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%I #41 Jan 07 2020 11:42:24
%S 1,1,2,8,86,1394,32134,957332,35154340,1537408202,78132541528,
%T 4528684996756,295011186006282,21345627856836734,1698954263159544138,
%U 147553846727480002824,13888244935445960871352,1408407905312396429259944,153105374581396386625831530
%N Number of magic series of order n.
%C Henry Bottomley's narrowing gap could be confirmed for 2 < n <= 64. - _Walter Trump_, Jan 21 2005
%C A new algorithm was found by _Robert Gerbicz_. Now the enumeration of magic series of orders greater than 100 is possible. - _Walter Trump_, May 05 2006
%D M. Kraitchik, Magic Series. Section 7.13.3 in Mathematical Recreations, New York, W. W. Norton, pp. 143 and 183-186, 1942.
%H T. D. Noe and Alois P. Heinz, <a href="/A052456/b052456.txt">Table of n, a(n) for n = 0..150</a> (from Gerbicz and Trump)
%H H. Bottomley, <a href="http://www.se16.info/js/partitionstest.htm">Partition and composition calculator</a>
%H H. Bottomley and W. Trump, <a href="/A052456/a052456.txt">First 36 terms</a>
%H Walter Trump, <a href="http://www.trump.de/magic-squares">Magic Squares</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MagicSeries.html">Magic Series</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MultimagicSeries.html">Multimagic Series</a>
%H Robert Gerbicz, Walter Trump, <a href="http://www.trump.de/magic-squares/magic-series/exact.htm">First 150 terms</a>
%H Robert Gerbicz, <a href="http://robert.gerbicz.googlepages.com/">C-program</a> to generate the sequence
%F a(n) = A067059(n, n*(n-1)) = r(n, n*(n-1), n^2*(n-1)/2) where r(n, m, k) is a restricted partition function giving the number of partitions of k into at most n positive parts each no more than m. - _Henry Bottomley_, Feb 25 2002.
%F It seems a(n) (at least for 2<n<=36) is in the narrowing gap between C(n^2, n)*1.381976597885.../n^(5/2) and C(n^2, n)*sqrt(6/(Pi*n^2*(n-1)*(n^2+1))): cf. A068606 and assuming the peak of a normal distribution = 1/sqrt(variance*2*Pi) - _Henry Bottomley_, Feb 25 2002.
%F a(n) ~ sqrt(3) * exp(n-1/2) * n^(n-3) / Pi. - _Vaclav Kotesovec_, Sep 05 2014
%e a(3) = 8 since a magic square of order 3 would require a row sum of 15=(1+2+...+9)/3 and there are 8 ways of writing 15 as the sum of three distinct positive numbers up to 9: 1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, 4+5+6.
%t $RecursionLimit = 1000; b[n_, i_, t_] /; i < t || n < t*((t + 1)/2) || n > t*((2*i - t + 1)/2) = 0; b[0, _, _] = 1; b[n_, i_, t_] := b[n, i, t] = b[n, i - 1, t] + If[n < i, 0, b[n - i, i - 1, t - 1]]; a[_, 0] = 1; a[0, _] = 0; a[n_, k_] := With[{s = k*(k*n + 1)}, If[Mod[s, 2] == 1, 0, b[s/2, k*n, k]]]; a[n_] := a[n] = a[n, n]; Table[Print[a[n]]; a[n], {n, 0, 18}] (* _Jean-François Alcover_, Aug 15 2013, after _Alois P. Heinz_ *)
%Y Cf. A007785, A052457, A052458. A100568 is the same sequence times n!.
%Y Main diagonal of A204459. - _Alois P. Heinz_, Jan 18 2012
%K nonn,nice
%O 0,3
%A _Eric W. Weisstein_
%E Edited and ten more terms from _Henry Bottomley_, Feb 16 2002
%E Terms through a(36) added to attached web page, Feb 04 2005
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