A052502 -id:A052502 - cp's OEIS frontend

A060079 Duplicate of A052502.

1, 2, 40, 2240, 246400, 44844800, 12197785600, 4635158528000, 2345390215168000

Offset: 0

dead

A060063 Triangle of coefficients of certain polynomials used for G.f.s of columns of triangle A060058.

1, 1, 1, 5, 26, 9, 61, 775, 1179, 225, 1385, 32516, 114318, 87156, 11025, 50521, 1894429, 11982834, 20371266, 9652725, 893025, 2702765, 148008446, 1472351967, 4417978068, 4546174779, 1502513550

Offset: 0

Wolfdieter Lang, Mar 16 2001

The row polynomials p(n,x) (rising powers of x) appear as numerators of the column g.f.s of triangle A060058.
First column (m=0) gives A000364 (Euler numbers). See A091742, A091743, A091744 for columns m=1..3.
The main diagonal gives A001818. The row sums give A052502. The alternating row sums give A091745.

			Triangle begins:
  {1};
  {1,1};
  {5,26,9};     <-- p(2,n)=5+26*x+9*x^2.
  {61,775,1179,225};
  ...

W. Lang, First 8 rows.

The row polynomials p(n, x) := Sum_{m=0..n} a(n, m)*x^m satisfy the differential equation: p(n, x) = x*((1-x)^2)*(d^2/dx^2)p(n-1, x) + (1+6*(n-1)*x+(5-6*n)*x^2)*(d/dx)p(n-1, x) + (3*n-2)*(1+(3*n-2)*x)*p(n-1, x), n >= 1, with input p(0, x)=1. - Wolfdieter Lang, Feb 13 2004

A261430 Number A(n,k) of permutations p of [n] without fixed points such that p^k = Id; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 2, 3, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 9, 0, 15, 0, 0, 1, 0, 0, 2, 0, 0, 40, 0, 0, 0, 1, 0, 1, 0, 3, 24, 105, 0, 105, 0, 0, 1, 0, 0, 0, 0, 20, 0, 0, 0, 0, 0, 0, 1, 0, 1, 2, 9, 0, 175, 0, 2625, 2240, 945, 0, 0

Offset: 0

Alois P. Heinz, Aug 18 2015

			Square array A(n,k) begins:
  1, 1,   1,  1,    1,  1,    1,   1,    1, ...
  0, 0,   0,  0,    0,  0,    0,   0,    0, ...
  0, 0,   1,  0,    1,  0,    1,   0,    1, ...
  0, 0,   0,  2,    0,  0,    2,   0,    0, ...
  0, 0,   3,  0,    9,  0,    3,   0,    9, ...
  0, 0,   0,  0,    0, 24,   20,   0,    0, ...
  0, 0,  15, 40,  105,  0,  175,   0,  105, ...
  0, 0,   0,  0,    0,  0,  210, 720,    0, ...
  0, 0, 105,  0, 2625,  0, 4585,   0, 7665, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0+1,2-10 give: A000007, A001147, A052502, A052503, A052504, A261317, A261427, A261428, A261429, A261381.
Main diagonal gives A261431.
Cf. A008307.

with(numtheory):
A:= proc(n, k) option remember; `if`(n<0, 0, `if`(n=0, 1,
      add(mul(n-i, i=1..j-1)*A(n-j, k), j=divisors(k) minus {1})))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..14);

A[0, 0] = A[0, 1] = 1; A[, 0|1] = 0; A[n, k_] := A[n, k] = If[n < 0, 0, If[n == 0, 1, Sum[Product[n - i, {i, 1, j - 1}]*A[n - j, k], {j, Rest @ Divisors[k]}]]]; Table[A[n, d - n], {d, 0, 14}, {n, 0, d}] // Flatten (* Jean-François Alcover, Jan 21 2017, after Alois P. Heinz *)

E.g.f. of column k: exp(Sum_{d|k, d>1} x^d/d).

A060706 For n >= 1 a(n) is the size of the conjugacy class in the symmetric group S_(4n) consisting of permutations whose cycle decomposition is a product of n disjoint 4-cycles.

Original entry on oeis.org

1, 6, 1260, 1247400, 3405402000, 19799007228000, 210384250804728000, 3692243601622976400000, 99579809935771673508000000, 3910499136177753618659160000000, 214428309633170941925556379440000000

Offset: 0

Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001

nonn

a(n) is the number of ways to seat 4n bridge players at n circular tables with four players at each table. - Geoffrey Critzer, Dec 17 2011

Harry J. Smith, Table of n, a(n) for n = 0..100
Bishal Deb and Alan D. Sokal, Higher-order Stirling cycle and subset triangles: Total positivity, continued fractions and real-rootedness, arXiv:2507.18959 [math.CO], 2025. See p. 7.

Cf. A000142. A001147, A052502, A052504.

for n from 0 to 20 do printf(`%d,`,(4*n)! / (n! * 4^n)) od:

nn = 40; a = x^4/4;f[list_] := Select[list, # > 0 &];
f[Range[0, nn]! CoefficientList[Series[Exp[a], {x, 0, nn}], x]]  (* Geoffrey Critzer, Dec 17 2011 *)

{ for (n=0, 100, write("b060706.txt", n, " ", (4*n)! / (n! * 4^n)); ) } \\ Harry J. Smith, Jul 09 2009

a(n) = (4n)! / (n! * 4^n). Recursion: a(0) = 1, a(1) = 6, for n >= 2 a(n) = a(n-1) * C(4n - 1, 3)* 6 = a(n-1)*(4n-1)*(4n-2)*(4n-3). Using Stirling's formula in A000142 we have a(n) ~ 2 * 64^n * (n/e)^(3n).
E.g.f.: exp(x^4/4). - Geoffrey Critzer, Dec 17 2011
Write the generating function for this sequence in the form A(x) = sum_{n>=0} a(n)* x^(3*n+1)/(3*n+1)!. Then A'(x)*( 1 - A(x)^3) = 1, consequently A(x) is a root of z^4 - 4*z + 4*x with A(0) = 0. Cf. A052502. - Peter Bala, Jan 02 2015

More terms from James Sellers, Apr 23 2001

A091752 Generalized Stirling2 array (-1,2)S2. Irregular triangle a(n, m) for n >= 1 and 2 <= m <= 2*n.

Original entry on oeis.org

1, 2, -2, 1, 40, -40, 20, -6, 1, 2240, -2240, 1120, -360, 80, -12, 1, 246400, -246400, 123200, -40320, 9520, -1680, 220, -20, 1, 44844800, -44844800, 22422400, -7392000, 1786400, -332640, 48720, -5600, 490, -30, 1, 12197785600, -12197785600, 6098892800, -2018016000, 493292800

Offset: 1

Wolfdieter Lang, Feb 27 2004

This a(n,m) array appears in the normal ordering formula ((1/x)*(d_x)^2)^n = Sum_{m=2..2*n} a(n,m)*x^(m-3*n)*(d_x)^m, n >= 1, with the derivative operator d_x := d/dx.
This is an extension of the generalized Stirling2 arrays S_{r,s}(n,k) (here k=m) considered for nonnegative r and s in the Blasiak et al. reference given in A078740. See also the Schork reference given there.
The sequence of row lengths for this array is [1,3,5,7,9,11,...] = A005408(n-1), n >= 1.
These generalized Stirling2 arrays have been treated in Carlitz's paper (with r = lambda + mu, s = mu), and a recurrence is given eq. (4). See the formula section for the present case mu = 2, lambda = -3, and Carlitz's a_{n,s-1} = a(n, s) (here s = m). - Wolfdieter Lang, Dec 16 2019
From Wolfdieter Lang and Werner Schulte, Jan 29 2020: (Start)
For the case of the (irregular) triangles (-1,s)S2, for s >= 1, W. Schulte conjectured that a(s; n, m) = T(s; (s+1)*n - m - 1, m - s), for n > = 1 and m = s, s+1, ..., s*n, with the row polynomials RT(s; n, x) = Sum_{m = 0..s*n} T(s; n, m)*x^m of the triangle T(s), n >= 0, defined by the Rodrigues-type formula exp(x^(s+1)/(s+1)) (d/dx)^n exp(-x^(s+1)/(s+1)) = (-1)^n*RT(s; n, x). Thus the rows of (-1,s)S2 are every (s+1)-th upwards antidiagonals of T(s), but with offset m = s instead of m = 0.
The proof of the conjecture follows by showing the Carlitz recurrence for (-1,s)S2, but with offset n = 0 and m = 0; that is, a(s; n+1, m+s) =: aHat(s; n, m) = Sum_{j = 0..s} binomial(s, j)*fallfac(s + m - j - (s+1)*n , s-j)*aHat(s; n-1, m-j), for n >= 0, m = 0, 1, ..., s*n, with aHat(s; 0, 0) = 1 and aHat(s; n, m) = 0 for m < 0 and m > s*n. The falling factorials are fallfac(x, n). With aHat(s; n, m) = T(s; (s+1)*n - m, m) this leads to a recurrence for T(s) which is equivalent to the recurrence for the row polynomials RT(s), namely RT(s; n, x) = Sum_{j=0 .. s} binomial(s, j)*x^j*fallfac(s - j - n, s - j)*RT(s; n - (s+1) + j, x), for n >= 1, and RT(s; 0, x) = 1. This, in turn, can be proved by induction over n >= 1 from the simpler recurrence for RT(s) obtained directly from the Rodrigues-type definition, namely RT(s; n, x) = x^s*RT(s; n-1, x) - (d/dx)RT(s; n-1, x), n >= 1,with RT(s; 0, x) = 1.
The e.g.f. of the triangle T(s), that is of the row polynomials {RT(s;n, x)}_{n>=0}, is E(s; t, x) = exp((x^(s+1) - (x - t)^(s+1))/(s+1)). This can be proved from the simple RT(s) recurrence, leading to (d/dt + d/dx)E(s; t, x) = x^s*E(s; t, x), with E(s; 0, x) = 1. After using E(s; t, x) = 1*exp(x^(s+1)/(s+1) + f(s; t, x)), with f(s; 0, x) = -x^(s+1)/(s+1), this becomes (d/dx - d/dt)f(s; t, x) = 0 meaning that f is a function of y = x - t, say, g(s; y) = -y^(s+1)/(s+1) because it has to become f(s; 0, x) for t = 0.
The explicit form for (-1,s)S2 is a(s; n, m) = (-1)^(n*s -m)*((s+1)*n - m-1)!/((s+1)^(n-1)*(n-1)!)*Sum_{j=0..floor((m-s)/(s+1))} (-1)^j* binomial(n-1, j)*binomial((s+1)*(n-1-j), m - s - (s+1)*j). One can prove the corresponding formula for T(s; n, m) by showing that it satisfies the T(s) recurrence T(s; n, m) = T(s; n-1, m-l) + (m+1)*T(s; n-1, m+1), for n >= 1, with T(s; 0, 0) = 1, and 0 for m < 0 or m > s*n.
The present entry is the instance s = 2, with the formulas given below. (End)

			Triangle starts:
{1},
{2, -2, 1},
{40, -40, 20, -6, 1},
{2240, -2240, 1120, -360, 80, -12, 1},
{246400, -246400, 123200, -40320, 9520, -1680, 220, -20, 1}.

Leonard Carlitz, On Arrays of Numbers, Am. J. Math., 54,4 (1932) 739-752.
Wolfdieter Lang, First 6 rows.

Column sequences for m=2..8 and n >= ceiling(m/2) are A052502(n-1), -A052502(n-1), A091535(n-1), -6*A091753(n), A091754(n), -12*A091755(n), A091756(n).

Cf. A078740, A331816.

w[n_, k_] := (Gamma[n-k/3] Gamma[1/3+n-k/3])/(Gamma[1/3-k/3] Gamma[-k/3]);
T[n_, k_] := 9^n Sum[(-1)^(k - j)  w[n, j]/((k - j)! j!), {j, 1, k}];
Table[Round[T[n,k]], {n,1,6}, {k, 2, 2 n}] (* Peter Luschny, Feb 05 2020 *)

a(n, m) = (((-1)^m)/m!)*Sum_{p=2..m} (-1)^p*binomial(m, p)*Product_{j=1..n} fallfac(p-3*(j-1), 2), n >= 1, 2 <= m <= 2*n, otherwise 0. From eq. (12) of the Blasiak et al. reference (see A078740) with r=-1, s=2, k=m.
Recurrence: a(n, m) = Sum_{p=0..2} binomial(2, p)*fallfac(-3*(n-1)+m-p, 2-p)*a(n-1, m-p), n >= 2, 2 <= m <= 2*n, a(1, 2) = 1, otherwise 0. Rewritten from eq. (19) of the Schork reference (see A078740) with r = -1, s = 2. fallfac(n, m) := A008279(n, m) (falling factorials triangle).
Recurrence (Carlitz): a(n m) = a(n-1, m-2) - 2*(3*n - (m +2))*a(n-1, m-1) + (3*n - (m + 3))*(3*n - (m + 2))*a(n-1, m), for n >= 2, m >= 1, and a(n, m) = 0 if m <= 1 or m > 2*n, and a(1, 2) = 1. - Wolfdieter Lang, Dec 16 2019
From Werner Schulte, Jan 29 2020: (Start)
a(n, m) = T(3*n - m - 1, m - 2), for n > = 1 and m = 2, 3, ..., 2*n, with the irregular triangle defined by (-1)^n*exp(x^3/3)*(d/dx)^n exp(-x^3/3) = RT(n, x) = Sum_{k=0..2*n} T(n, k)*x^k, for n >= 0. For T(n, k) see A331816.
The recurrence RT(n, x) = x^2*RT(n-1, x) - (d/dx)RT(n-1, x), n >= 1, with RT(0, x) = 1, implies the T recurrence T(n, k) = T(n-1, k-2) - (k+1)*T(n-1, k+1), for n >= 1, with T(0, 0) = 1, and T(n, m) = 0 for m < 0 and m > 2*n. Also, by induction over n: RT(n, x) = x^2*RT(n-1, x) - 2*(n-1)*x*RT(n-2, x) + (n-1)*(n-2)*RT(n-3, x), using the former recurrence and inserting the derivatives. This translates to an obvious further recurrence for the irregular triangle T. It is used in order to prove the Carlitz recurrence for the index shifted aHat(n, m) = a(n+1, m + 2).
The e.g.f. of the irregular triangle, that is of the row polynomials RT, is E(t, x) = exp((x^3 - (x - t)^3)/3). See the comment above for a proof (setting s=2 there).
The explicit form is a(n, m) = (-1)^m*(3*n - m - 1)!/(3^(n-1)*(n-1)!)*Sum_{j=0..floor((m-2)/3)} (-1)^j*binomial(n-1, j)*binomial(3*(n-1-j), m -2 - 3*j), for n >= 1, and 2 <= m <= 2*n.
(End)
T(n, k) = 9^n*Sum_{j=1..k} (-1)^(k-j)*w(n,j)/((k-j)!*j!) where w(n,k) = (Gamma(n-k/3)*Gamma((1-k)/3+n))/(Gamma((1-k)/3)*Gamma(-k/3)). - Peter Luschny, Feb 05 2020

A052504 Number of permutations sigma of [5n] without fixed points such that sigma^5 = Id.

Original entry on oeis.org

1, 24, 72576, 1743565824, 162193467211776, 41363226782215962624, 23578031983305871878782976, 26242915470187034742010543079424, 51804144968120491069562620291816882176, 168779147605615794796420686413626405734580224, 858246016274098851318874304509764200194078068965376

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_(5n) consisting of permutations whose cycle decomposition is a product of n disjoint 5-cycles.

G. C. Greubel, Table of n, a(n) for n = 0..100
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 29

Cf. A001147, A052502, A060706, A261317, A261381.

Quintisection of column k=5 of A261430.

List([0..15], n-> Factorial(5*n)/(5^n*Factorial(n))) # G. C. Greubel, May 14 2019

[Factorial(5*n)/(5^n*Factorial(n)): n in [0..15]]; // G. C. Greubel, May 14 2019

spec := [S,{S=Set(Union(Cycle(Z,card=5)))},labeled]: seq(combstruct[count](spec,size=n), n=0..20);

nn = 50; Select[Range[0, nn]! CoefficientList[Series[Exp[x^5/5], {x, 0, nn}], x], # > 0 &]  (* Geoffrey Critzer, Aug 19 2012 *)

{a(n) = (5*n)!/(5^n*n!)}; \\ G. C. Greubel, May 14 2019

[factorial(5*n)/(5^n*factorial(n)) for n in (0..15)] # G. C. Greubel, May 14 2019

a(n) = (5n)! * [x^(5n)] exp(x^5/5).
From Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001: (Start)
a(n) = (5*n)! / (n! * 5^n).
a(0) = 1, a(1) = 24, for n >= 2 a(n) = a(n-1) * C(5*n - 1, 4)* 24 = (5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*a(n-1).
a(n) ~ sqrt(5) * 625^n * (n/e)^(4n). (End)
Write the generating function for this sequence in the form A(x) = Sum_{n >= 0} a(n)* x^(4*n+1)/(4*n+1)!. Then A'(x)*( 1 - A(x)^4) = 1. Cf. A052502. - Peter Bala, Jan 02 2015

A157702 G.f.s of the z^p coefficients of the polynomials in the GF1 denominators of A156921.

Original entry on oeis.org

1, 1, 1, 7, 26, 7, 3, 166, 951, 951, 166, 3, 263, 8999, 59637, 108602, 59637, 8999, 263, 174, 33124, 848555, 6062651, 15477896, 15477896, 6062651, 848555, 33124, 174, 45, 66963, 5856626, 122966782, 920090513

Offset: 0

Johannes W. Meijer, Mar 07 2009

The formula for the PDGF1(z;n) polynomials in the GF1 denominators of A156921 can be found below.
The general structure of the GFKT1(z;p) that generate the z^p coefficients of the PDGF1(z; n) polynomials can also be found below. The KT1(z;p) polynomials in the numerators of the GFKT1(z;p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT1(z;p) polynomials is: 1, 2, 3, 6, 7, 10, 13, 14, 17, 20, 23, 24, 27, 30, 33, 36, 37, 40. The first differences follow a simple pattern. The positions of the 1's follow the Lazy Caterer's sequence A000124.
A Maple algorithm that generates relevant GFKT1(z;p) information can be found below.

			Some PDGF1 (z;n) are:
  PDGF1(z;n=3) = (1-5*z)*(1-3*z)^2*(1-z)^3
  PDGF1(z;n=4) = ((1-7*z)*(1-5*z)^2*(1-3*z)^3*(1-z)^4)
The first few GFKT1's are:
  GFKT1(z;p=0) = 1/(1-z)
  GFKT1(z;p=1) = -z*(1+z)/(1-z)^4
  GFKT1(z;p=2) = z^2*(7+26*z+7*z^2)/(1-z)^7
Some KT1(z;p) polynomials are:
  KT1(z;p=2) = 7+26*z+7*z^2
  KT1(z;p=3) = 3+166*z+951*z^2+951*z^3+166*z^4+3*z^5
  KT1(z;p=4) = 263+8999*z+59637*z^2+108602*z^3+59637*z^4+8999*z^5+263*z^6

Originator sequence A156921.
See A000330 for the z^1 coefficients and A157706 for the z^2 coefficients.
Row sums equal A052502.
Cf. A157703, A157704, A157705.

p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) = fn, a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-(2*m-1)*z)^(n2+1-m),m=1..n2); a(n2):= coeff(fz(n2),z,p); end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)=sort(simplify(fk)); GFKT1(p):=(sum(fk*z^k,k=0..infinity)); q1:=ldegree((numer(GFKT1(p)))): KT1(p):=sort((-1)^p*simplify((GFKT1(p))*(1-z)^(3*p+1)/z^q1),z, ascending);

PDGF1(z;n) = Product_{m=1..n} (1-(2*m-1)*z)^(n+1-m) with n = 1, 2, 3, ... .
GFKT1(z;p) = (-1)^(p)*(z^q1)*KT1(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ... .
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .

A261317 Number of permutations sigma of [n] without fixed points such that sigma^6 = Id.

Original entry on oeis.org

1, 0, 1, 2, 3, 20, 175, 210, 4585, 24920, 101745, 1266650, 13562395, 48588540, 1082015935, 9135376250, 63098660625, 1069777108400, 13628391601825, 88520971388850, 2134604966569075, 23945393042070500, 236084869688242575, 4893567386193135650, 72576130763294383225

Offset: 0

Alois P. Heinz, Aug 14 2015

nonn

			a(4) = 3: 2143, 3412, 4321.
a(5) = 20: 21453, 21534, 23154, 24513, 25431, 31254, 34152, 34521, 35124, 35412, 41523, 43251, 43512, 45132, 45213, 51432, 53214, 53421, 54123, 54231.

Alois P. Heinz, Table of n, a(n) for n = 0..500

Column k=6 of A261430.

Cf. A001147, A052502, A052503, A052504, A053496, A261381.

a:= proc(n) option remember; `if`(n<0, 0, `if`(n=0, 1,
       add(mul(n-i, i=1..j-1)*a(n-j), j=[2, 3, 6])))
    end:
seq(a(n), n=0..30);

a[n_] := a[n] = If[n < 0, 0, If[n == 0, 1, Sum[Product[n - i, {i, 1, j - 1}]*a[n - j], {j, {2, 3, 6}}]]];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 10 2018, from Maple *)

E.g.f.: exp(x^2*(x^4+2*x+3)/6).

D-finite with recurrence a(n) +(-n+1)*a(n-2) -(n-1)*(n-2)*a(n-3) -(n-5)*(n-1)*(n-2)*(n-3)*(n-4)*a(n-6)=0. - R. J. Mathar, Jul 04 2023

A193535 Decimal expansion of log(2)/3.

Original entry on oeis.org

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

Offset: 0

Alonso del Arte, Jul 29 2011

This number is involved as an addend or subtrahend in the closed forms of certain series of reciprocals of integers (see for example A113476).

			0.231049060186648...

L. B. W. Jolley, Summation of Series, Dover (1961).
Murray R. Spiegel, Seymour Lipschutz, John Liu. Mathematical Handbook of Formulas and Tables, 3rd Ed. Schaum's Outline Series. New York: McGraw-Hill (2009): p. 135, equations 21.16 and 21.18.

Index entries for transcendental numbers

Cf. A052502, A113476.

Mathematica
```
RealDigits[(Log[2]/3), 10, 100][[1]]
```

log(2)/3 \\ Charles R Greathouse IV, Jul 29 2011

Equals lim_{n->oo} [Sum_{i = 1..n} i^2/(n^3 + i^3)]. [Jolley eq 292, p.52]
Equals Sum_{n>=1} (-1)^(n-1)/(n*2^n*binomial(2*n, n)). - Arkadiusz Wesolowski, Jan 20 2013
From Amiram Eldar, Aug 05 2020: (Start)
Equals Integral_{x=1..oo} 1/(x^4 + x) dx.
Equals Integral_{x=0..oo} 1/(exp(2*x) + 3) dx. (End)
From Peter Bala, Feb 27 2024: (Start)
Equals (1/2)*Sum_{k >= 0} (-1)^k/((3*k + 1)*(3*k + 2)) = (1/2)*(1/(2 + (1*2)^2/(18 + (4*5)^2/(2*18 + (7*8)^2/(3*18 + (10*11)^2/(4*18 +  ... )))))) (continued fraction). See A052502.
Equals 7/32 + (3/2)*Sum_{k >= 0} (-1)^k/((3*k + 1)*(3*k + 2)*(3*k + 3)*(3*k + 4)*(3*k + 5)). (End)

A261381 Number of permutations sigma of [n] without fixed points such that sigma^10 = Id.

Original entry on oeis.org

1, 0, 1, 0, 3, 24, 15, 504, 105, 9072, 436401, 166320, 28750491, 3243240, 1307809503, 27965161224, 52309001745, 3795543015264, 2000776242465, 324424646818272, 17268536366932851, 22708075360010040, 3974396337125445231, 1436250980764880280, 548178165969608527353

Offset: 0

Alois P. Heinz, Aug 17 2015

nonn

			a(4) = 3: 2143, 3412, 4321:
a(5) = 24: 23451, 23514, 24153, 24531, 25134, 25413, 31452, 31524, 34251, 34512, 35214, 35421, 41253, 41532, 43152, 43521, 45123, 45231, 51234, 51423, 53124, 53412, 54132, 54213.
a(6) = 15: 214365, 215634, 216543, 341265, 351624, 361542, 432165, 456123, 465132, 532614, 546213, 564312, 632541, 645231, 654321.

Alois P. Heinz, Table of n, a(n) for n = 0..490

Column k=10 of A261430.

Cf. A001147, A052502, A052503, A052504, A053496, A261317.

a:= proc(n) option remember; `if`(n<0, 0, `if`(n=0, 1,
       add(mul(n-i, i=1..j-1)*a(n-j), j=[2, 5, 10])))
    end:
seq(a(n), n=0..30);

E.g.f.: exp(x^2/2+x^5/5+x^10/10).

cp's OEIS Frontend

A060079 Duplicate of A052502.

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A261430 Number A(n,k) of permutations p of [n] without fixed points such that p^k = Id; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A060706 For n >= 1 a(n) is the size of the conjugacy class in the symmetric group S_(4n) consisting of permutations whose cycle decomposition is a product of n disjoint 4-cycles.

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A091752 Generalized Stirling2 array (-1,2)S2. Irregular triangle a(n, m) for n >= 1 and 2 <= m <= 2*n.

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A052504 Number of permutations sigma of [5n] without fixed points such that sigma^5 = Id.

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A157702 G.f.s of the z^p coefficients of the polynomials in the GF1 denominators of A156921.

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A261317 Number of permutations sigma of [n] without fixed points such that sigma^6 = Id.

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