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A056923 Write the integers in groups: 0; 1,2; 3,4,5; 6,7,8,9; ... and form the product of the members of each group.

%I A056923 #25 Feb 13 2024 11:37:26
%S A056923 0,2,60,3024,240240,27907200,4475671200,948964262400,257256702743040,
%T A056923 86839771951296000,35728290125079552000,17602963463032472448000,
%U A056923 10233395250958706770944000,6932022668773077815267328000
%N A056923 Write the integers in groups: 0; 1,2; 3,4,5; 6,7,8,9; ... and form the product of the members of each group.
%C A056923 Each group begins with a triangular number and proceeds until one short of the next triangular number.
%C A056923 Also, the number under the radical using Brahmagupta's formula for an n-sided cyclic quadrilateral with sides 1..n. - _Ben Paul Thurston_, Dec 05 2006
%H A056923 Vincenzo Librandi, <a href="/A056923/b056923.txt">Table of n, a(n) for n = 0..100</a>
%H A056923 Nick Hobson, <a href="https://web.archive.org/web/20160327211231/http://www.qbyte.org/puzzles/p010s.html">Solution to puzzle 10: Farmer's enclosure</a>
%F A056923 a(n) = (n (n + 3)/2)!/((n - 1)(n + 2)/2)!.
%F A056923 a(n) = Product_{j=1..n+1} ((n+2)*(n+1)/2-j). - _Ben Paul Thurston_, Dec 05 2006
%p A056923 a:= n-> mul(n*(n+1)/2+j, j=0..n):
%p A056923 seq(a(n), n=0..15);  # _Alois P. Heinz_, Feb 02 2019
%t A056923 Table[(n (n + 3)/2)!/((n - 1)(n + 2)/2)!, {n, 0, 15}]
%t A056923 Times@@Range[First[#],Last[#]-1]&/@Partition[Accumulate[Range[0,15]],2,1] (* _Harvey P. Dale_, Apr 25 2014 *)
%Y A056923 Cf. A000217, A027480.
%K A056923 nonn
%O A056923 0,2
%A A056923 _Robert G. Wilson v_, Sep 09 2000