A167568 -id:A167568 - cp's OEIS frontend

A005990 a(n) = (n-1)*(n+1)!/6.

0, 1, 8, 60, 480, 4200, 40320, 423360, 4838400, 59875200, 798336000, 11416204800, 174356582400, 2833294464000, 48819843072000, 889218570240000, 17072996548608000, 344661117825024000, 7298706024529920000, 161787983543746560000

Offset: 1

N. J. A. Sloane

Coefficients of Gandhi polynomials.
a(n) = Sum_{pi in Symm(n)} Sum_{i=1..n} max(pi(i)-i,0), i.e., the total positive displacement of all letters in all permutations on n letters. - Franklin T. Adams-Watters, Oct 25 2006
a(n) is also the sum of the excedances of all permutations of [n]. An excedance of a permutation p of [n] is an i (1 <= i <= n-1) such that p(i) > i. Proof: i is an excedance if p(i) = i+1, i+2, ..., n (n-i possibilities), with the remaining values of p forming any permutation of [n]\{p(i)} in the positions [n]\{i} ((n-1)! possibilities). Summation of i(n-i)(n-1)! over i from 1 to n-1 completes the proof. Example: a(3)=8 because the permutations 123, 132, 213, 231, 312, 321 have excedances NONE, {2}, {1}, {1,2}, {1}, {1}, respectively. - Emeric Deutsch, Oct 26 2008
a(n) is also the number of doubledescents in all permutations of {1,2,...,n-1}. We say that i is a doubledescent of a permutation p if p(i) > p(i+1) > p(i+2). Example: a(3)=8 because each of the permutations 1432, 4312, 4213, 2431, 3214, 3421 has one doubledescent, the permutation 4321 has two doubledescents and the remaining 17 permutations of {1,2,3,4} have no doubledescents. - Emeric Deutsch, Jul 26 2009
Equals the second right hand column of A167568 divided by 2. - Johannes W. Meijer, Nov 12 2009
Half of sum of abs(p(i+1) - p(i)) over all permutations on n, e.g., 42531 = 2 + 3 + 2 + 2 = 9, and the total over all permutations on {1,2,3,4,5} is 960. - Jon Perry, May 24 2013
a(n) gives the number of non-occupied corners in tree-like tableaux of size n+1 (see Gao et al. link). - Michel Marcus, Nov 18 2015
a(n) is the number of sequences of n+2 balls colored with at most n colors such that exactly three balls are the same color as some other ball in the sequence. - Jeremy Dover, Sep 26 2017
a(n) is the number of triangles (3-cycles) in the (n+1)-alternating group graph. - Eric W. Weisstein, Jun 09 2019

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..300
D. Dumont, Interpretations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318. (Annotated scanned copy)
Alice L. L. Gao, Emily X. L. Gao, and Brian Y. Sun, Zubieta's Conjecture on the Enumeration of Corners in Tree-like Tableaux, arXiv:1511.05434 [math.CO], 2015. The second version of this paper has a different title and different authors: A. L. L. Gao, E. X. L. Gao, P. Laborde-Zubieta, and B. Y. Sun, Enumeration of Corners in Tree-like Tableaux and a Conjectural (a,b)-analogue, arXiv preprint arXiv:1511.05434v2, 2015.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Eric Weisstein's World of Mathematics, Alternating Group Graph.
Eric Weisstein's World of Mathematics, Graph Cycle.

Cf. A001715, A090672, A167568.

Cf. A001113, A001620, A091725, A099285.

[(n-1)*Factorial(n+1)/6: n in [1..25]]; // Vincenzo Librandi, Oct 11 2011

[ seq((n-1)*(n+1)!/6,n=1..40) ];
a:=n->sum(sum(sum(n!/6, j=1..n),k=-1..n),m=0..n): seq(a(n), n=0..19); # Zerinvary Lajos, May 11 2007
seq(sum(mul(j,j=3..n), k=3..n)/3, n=2..21); # Zerinvary Lajos, Jun 01 2007
restart: G(x):=x^3/(1-x)^2: f[0]:=G(x): for n from 1 to 21 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n]/3!,n=2..21); # Zerinvary Lajos, Apr 01 2009

Table[Sum[n!/6, {i, 3, n}], {n, 2, 21}] (* Zerinvary Lajos, Jul 12 2009 *)
Table[(n - 1) (n + 1)!/6, {n, 20}] (* Harvey P. Dale, Apr 07 2019 *)
Table[(n - 1) Pochhammer[4, n - 2], {n, 20}] (* Eric W. Weisstein, Jun 09 2019 *)
Table[(n - 1) Gamma[n + 2]/6, {n, 20}] (* Eric W. Weisstein, Jun 09 2019 *)
Range[0, 20]! CoefficientList[Series[x/(1 - x)^4, {x, 0, 20}], x] (* Eric W. Weisstein, Jun 09 2019 *)

a(n)=(n-1)*(n+1)!/6 \\ Charles R Greathouse IV, May 24 2013

a(n) = A090672(n)/2.
a(n) = A052571(n+2)/6. - Zerinvary Lajos, May 11 2007
a(n) = Sum_{m=0..n} Sum_{k=-1..n} Sum_{j=1..n} n!/6, n >= 0. - Zerinvary Lajos, May 11 2007
If we define f(n,i,x) = Sum_{k=i..n} (Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)*Stirling2(j,i)*x^(k-j)) then a(n+1) = (-1)^(n-1)*f(n,1,-4), (n >= 1). - Milan Janjic, Mar 01 2009
E.g.f.: (-1+3*x)/(3!*(1-x)^3), a(0) = -1/3!. Such e.g.f. computations resulted from e-mail exchange with Gary Detlefs. - Wolfdieter Lang, May 27 2010
a(n) = ((n+3)!/2) * Sum_{j=i..k} (k+1)!/(k+3)!, with offset 0. - Gary Detlefs, Aug 05 2010
a(n) = (n+2)!*Sum_{k=1..n-1} 1/((2*k+4)*(k+3)). - Gary Detlefs, Oct 09 2011
a(n) = (n+2)!*(1 + 3*(H(n+1) - H(n+2)))/6, where H(n) is the n-th harmonic number. - Gary Detlefs, Oct 09 2011
With offset = 0, e.g.f.: x/(1-x)^4. - Geoffrey Critzer, Aug 30 2013
From Amiram Eldar, May 06 2022: (Start)
Sum_{n>=2} 1/a(n) = 3*(Ei(1) - gamma) - 6*e + 27/2, where Ei(1) = A091725, gamma = A001620, and e = A001113.
Sum_{n>=2} (-1)^n/a(n) = 3*(gamma - Ei(-1)) - 3/2, where Ei(-1) = -A099285. (End)

Better definition from Robert Newstedt

A005359 a(n) = n! if n is even, otherwise 0 (from Taylor series for cos x).

Original entry on oeis.org

1, 0, 2, 0, 24, 0, 720, 0, 40320, 0, 3628800, 0, 479001600, 0, 87178291200, 0, 20922789888000, 0, 6402373705728000, 0, 2432902008176640000, 0, 1124000727777607680000, 0, 620448401733239439360000, 0

Offset: 0

N. J. A. Sloane, Russ Cox

nonn

Normally sequences like this are not included, since with the alternating 0's deleted it is already in the database.
Stirling transform of a(n)=[0,2,0,24,0,720,...] is A052841(n)=[0,2,6,38,270,...]. - Michael Somos, Mar 04 2004
Stirling transform of a(n-1)=[1,0,2,0,24,0,...] is A000670(n-1)=[1,1,3,13,75,...]. - Michael Somos, Mar 04 2004
Stirling transform of a(n-1)=[0,0,2,0,24,0,...] is A052875(n-1)=[0,0,2,12,74,...]. - Michael Somos, Mar 04 2004
Stirling transform of (-1)^n*A052811(n)=[0,2,-6,46,-340,...] is a(n)=[0,2,0,24,0,...]. - Michael Somos, Mar 04 2004
Also n-th derivative of arctanh(x) at x=0. - Michel Lagneau, Aug 13 2012
Binomial convolution square of A177145 (with offset 0) because each permutation in S_{2n} uniquely determines a bi-partition of its elements into even and odd cycles and these are both enumerated by A177145. - Michael Somos, Mar 19 2019

Douglas Hofstadter, "Fluid Concepts and Creative Analogies: Computer Models of the Fundamental Mechanisms of Thought".

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Michael Somos, Number of permutations with all cycles of even length, answer to Mathematics Stack Exchange question 3152701, Mar 18 2019.
Index entries for sequences related to factorial numbers

From Johannes W. Meijer, Nov 12 2009: (Start)
Equals the first right hand column of A167565.
Equals the first left hand column of A167568.
(End)
Cf. A177145.
Bisection (even part) gives A010050.

BB:={E=Prod(Z,Z),S=Union(Epsilon,Prod(S,E))}: ZL:=[S,BB, labeled]: > seq(count(ZL,size=n),n=0..25); # Zerinvary Lajos, Apr 22 2007
a:=n->n!+(-1)^n*n!: seq(a(n)/2, n=0..25); # Zerinvary Lajos, Mar 25 2008

Riffle[Range[0,30,2]!,0] (* Harvey P. Dale, Nov 16 2011 *)
a[ n_] := If[n >= 0 && EvenQ[n], n!, 0]; (* Michael Somos, Mar 19 2019 *)

{a(n) = if(n<0, 0, if(n%2, 0, n!))}; /* Michael Somos, Mar 04 2004 */

E.g.f. 1/(1-x^2) = d/dx log(sqrt((1+x)/(1-x))). a(2n)=(2n)!, a(2n+1)=0. - Michael Somos, Mar 04 2004
a(n) = Product_{k=0..n/2-1} binomial(n-2k,2)*2^(n/2) for even n. - Geoffrey Critzer, Jun 05 2016
From Ilya Gutkovskiy, Jun 05 2016: (Start)
D-finite with recurrence a(n) = n*(n - 1)*a(n-2), a(0)=1, a(1)=0.
a(n) = n!*((-1)^n + 1)/2. (End)

A167560 The ED2 array read by ascending antidiagonals.

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 24, 16, 6, 1, 120, 80, 32, 8, 1, 720, 480, 192, 54, 10, 1, 5040, 3360, 1344, 384, 82, 12, 1, 40320, 26880, 10752, 3072, 680, 116, 14, 1, 362880, 241920, 96768, 27648, 6144, 1104, 156, 16, 1

Offset: 1

Johannes W. Meijer, Nov 10 2009

The coefficients in the upper right triangle of the ED2 array (m>n) were found with the a(n,m) formula while the coefficients in the lower left triangle of the ED2 array (m<=n) were found with the recurrence relation, see below. We use for the array rows the letter n (>=1) and for the array columns the letter m (>=1).
The ED2 array is related to the EG1 matrix, see A162005, because sum(EG1(2*m-1,n) * z^(2*m-1), m=1..infinity) = ((2*n-1)!/(4^(n-1)*(n-1)!^2))*int(sinh(y*(2*z))/cosh(y)^(2*n), y=0..infinity).
For the ED1, ED3 and ED4 arrays see A167546, A167572 and A167584.

			The ED2 array begins with:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1
2, 4, 6, 8, 10, 12, 14, 16, 18, 20
6, 16, 32, 54, 82, 116, 156, 202, 254, 312
24, 80, 192, 384, 680, 1104, 1680, 2432, 3384, 4560
120, 480, 1344, 3072, 6144, 11160, 18840, 30024, 45672, 66864
720, 3360, 10752, 27648, 61440, 122880, 226800, 392832, 646128, 1018080

Johannes W. Meijer, The four Escher-Droste arrays, Mar 08 2013.

A000012, A005843 (n>=1), 2*A104249 (n>=1), A167561, A167562 and A167563 equal the first sixth rows of the array.
A000142 equals the first column of the array.
A047053 equals the a(n, n) diagonal of the array.
2*A034177 equals the a(n+1, n) diagonal of the array.
A167570 equals the a(n+2, n) diagonal of the array,
A167564 equals the row sums of the ED2 array read by antidiagonals.
A167565 is a triangle related to the a(n) formulas of the rows of the ED2 array.
A167568 is a triangle related to the GF(z) formulas of the rows of the ED2 array.
A167569 is the lower left triangle of the ED2 array.
Cf. A162005 (EG1 triangle).
Cf. A167546 (ED1 array), A167572 (ED3 array), A167584 (ED4 array).

nmax:=10; mmax:=10; for n from 1 to nmax do for m from 1 to n do a(n,m) := 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! od; for m from n+1 to mmax do a(n,m):= n! + sum((-1)^(k-1)*binomial(n-1,k)*a(n,m-k),k=1..n-1) od; od: for n from 1 to nmax do for m from 1 to n do d(n,m):=a(n-m+1,m) od: od: T:=1: for n from 1 to nmax do for m from 1 to n do a(T):= d(n,m): T:=T+1: od: od: seq(a(n),n=1..T-1);
# alternative
A167560 := proc(n,m)
    option remember ;
    if m > n then
        n!+add( (-1)^(k-1)*binomial(n-1,k)*procname(n,m-k),k=1..n-1) ;
    else
        4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! ;
    end if;
end proc:
seq( seq(A167560(d-m,m),m=1..d-1),d=2..12) ; # R. J. Mathar, Jun 28 2024

nmax = 10; mmax = 10; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[n, m] = 4^(m - 1)*(m - 1)!*((n + m - 1)!/(2*m - 1)!)]; For[m = n + 1, m <= mmax, m++, a[n, m] = n! + Sum[(-1)^(k - 1)*Binomial[n - 1, k]*a[n, m - k], {k, 1, n - 1}]]; ]; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, d[n, m] = a[n - m + 1, m]]; ]; t = 1; For[n = 1, n <= nmax, n++, For[m = 1, m <= n, m++, a[t] = d[n, m]; t = t + 1]]; Table[a[n], {n, 1, t - 1}] (* Jean-François Alcover, Dec 20 2011, translated from Maple *)

a(n,m) = ((m-1)!/((m-n-1)!))*int(sinh(y*(2*n))/(cosh(y))^(2*m),y=0..infinity) for m>n.
The (n-1)-differences of the n-th array row lead to the recurrence relation
sum((-1)^k*binomial(n-1,k)*a(n-1,m-k),k=0..n-1) = n!
which in its turn leads to, see A167569,
a(n,m) = 4^(m-1)*(m-1)!*(n+m-1)!/(2*m-1)! if m<=n.

cp's OEIS Frontend

A005990 a(n) = (n-1)*(n+1)!/6.

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A005359 a(n) = n! if n is even, otherwise 0 (from Taylor series for cos x).

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A167560 The ED2 array read by ascending antidiagonals.

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