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%I #29 Nov 09 2019 16:27:02
%S 1,1,4,84,8008,3268760,5567902560,39049918716424,1118770292985239888,
%T 130276394656770614583240,61448471214136179596720592960,
%U 117118180539414377821494470432491764,900390992257782351906806257139068209113040,27883369051325994219981405855549095749234579210080
%N a(n) = binomial(n^2, n*(n+1)/2).
%C 8*a(2*n+1)^4 = A182010(n) = number of potential group developed cocyclic Hadamard matrices over (the group) Z_{(2*n+1)^2} X Z^2_2 [Baliga, et al., p. 130]. - _L. Edson Jeffery_, Apr 10 2012
%H Andrew Howroyd, <a href="/A109901/b109901.txt">Table of n, a(n) for n = 0..50</a>
%H A. Baliga and K. J. Horadam, <a href="http://ajc.maths.uq.edu.au/pdf/11/ocr-ajc-v11-p123.pdf">Cocyclic Hadamard matrices over Z_t X Z^2_2</a>, Australas. J. Combin. 11(1995), 123-134.
%F a(n) = C(n^2, n*(n+1)/2) = (n^2!)/((n(n+1)/2)!*(n(n-1)/2)!).
%F a(n) = C(n^2, n*(n-1)/2).
%e a(6) = 36!/(21!*15!) = 5567902560.
%p seq(binomial(n^2,n*(n+1)/2),n=0..12); # _Emeric Deutsch_, Jul 16 2005
%t Table[Binomial[n^2,(n(n+1))/2],{n,20}] (* _Harvey P. Dale_, Jun 04 2011 *)
%o (PARI) a(n)=binomial(n^2,n*(n+1)/2)
%Y Cf. variants: A014062 (C(n^2,n*(n-1)), A135757 (C(n*(n+1),n*(n+1)/2)).
%Y Cf. A182010.
%K easy,nonn
%O 0,3
%A _Amarnath Murthy_, Jul 14 2005
%E More terms from _Emeric Deutsch_, Jul 16 2005
%E Offset changed to 0 (with a(0)=1), and name changed slightly by _Paul D. Hanna_, Jun 24 2011
%E Terms a(12) and beyond from _Andrew Howroyd_, Nov 09 2019
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