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%I #8 Mar 07 2023 08:49:10
%S 1,0,1,2,0,1,0,8,3,0,1,24,0,0,20,15,0,1,0,144,90,40,0,0,0,40,45,0,1,
%T 720,0,0,0,504,630,280,210,0,0,0,70,105,0,1,0,5760,3360,2688,1260,0,0,
%U 0,0,0,1344,2520,1120,1680,105,0,0,0,112,210,0,1
%N Coefficients of the cycle index polynomial for the alternating group A_n multiplied by n!/2, n>=1, read as partition polynomial.
%C The row lengths sequence is A000041.
%C The partitions are ordered like in Abramowitz-Stegun (for the reference see A036036, where also a link to a work by C. F. Hindenburg from 1779 is found where this order has been used).
%C The row sums are A001710(n-1), n>=1.
%C The cycle index (multivariate polynomial) for the alternating group A_n, called Z(A_n), is
%C Z(S_n) + Z(S_n;x[1],-x[2],x[3],-x[4],... ), n>=1,
%C with the cycle index Z(S_n) for the symmetric group S_n, in the variables x[1],...,x[n]. See the Harary and Palmer reference. The coefficients of n!*Z(S_n) are the M_2 numbers of Abramowitz-Stegun, pp. 831-2. See A036039 and A102189, also for the Abramowitz-Stegun reference.
%D F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 36, (2.2.6).
%H Wolfdieter Lang, <a href="/A212358/a212358.pdf">Rows n=1..10, Z(A_n)) for n=1..13.</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AlternatingGroup.html">Alternating Group</a>.
%F The cycle index polynomial for the alternating group A_n is Z(A_n) = (2*a(n,k)*x[1]^(e[k,1])*x[2]^(e[k,2])*...*x[n]^(e[k,n]))/n!, n>=1, if the k-th partition of n in Abramowitz-Stegun order is 1^(e[k,1]) 2^(e[k,2]) ... n^(e[k,n]), where a part j with vanishing exponent e[k,j] has to be omitted. The n dependence of the exponents has been suppressed. See the comment above for the Z(A_n) formula, and the link for these polynomials for n=1..13.
%F a(n,k) is the coefficient the term of (n!/2)*Z(A_n) corresponding to the k-th partition of n in Abramowitz-Stegun order. a(n,k) = 0 if there is no such term in Z(A_n).
%e n\k 1 2 3 4 5 6 7 8 9 10 11 ...
%e 1: 1
%e 2: 0 1
%e 3: 2 0 1
%e 4: 0 8 3 0 1
%e 5: 24 0 0 20 15 0 1
%e 6: 0 144 90 40 0 0 0 40 45 0 1
%e ...
%e See the link for rows n=1..10 and the Z(A_n) polynomials for n=1..13.
%e n=6: Z(A_6) = 2*(144*x[1]*x[5] + 90*x[2]*x[4] + 40*x[3]^2 + 40*x[1]^3*x[3] + 45*x[1]^2*x[2]^2 + 1*x[1]^6)/6!, because the relevant partitions of 6 appear for k=2: 1,5; k=3: 2,4; k=4: 3^2; k=8: 1^3,3; k=9: 1^2,2^2 and k=11: 1^6. Thus, Z(A_6) = (2/5)*x[1]*x[5] + (1/4)*x[2]*x[4] + (1/9)*x[3]^2 + (1/9)*x[1]^3*x[3] + (1/8)*x[1]^2*x[2]^2 + (1/360)*x[1]^6.
%Y Cf. A036039 or A102189 for Z(S_n), A212355 for Z(D_n), and A212357 for Z(C_n).
%K nonn,tabf
%O 1,4
%A _Wolfdieter Lang_, Jun 12 2012
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