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 A000198 Largest order of automorphism group of a tournament with n nodes. (Formerly M2280 N0902) 1

%I M2280 N0902

%S 1,1,3,3,5,9,21,21,81,81,81,243,243,441,1215,1701,1701,6561,6561,6561,

%T 45927,45927,45927,137781,137781,229635,1594323,1594323,1594323,

%U 4782969,4782969,7971615,14348907,33480783,33480783,129140163,129140163,129140163

%N Largest order of automorphism group of a tournament with n nodes.

%C It appears that all terms except a(5) = 5 are divisible by a power of 3. [Jonathan Vos Post, Apr 20, 2011]

%D J. W. Moon, Topics on Tournaments. Holt, NY, 1968, p. 81.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

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

%H Joseph Myers, <a href="/A000198/b000198.txt">Table of n, a(n) for n = 1..1000</a>

%H B. Alspach, <a href="http://dx.doi.org/10.4153/CMB-1968-078-3">A combinatorial proof of a conjecture of Goldberg and Moon</a>, Canad. Math. Bull. 11 (1968), 655-661.

%H B. Alspach and J. L. Berggren, <a href="http://dx.doi.org/10.4153/CMB-1973-003-7">On the determination of the maximum order of the group of a tournament</a>, Canad. Math. Bull 16 (1973), 11-14.

%H J. D. Dixon, <a href="http://dx.doi.org/10.4153/CMB-1967-048-9">The maximum order of the group of a tournament</a>, Canad. Math. Bull. 10 (1967), 503-505.

%H <a href="/index/To#tournament">Index entries for sequences related to tournaments</a>

%F a(3^k) = 3^((3^k - 1)/2), a(5*3^k) = 5*3^((5*3^k - 5)/2), a(7*3^k) = 7*3^((7*3^k - 5)/2), and, for all other n, a(n) = max (a(i)a(n-i)) where the max is taken over 1 <= i <= n-1 (from Alspach and Berggren (1973) Theorem 4).

%F a(3r) = (3^r)a(r), a(n) = a(n-1) for n = 1, 2 or 4 mod 9, a(9k+8) = max(a(9k+7), a(5)a(9k+3)), a(9k+5) = max(a(2)a(9k+3), a(5)a(9k), a(7)a(9k-2)), a(9k+7) = a(7)a(9k) (from Alspach and Berggren (1973) Theorem 5).

%p a:= proc(n) local t, r; t:= irem(n, 9);

%p `if`(3^ilog[3](n)=n, 3^((3^ilog[3](n)-1)/2),

%p `if`(irem(n, 5, 'r')=0 and 3^ilog[3](r)=r, 5*3^((5*3^ilog[3](r)-5)/2),

%p `if`(irem(n, 7, 'r')=0 and 3^ilog[3](r)=r, 7*3^((7*3^ilog[3](r)-5)/2),

%p `if`(irem(n, 3, 'r')=0, 3^r*a(r),

%p `if`(t in {1, 2, 4}, a(n-1),

%p `if`(t = 8, max (a(n-1), a(5)*a(n-5)),

%p `if`(t = 5, max (a(2)*a(n-2), a(5)*a(n-5), a(7)*a(n-7)),

%p a(7)*a(n-7) )))))))

%p end:

%p seq (a(n), n=1..50); # _Alois P. Heinz_, Jun 29 2012

%t a[n_] := a[n] = With[{t = Mod[n, 9]}, Which[ IntegerQ[Log[3, n]], 3^((1/2)*(n-1)),{q, r} = QuotientRemainder[n, 5]; r == 0 && IntegerQ[Log[3, q]], 5*3^((1/2)*(n-5)),{q, r} = QuotientRemainder[n, 7];r == 0 && IntegerQ[Log[3, q]], 7*3^((1/2)*(n-5)), {q, r} = QuotientRemainder[n, 3]; r == 0, 3^q*a[q],MemberQ[{1, 2, 4}, t], a[n-1],t == 8, Max[a[n-1], a[5]*a[n-5]], t == 5, Max[a[2]*a[n-2],a[5]*a[n-5], a[7]*a[n-7]],True, a[7]*a[n-7]]]; Table[a[n], {n, 1, 38}] (* _Jean-François Alcover_, Nov 12 2012, after _Alois P. Heinz_ *)

%K nonn,nice

%O 1,3

%A _N. J. A. Sloane_.

%E Edited and extended by _Joseph Myers_, Jun 28 2012

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