*C(n,7) +24430*C(n,8) +7245*C(n,9) +945*C(n,10). And then substituting for the 10th number of such a S1(n+p,n) gives S1(15,10) = 37312275.

Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964, 9th Printing (1970), pp. 833-834.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Francis L. Miksa (1901-1975), Stirling numbers of the first kind, "27 leaves reproduced from typewritten manuscript on deposit in the UMT File", Mathematical Tables and Other Aids to Computation, vol. 10, no. 53, January 1956, pp. 37-38 (Reviews and Descriptions of Tables and Books, 7[I]).
Dragoslav S. Mitrinovic (1908-1995), Sur les nombres de Stirling de première espèce et les polynômes de Stirling, AMS 11B73_05A19, Publications de la Faculté d'Electrotechnique de l'Université de Belgrade, Série Mathématiques et Physique (ISSN 0522-8441), no. 23, 1959 (5.V.1959), pp. 1-20.
John J. O'Connor and Edmund F. Robertson, James Stirling (1692-1770), (September 1998).
Eric Weisstein's World of Mathematics, Stirling numbers of the first kind.
Stephen Wolfram, Wolfram Research, Mathematica 5.2, webMathematica 2.

Cf. A000914, A001303, A000915, A053567, A008275, A008276.

Cf. A000012, A000217, A001147, A000142, A094262.

row[m_] := Module[{eq, t}, eq[n_] := Array[t, 2 m].Table[Binomial[n, k], {k, 1, 2 m}] == Abs[StirlingS1[n + m, n]]; Array[t, 2 m] /. Solve[ Array[ eq, 2 m]] // First];
Array[row, 7] // Flatten (* Jean-François Alcover, Nov 14 2019 *)

a(1,k) = k!
...
a(2*k-5,k) = a(2*k,k) * (175000*k^8 -2117500*k^7 +10856650*k^6 -30743377*k^5 +52511770*k^4 -55386931*k^3 +35321832*k^2 -12560580*k+1944000) / (1632960*k^3 -7348320*k^2 +9389520*k -3061800).
a(2*k-4,k) = a(2*k,k) * (2500*k^6 -17400*k^5 +48511*k^4 -69378*k^3 +53929*k^2 -21906*k +3744) / (7776*k^2-15552*k+5832).
a(2*k-3,k) = a(2*k,k) * (1250*k^4-4225*k^3+5023*k^2-2600*k+528) / (1620*k-810).
a(2*k-2,k) = a(2*k,k) * (50*k^3-93*k^2+55*k-12) / (36*k-18).
a(2*k-1,k) = a(2*k,k) * (5*k-2) / 3.
a(2*k,k) = (2*k)! / (k!*2^k).

A024199 a(n) = (2n-1)!! * Sum_{k=0..n-1}(-1)^k/(2k+1).

Original entry on oeis.org

0, 1, 2, 13, 76, 789, 7734, 110937, 1528920, 28018665, 497895210, 11110528485, 241792844580, 6361055257725, 163842638377950, 4964894559637425, 147721447995130800, 5066706567801827025, 171002070002301095250, 6548719685561840296125, 247199273204273879989500

Offset: 0

Clark Kimberling

(2*n + 1)!!/a(n+1), n>=0, is the n-th approximant for William Brouncker's continued fraction for 4/Pi = 1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... ))) See the C. Brezinski and J.-P. Delahaye references given under A142969 and A142970, respectively. The double factorials (2*n + 1)!! = A001147(n+1) enter. - Wolfdieter Lang, Oct 06 2008

			a(3) = (2*3 - 1)!! * Sum_{k=0..2} (-1)^k/(2k + 1) = 5!! * (1/(2*0 + 1) - 1/(2*1 + 1) + 1/(2*2 + 1)) = 5*3*1*(1/1 - 1/3 + 1/5) = 15 - 5 + 3 = 13. Notice that the first factor always cancels the common denominator of the sum. - _Michael B. Porter_, Jul 22 2016

A. E. Jolliffe, Continued Fractions, in Encyclopaedia Britannica, 11th ed., pp. 30-33; see p. 31.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Peter Luschny, Is the Gamma function misdefined? Hadamard versus Euler - Who found the better Gamma function?
Wikipedia, William Brouncker, 2nd Viscount Brouncker

Cf. A004041, A000407.
From Johannes W. Meijer, Nov 12 2009: (Start)
Cf. A007509 and A025547.
Equals first column of A167584.
Equals row sums of A167591.
Equals first right hand column of A167594.
(End)
Cf. A167576 and A135457.

[0] cat [ n le 2 select (n) else 2*Self(n-1)+(2*n-3)^2*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 17 2015

a := proc(n) option remember; if n=0 then 0 elif n=1 then 1 else 2*a(n-1)+(2*n-3)^2* a(n-2) fi end: seq(a(n), n=0..20); # Peter Luschny, Nov 16 2016 after N. J. A. Sloane

f[k_] := (2 k - 1) (-1)^(k + 1)
t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 22}]    (* A024199 signed *)
(* Clark Kimberling, Dec 30 2011 *)
RecurrenceTable[{a[n+1] == 2*a[n] + (2*n-1)^2*a[n-1],a[0] == 0, a[1] == 1},a,{n,0,20}] (* Vaclav Kotesovec, Mar 18 2014 *)
CoefficientList[Series[Pi/4/Sqrt[1-2*x] - 1/2*Log[2*x+Sqrt[4*x^2-1]]/Sqrt[2*x-1], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Mar 18 2014 *)

a(n) = s(1)s(2)...s(n)(1/s(1) - 1/s(2) + ... + c/s(n)) where c=(-1)^(n+1) and s(k) = 2k-1 for k = 1, 2, 3, ... (was previous definition). - Clark Kimberling
D-finite with recurrence a(0) = 0, a(1) = 1, a(n+1) = 2*a(n) + (2*n-1)^2*a(n-1). - N. J. A. Sloane, Jul 19 2002
a(n) + A024200(n) = A001147(n) = (2n-1)!!. - Max Alekseyev, Sep 23 2007
a(n)/A024200(n) -> Pi/(4-Pi) as n -> oo. - Max Alekseyev, Sep 23 2007
From Wolfdieter Lang, Oct 06 2008: (Start)
E.g.f. for a(n+1), n>=0: (sqrt(1-2*x)+arcsin(2*x)*sqrt(1+2*x)/2)/((1-4*x^2)^(1/2)*(1-2*x)). From the recurrence, solving (1-4*x^2)y''(x)-2*(8*x+1)*y'(x)-9*y=0 with inputs y(0)=1, y'(0)=2.
a(n+1) = A003148(n) + A143165(n), n>=0 (from the two terms of the e.g.f.). (End)
From Johannes W. Meijer, Nov 12 2009: (Start)
a(n) = (-1)^(n-1)*(2*n-3)!! + (2*n-1)*a(n-1) with a(0) = 0.
a(n) = (2*n-1)!!*sum((-1)^(k)/(2*k+1), k=0..n-1)
(End)
E.g.f.: Pi/4/sqrt(1-2*x) - 1/2*log(2*x+sqrt(4*x^2-1))/sqrt(2*x-1). - Vaclav Kotesovec, Mar 18 2014
a(n) ~ Pi * 2^(n-3/2) * n^n / exp(n). - Vaclav Kotesovec, Mar 18 2014
a(n) = (2*H(n+1/2)-Gamma(n+1/2))*2^(n-2)*sqrt(Pi) with H(x) the Hadamard factorial (see the link section). - Cyril Damamme, Jul 19 2015
a(n) = A135457(n) - (-1)^n A001147(n-1). - Cyril Damamme, Jul 19 2015
a(n) = (Pi + (-1)^n*(Psi(n/2 + 1/4) - Psi(n/2 + 3/4)))*Gamma(n+1/2)*2^(n-2)/sqrt(Pi). - Robert Israel, Jul 20 2015
a(n) = A167576(n) - A135457(n). - Cyril Damamme, Jul 22 2015
a(n)/A001147(n) -> Pi/4 as n -> oo. - Daniel Suteu, Jul 21 2016
From Peter Bala, Nov 15 2016: (Start)
Conjecture: a(n) = 1/2*Sum_{k = 0..2*n-1} (-1)^(n-k+1)*k!*(2*n - 2*k - 3)!!, where the double factorial of an odd integer (positive or negative) may be defined in terms of the gamma function as (2*N - 1)!! = 2^((N+1)/2)*Gamma(N/2 + 1)/sqrt(Pi).
E.g.f. 1/2*arcsin(2*x)/sqrt(1 - 2*x) = x + 2*x^2/2! + 13*x^3/3! + 76*x^4/4! + .... (End)

Edited by N. J. A. Sloane, Jul 19 2002

New name from Cyril Damamme, Jul 19 2015

A049209 a(n) = -Product_{k=0..n} (7*k-1); sept-factorial numbers.

Original entry on oeis.org

1, 6, 78, 1560, 42120, 1432080, 58715280, 2818333440, 155008339200, 9610517030400, 663125675097600, 50397551307417600, 4182996758515660800, 376469708266409472000, 36517561701841718784000, 3797826416991538753536000, 421558732286060801642496000

Offset: 0

Wolfdieter Lang

G. C. Greubel, Table of n, a(n) for n = 0..330
Index entries for sequences related to factorial numbers.

Cf. A034833, A045754, A048994, A051188, A084947, A144739, A144827, A147585.
Row sums of triangle A051186 (scaled Stirling1 triangle).
Sequences of the form m^n*Pochhammer((m-1)/m, n): A000007 (m=1), A001147 (m=2), A008544 (m=3), A008545 (m=4), A008546 (m=5), A008543 (m=6), this sequence (m=7), A049210 (m=8), A049211 (m=9), A049212 (m=10), A254322 (m=11), A346896 (m=12).

[ -&*[ (7*k-1): k in [0..n-1] ]: n in [1..15] ]; // Klaus Brockhaus, Nov 10 2008

CoefficientList[Series[(1-7*x)^(-6/7),{x,0,20}],x] * Range[0,20]! (* Vaclav Kotesovec, Jan 28 2015 *)
With[{m=7}, Table[m^n*Pochhammer[(m-1)/m, n], {n, 0, 30}]] (* G. C. Greubel, Feb 16 2022 *)

m=7; [m^n*rising_factorial((m-1)/m, n) for n in (0..30)] # G. C. Greubel, Feb 16 2022

a(n) = 6*A034833(n) = (7*n-1)*(!^7), n >= 1, a(0) := 1.
a(n) = Product_{k=1..n} (7*k - 1). a(0) = 1; a(n) = (7*n - 1)*a(n-1) for n > 0. - Klaus Brockhaus, Nov 10 2008
G.f.: 1/(1-6*x/(1-7*x/(1-13*x/(1-14*x/(1-20*x/(1-21*x/(1-27*x/(1-28*x/(1-...(continued fraction). - Philippe Deléham, Jan 08 2012
a(n) = (-1)^n*Sum_{k=0..n} 7^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n) = 7^n * Gamma(n+6/7) / Gamma(6/7). - Vaclav Kotesovec, Jan 28 2015
E.g.f.: (1-7*x)^(-6/7). - Vaclav Kotesovec, Jan 28 2015
From Nikolaos Pantelidis, Dec 19 2020: (Start)
G.f.: 1/G(0) where G(k) = 1 - (14*k+6)*x - 7*(k+1)*(7*k+6)*x^2/G(k+1); (continued fraction).
which starts as 1/(1-6*x-42*x^2/(1-20*x-182*x^2/(1-34*x-420*x^2/(1-48*x-756*x^2/(1-62*x-1190*x^2/(1-... )))))) (Jacobi continued fraction).
G.f.: 1/Q(0) where Q(k) = 1 - (7*k+6)*x/(1 - (7*k+7)*x/Q(k+1) ); (continued fraction). (End)
Sum_{n>=0} 1/a(n) = 1 + (e/7)^(1/7)*(Gamma(6/7) - Gamma(6/7, 1/7)). - Amiram Eldar, Dec 19 2022

A052502 Number of permutations sigma of [3n] without fixed points such that sigma^3 = Id.

Original entry on oeis.org

1, 2, 40, 2240, 246400, 44844800, 12197785600, 4635158528000, 2345390215168000, 1524503639859200000, 1237896955565670400000, 1227993779921145036800000, 1461312598106162593792000000, 2054605512937264606871552000000

Offset: 0

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

For n >= 1 a(n) is the size of the conjugacy class in the symmetric group S_(3n) consisting of permutations whose cycle decomposition is a product of n disjoint 3-cycles.

F. W. J. Olver, Asymptotics and special functions, Academic Press, 1974, pages 336-344.

G. C. Greubel, Table of n, a(n) for n = 0..210
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 27
F. W. J. Olver et al., NIST Digital Library of Mathematical Functions, eq. 9.10.17.

Cf. A000142. Row sums of triangle A060063.
First column of array A091752 (also negative of second column).
Equals row sums of A157702. - Johannes W. Meijer, Mar 07 2009
Karol A. Penson suggested that the row sums of A060063 coincide with this entry.
Cf. A001147, A052504, A060706, A261317, A261381.
Trisection of column k=3 of A261430.

List([0..20], n-> Factorial(3*n)/(3^n*Factorial(n))) # G. C. Greubel, May 14 2019

[Factorial(3*n)/(3^n*Factorial(n)): n in [0..20]]; // G. C. Greubel, May 14 2019

spec := [S,{S=Set(Union(Cycle(Z,card=3)))},labeled]: seq(combstruct[count](spec,size=n), n=0..20);

Table[(3*n)!/(3^n*n!), {n, 0, 20}] (* G. C. Greubel, May 14 2019 *)

{a(n) = (3*n)!/(3^n*n!)}; \\ G. C. Greubel, May 14 2019

[factorial(3*n)/(3^n*factorial(n)) for n in (0..20)] # G. C. Greubel, May 14 2019

From Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001: (Start)
a(n) = (3*n)!/(3^n * n!).
a(n) ~ sqrt(3) * 9^n * (n/e)^(2n). (End)
E.g.f.: (every third coefficient of) exp(x^3/3).
G.f.: hypergeometric3F0([1/3, 2/3, 1], [], 9*x).
D-finite with recurrence a(n) = (3*n-1)*(3*n-2)*a(n-1) for n >= 1, with a(0) = 1.
Write the generating function for this sequence in the form A(x) = Sum_{n >= 0} a(n)* x^(2*n+1)/(2*n+1)!. The g.f. A(x) satisfies A'(x)*( 1 - A(x)^2) = 1. Robert Israel remarks that consequently A(x) is a root of z^3 - 3*z + 3*x with A(0) = 0. Cf. A001147, A052504 and A060706. - Peter Bala, Jan 02 2015
From Peter Bala, Feb 27 2024: (Start)
u(n) := a(n+1) satisfies the second-order recurrence u(n) = 18*n*u(n-1) + (3*n - 1)^2*(3*n - 2)^2*u(n-2) with u(0) = 2 and u(1) = 40.
A second solution to the recurrence is given by v(n) := u(n)*Sum_{k = 0..n} (-1)^k/((3*k + 1)*(3*k + 2)) with v(0) = 1 and v(1) = 18.
This leads to the continued fraction expansion (2/3)*log(2) = Sum_{k = 0..n} (-1)^k/((3*k + 1)*(3*k + 2)) = Limit_{n -> oo} v(n)/u(n) = 1/(2 + (1*2)^2/(18 + (4*5)^2/(2*18 + (7*8)^2/(3*18 + (10*11)^2/(4*18 +  ... ))))). (End)
From Gabriel B. Apolinario, Jul 30 2024: (Start)
a(n) = 3 * Integral_{t=0..oo} Ai(t)*t^(3*n) dt, where Ai(t) is the Airy function.
a(n) = Integral_{t=-oo..oo} Ai(t)*t^(3*n) dt. (End)

Edited by Wolfdieter Lang, Feb 13 2004

Title improved by Geoffrey Critzer, Aug 14 2015

A104443 Square of P(n,t) read by antidiagonals. P(n,t) = number of ways to split [t*n] into n arithmetic progressions each with t terms.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 2, 15, 1, 1, 2, 5, 105, 1, 1, 2, 4, 15, 945, 1, 1, 2, 4, 11, 55, 10395, 1, 1, 2, 4, 10, 23, 232, 135135, 1, 1, 2, 4, 10, 21, 68, 1161, 2027025, 1, 1, 2, 4, 10, 20, 59, 161, 6643, 34459425, 1, 1, 2, 4, 10, 20, 57, 125, 488, 44566, 654729075, 1

Offset: 1

Jonas Wallgren, Mar 17 2005

			Square array begins:
  1,        1,     1,    1,   1,   1,   1,   1,   1, ...
  1,        3,     2,    2,   2,   2,   2,   2,   2, ...
  1,       15,     5,    4,   4,   4,   4,   4,   4, ...
  1,      105,    15,   11,  10,  10,  10,  10,  10, ...
  1,      945,    55,   23,  21,  20,  20,  20,  20, ...
  1,    10395,   232,   68,  59,  57,  56,  56,  56, ...
  1,   135135,  1161,  161, 125, 119, 117, 116, 116, ...
  1,  2027025,  6643,  488, 349, 329, 323, 321, 320, ...
  1, 34459425, 44566, 1249, 848, 760, 745, 739, 737, ...
  ...

Cf. A104429-A104442. P(1, )=P(, 1) = A000012, P(_, 2) = A001147.

More terms from Alois P. Heinz, Nov 18 2020

A002135 Number of terms in a symmetrical determinant: a(n) = na(n-1) - (n-1)(n-2)*a(n-3)/2.

Original entry on oeis.org

1, 1, 2, 5, 17, 73, 388, 2461, 18155, 152531, 1436714, 14986879, 171453343, 2134070335, 28708008128, 415017867707, 6416208498137, 105630583492969, 1844908072865290, 34071573484225549, 663368639907213281, 13580208904207073801

Offset: 0

N. J. A. Sloane

a(n) is the number of collections of necklaces created by using exactly n different colored beads (to make the entire collection). - Geoffrey Critzer, Apr 19 2009
a(n) is the number of ways that a deck with 2 cards of each of n types may be dealt into n hands of 2 cards each, assuming that the order of the hands and the order of the cards in each hand are irrelevant. See the Art of Problem Solving link for proof. - Joel B. Lewis, Sep 30 2012
From Bruce Westbury, Jan 22 2013: (Start)
It follows from the respective exponential generating functions that A002135 is the binomial transform of A002137:
A002135(n) = Sum_{k=0..n} binomial(n,k)*A002137(k),
2 = 1.1 + 2.0 + 1.1,
5 = 1.1 + 3.0 + 3.1 + 1.1,
17 = 1.1 + 4.0 + 6.1 + 4.1 + 1.6, ...
A002137 arises from looking at the dimension of the space of invariant tensors of the r-th tensor power of the adjoint representation of the symplectic group Sp(2n) (for n large compared to r).
(End)
a(n) is the number of representations required for the symbolic central moments of order 2 for the multivariate normal distribution, that is,   E[X1^2 X2^2 .. Xn^2|mu=0, Sigma] (Phillips 2010). These representations are the upper-triangular, positive integer matrices for which for each i, the sum of the i-th row and i-th column equals 2, the power of each component. This can be shown starting from the formulation by Joel B Lewis. See "Proof for multivariate normal moments" link below for a proof. - Kem Phillips, Aug 20 2014
Equivalent to Critzer's comment, a(n) is the number of ways to cover n labeled vertices by disjoint undirected cycles, hence the exponential transform of A001710(n - 1). - Gus Wiseman, Oct 20 2018

			For n = 3, the a(3) = 5 ways to deal out the deck {1, 1, 2, 2, 3, 3} into three two-card hands are {11, 22, 33}, {12, 12, 33}, {13, 13, 22}, {11, 23, 23}, {12, 13, 23}. - _Joel B. Lewis_, Sep 30 2012

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 260, #12, a_n.
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).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 5.2.9 and Problem 5.22.

T. D. Noe, Table of n, a(n) for n=0..100
A. C. Aitken, On the number of distinct terms in the expansion of symmetric and skew determinants, Edinburgh Math. Notes, No. 34 (1944), 1-5.
A. C. Aitken, On the number of distinct terms in the expansion of symmetric and skew determinants, Edinburgh Math. Notes, No. 34 (1944), 1-5. [Annotated scanned copy]
Art of Problem Solving, Partitioning a deck with 2 cards in n types into pairs
A. Cayley, On the number of distinct terms in a symmetrical or partially symmetrical determinant, Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 9, pp. 185-190. [Annotated scanned copy]
Tomislav Došlic and Darko Veljan, Logarithmic behavior of some combinatorial sequences, Discrete Math. 308 (2008), no. 11, 2182--2212. MR2404544 (2009j:05019) - _N. J. A. Sloane_, May 01 2012
Rui-Li Liu and Feng-Zhen Zhao, New Sufficient Conditions for Log-Balancedness, With Applications to Combinatorial Sequences, J. Int. Seq., Vol. 21 (2018), Article 18.5.7.
P. A. MacMahon, Combinations derived from m identical sets of n different letters and their connexion with general magic squares, Proc. London Math. Soc., 17 (1917), 25-41.
Toufik Mansour, The length of the initial longest increasing sequence in a permutation, Art Disc. Appl. Math. (2023).
T. Muir, The Theory of Determinants in the Historical Order of Development, 4 vols., Macmillan, NY, 1906-1923. [Annotated scans of selected pages]
K. Phillips, R functions to symbolically compute the central moments of the multivariate normal distribution, Journal of Statistical Software, Feb 2010.
Kem Phillips, Proof for multivariate normal moments
Richard Stanley and J. Riordan, Problem E2297, Amer. Math. Monthly, 79 (1972), 519-520.

A diagonal of A260338.
Row sums of A215771.
Column k=2 of A257463 and A333467.
Cf. A002137, A059422, A059423, A059424.
Cf. A001147, A007717, A037143, A001710, A215771, A319225, A319226, A320655, A320656.

G:=proc(n) option remember; if n <= 1 then 1 elif n=2 then
2 else n*G(n-1)-binomial(n-1,2)*G(n-3); fi; end;

a[x_]:=Log[1/(1-x)^(1/2)]+x/2+x^2/4;Range[0, 20]! CoefficientList[Series[Exp[a[x]], {x, 0, 20}], x]
RecurrenceTable[{a[0]==a[1]==1,a[2]==2,a[n]==n*a[n-1]-(n-1)(n-2)* a[n-3]/2}, a,{n,30}] (* Harvey P. Dale, Dec 16 2011 *)
Table[Sum[Binomial[k, i] Binomial[i - 1/2, n - k] (3^(k - i) n!)/(4^k k!) (-1)^(n - k - i), {k, 0, n}, {i, 0, k}], {n, 0, 12}] (* Emanuele Munarini, Aug 25 2017 *)

a(n):=sum(sum(binomial(k,i)*binomial(i-1/2,n-k)*(3^(k-i)*n!)/(4^k*k!)*(-1)^(n-k-i),i,0,k),k,0,n);
makelist(a(n),n,0,12); /* Emanuele Munarini, Aug 25 2017 */

a(n) = if(n<3, [1,1,2][n+1], n*a(n-1) - (n-1)*(n-2)*a(n-3)/2 ); /* Joerg Arndt, Apr 07 2013 */

E.g.f.: (1-x)^(-1/2)*exp(x/2+x^2/4).
D-finite with recurrence a(n+1) = (n+1)*a(n) - binomial(n, 2)*a(n-2). See Comtet.
Asymptotics: a(n) ~ sqrt(2)*exp(3/4-n)*n^n*(1+O(1/n)). - Pietro Majer, Oct 27 2009
E.g.f.: G(0)/sqrt(1-x) where G(k) = 1 + x*(x+2)/(4*(2*k+1) - 4*x*(x+2)*(2*k+1)/(x*(x+2) + 8*(k + 1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 31 2013
a(n) = Sum_{k=0..n} Sum_{i=0..k} binomial(k,i)*binomial(i-1/2,n-k)*(3^(k-i)*n!)/(4^k*k!)*(-1)^(n-k-i). - Emanuele Munarini, Aug 25 2017

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A079484 a(n) = (2n-1)!! * (2n+1)!!, where the double factorial is A006882.

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A156919 Table of coefficients of polynomials related to the Dirichlet eta function.

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A000140 Kendall-Mann numbers: the most common number of inversions in a permutation on n letters is floor(n*(n-1)/4); a(n) is the number of permutations with this many inversions.

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A039683 Signed double Pochhammer triangle: expansion of x(x-2)(x-4)..(x-2n+2).

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A094216 Triangle read by rows giving the coefficients of formulas generating each variety of S1(n,k) (unsigned Stirling numbers of first kind). The p-th row (p>=1) contains T(i,p) for i=1 to 2*p, where T(i,p) satisfies Sum_{i=1..2*p} T(i,p) * C(n,i).

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A024199 a(n) = (2n-1)!! * Sum_{k=0..n-1}(-1)^k/(2k+1).

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A094216 Triangle read by rows giving the coefficients of formulas generating each variety of S1(n,k) (unsigned Stirling numbers of first kind). The p-th row (p>=1) contains T(i,p) for i=1 to 2p, where T(i,p) satisfies Sum_{i=1..2p} T(i,p) * C(n,i).

A002135 Number of terms in a symmetrical determinant: a(n) = na(n-1) - (n-1)(n-2)*a(n-3)/2.