

A067370


The weight of the periphery of the alternating group, denoted v(P_N).


3



0, 1, 3, 24, 160, 1290, 11046, 106848, 1117152, 12849840, 159089040, 2132602560, 30554297280, 468754715520, 7634862748800, 132058767052800, 2410986506342400, 46443330717235200, 939668036761036800, 19955747250238464000, 443271664862659584000, 10290986066890045440000
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OFFSET

1,3


COMMENTS

Sequences A067369, A067370 and A067318 are related. A067318 counts transpositions in the symmetric group, denoted S_n. One can think of the transpositions in S_n as being split between the alternating group A_n and its complement, which we call the periphery and denote P_N. For n >= 3, A067369 v(P_N) and A067370 v(A_n) always differ by (n2)!. When n is odd, v(A_n) is larger; when n is even, v(P_N) is larger. This gives new meaning to the name alternating group. The average weight of a permutation in A_n converges with the average weight for a permutation in P_N at infinity.


LINKS

Charlie Neder and Muniru A Asiru, Table of n, a(n) for n = 1..446


FORMULA

v(P_N) = p(n) = p(n1) + floor((n1)!/2)*(vbar(A_n1)+1)*((n1)) where vbar(A_n) is the average weight of a permutation in A_n, the alternating group. vbar(A_n1) is a(n1)/(n1)!/2 where a(n) is from the sequence A067369.
From Vladeta Jovovic, Feb 02 2003: (Start)
a(n) = (1/2)*((1)^n*(n2)! + n*n!  abs(Stirling1(n+1, 2))), n > 1.
E.g.f.: (1/2)*((1+x)*log(1+x)  x + x/(1x)^2 + log(1x)/(1x)). (End)


EXAMPLE

Let n=4. v(S_n)=46, see A067318. (n2)! = 2! = 2. n is even so P_N is larger than A_n. v(P_N) = 23 + 1 = 24. v(A_n) = 23  1 = 22, see A067369. Let n=5. v(S_n)=326. (n2)! = 3! = 6. n is odd so A_n is larger than P_N. v(P_N) = 163  3 = 160. v(A_n) = 163 + 3 = 166.


MAPLE

seq(coeff(series(factorial(n)*(1/2)*((1+x)*log(1+x)x+x/(1x)^2+log(1x)/(1x)), x, n+1), x, n), n = 1 .. 25); # Muniru A Asiru, Dec 15 2018


MATHEMATICA

a[n_] := (n*n! + (1)^n*((n2)! + StirlingS1[n+1, 2]))/2; a[1] = 0; Table[a[n], {n, 1, 19}] (* JeanFrançois Alcover, May 23 2012, after Vladeta Jovovic *)


PROG

(PARI) a(n)={if(n < 2, 0, (1/2)*((1)^n*(n2)! + n*n!  abs(stirling(n+1, 2, 1))))} \\ Andrew Howroyd, Dec 14 2018
(GAP) Concatenation([0], List([2..25], n>(1/2)*((1)^n*Factorial(n2)+n*Factorial(n)AbsInt(Stirling1(n+1, 2))))); # Muniru A Asiru, Dec 15 2018


CROSSREFS

Cf. A067369 A067318.
Sequence in context: A119581 A240916 A006292 * A322237 A289795 A094432
Adjacent sequences: A067367 A067368 A067369 * A067371 A067372 A067373


KEYWORD

easy,nice,nonn


AUTHOR

Nick Hann (nickhann(AT)aol.com), Jan 20 2002


EXTENSIONS

Corrected and extended by Vladeta Jovovic, Feb 02 2003
a(20)a(22) from Charlie Neder, Dec 14 2018


STATUS

approved



