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%I #68 Jul 03 2018 17:19:38
%S 1,1,1,0,1,1,3,4,1,15,10,0,6,1,45,40,45,36,1,105,175,315,126,105,120,
%T 1,315,616,1890,336,2520,960,0,0,0,0,0,0,0,336,1,1323,2884,9450,756,
%U 18900,4320,0,6720,2268,0,3780,0,0,3024,1,5355,15520,47250,19656
%N Triangle read by rows: T(n,k) is the number of labeled cyclic subgroups of order k in the alternating group A_n, 1 <= k <= A051593(n).
%e Triangle begins:
%e 1;
%e 1;
%e 1, 0, 1;
%e 1, 3, 4;
%e 1, 15, 10, 0, 6;
%e 1, 45, 40, 45, 36;
%e 1, 105, 175, 315, 126, 105, 120;
%e 1, 315, 616, 1890, 336, 2520, 960, 0, 0, 0, 0, 0, 0, 0, 336;
%e ...
%o (PARI)
%o permcount(v) = {my(m=1,s=0,k=0,t); for(i=1,#v,t=v[i]; k=if(i>1&&t==v[i-1],k+1,1); m*=t*k;s+=t); s!/m}
%o G(n)={my(s=0); forpart(p=n, if(sum(i=1,#p,p[i]-1)%2==0, my(d=lcm(Vec(p))); s+=x^d*permcount(p)/eulerphi(d))); s}
%o for(n=1, 10, print(Vecrev(G(n)/x)))
%Y Row sums are A051636.
%Y Cf. A051593, A074881, A181950.
%K nonn,tabf
%O 1,7
%A _Andrew Howroyd_, Jul 03 2018
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