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%I #25 Dec 10 2017 04:55:49
%S 1,759,17296,249849,3217056,39703755,481008528,5776211364,69065734464,
%T 824142912363,9826364199840,117145945726810,1396918583188128,
%U 16665451879695801,198937019774252928
%N Leading term in extremal weight enumerator of doubly-even binary self-dual code of length 24n.
%C The term after the leading nonzero term eventually becomes negative and so for large n the extremal codes do not exist (see references, also A034415).
%D F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, see Theorem 13, p. 624.
%D C. L. Mallows and N. J. A. Sloane, An Upper Bound for Self-Dual Codes, Information and Control, 22 (1973), 188-200.
%H Seiichi Manyama, <a href="/A034414/b034414.txt">Table of n, a(n) for n = 0..919</a> (terms 0..250 from N. J. A. Sloane)
%H G. Nebe, E. M. Rains and N. J. A. Sloane, <a href="http://neilsloane.com/doc/cliff2.html">Self-Dual Codes and Invariant Theory</a>, Springer, Berlin, 2006.
%H E. M. Rains and N. J. A. Sloane, Self-dual codes, pp. 177-294 of Handbook of Coding Theory, Elsevier, 1998 (<a href="http://neilsloane.com/doc/self.txt">Abstract</a>, <a href="http://neilsloane.com/doc/self.pdf">pdf</a>, <a href="http://neilsloane.com/doc/self.ps">ps</a>).
%H N. J. A. Sloane, <a href="http://neilsloane.com/doc/sg.txt">My favorite integer sequences</a>, in Sequences and their Applications (Proceedings of SETA '98).
%F a(24n) = C(24n, 5)*C(5n-2, n-1)/C(4n+4, 5).
%e At length 24, the extremal weight enumerator is 1+759*x^8+2576*x^12+..., with leading coefficient 759; this is the weight enumerator of the binary Golay code.
%p # Extremal weight enumerators:
%p kernelopts(printbytes=false): interface(screenwidth=200);
%p W0:=1; f:=1+14*x+x^2; f:=f^3; g:=x*(1-x)^4;
%p for mu from 1 to 100 do
%p # set max deg
%p md:=mu+3; W0:=series(f^mu,x,md): h:=series(g/f,x,md): A:=series(W0,x,md): Z:=A:
%p for i from 1 to mu do
%p Z:=series(Z*h,x,md); A:=series(A-coeff(A,x,i)*Z,x,md); od: lprint(A);
%p od:
%t a[n_] := 18(6n-1)(8n-1)(12n-1)(24n-1)Binomial[5n-2, n-1]/((n+1)(2n+1)(4n+1)(4n+3)); a[0] = 1; Table[a[n], {n, 0, 14}](* _Jean-François Alcover_, Oct 06 2011, after formula *)
%Y Cf. A034415 (second coefficient, which becmes negative), A001380, A034597.
%K nonn,easy,nice
%O 0,2
%A _N. J. A. Sloane_
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