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Number of unrooted steric quartic trees with n (unlabeled) nodes and possessing a centroid; number of n carbon alkanes C(n)H(2n +2) with a centroid when stereoisomers are regarded as different.
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%I #9 Dec 29 2014 08:26:05

%S 1,0,1,1,3,2,11,9,55,70,345,494,2412,3788,18127,30799,143255,256353,

%T 1173770,2190163,9892302,19130814,85289390,169923748,749329719,

%U 1531701274,6688893605,13984116304,60526543480,129073842978

%N Number of unrooted steric quartic trees with n (unlabeled) nodes and possessing a centroid; number of n carbon alkanes C(n)H(2n +2) with a centroid when stereoisomers are regarded as different.

%C The degree of each node is <= 4.

%C A centroid is a node with less than n/2 nodes in each of the incident subtrees, where n is the number of nodes in the tree. If a centroid exists it is unique.

%C Regarding stereoisomers as different means that only the alternating group A_4 acts at each node, not the full symmetric group S_4. See A010372 for the analogous sequence when stereoisomers are not counted as different.

%H <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>

%H <a href="/index/Tra#trees">Index entries for sequences related to trees</a>

%F Let r(x) = g.f. A(x) for A000625 truncated after the x^n term (x^0 through x^n terms only). Then coefficients of x^(2n) and x^(2n+1) in [r(x)^4 + 8 r(x^3) r(x) + 3 r(x^2)^2]/12 are terms 2n+1 and 2n+2 in current sequence..

%t c[0] = 1; f[x_, m_] := Sum[c[k] x^k, {k, 0, m}]; coes[m_] := CoefficientList[Series[f[x, m] - 1 - (x*(f[x, m]^3 + 2*f[x^3, m])/3), {x, 0, m}], x] // Rest; r[x_, m_] := r[x, m] = (f[x, m] /. Solve[Thread[coes[m] == 0]] // First); b[m_] := CoefficientList[(1/12)*(r[x, m]^4 + 3*r[x^2, m]^2 + 8*r[x, m]*r[x^3, m]), x]; a[1]=1; a[2]=0; a[n_] := b[Quotient[n-1, 2]][[n]]; Table[Print["a(", n, ") = ", an = a[n]]; an, {n, 1, 30}] (* _Jean-François Alcover_, Dec 29 2014 *)

%Y Cf. A000598, A000602, A010732, A010733, A000625, A000628, A086200.

%Y For even n A000628(n) = a(n) + A086200(n/2), for odd n A000628(n) = a(n), since every tree has either a centroid or a bicentroid but not both.

%K nonn

%O 1,5

%A Steve Strand (snstrand(AT)comcast.net), Aug 28 2003