%I #15 Mar 30 2021 12:03:00
%S 3,12,12,30,120,30,60,600,600,60,105,2100,5250,2100,105,168,5880,
%T 29400,29400,5880,168,252,14112,123480,246960,123480,14112,252,360,
%U 30240,423360,1481760,1481760,423360,30240,360,495,59400,1247400,6985440,12224520,6985440,1247400,59400,495
%N Triangular array: T(n,k) = sqrt(C(n-1,k-1)*C(n-1,k)*C(n,k+1)* C(n+1,k+1)*C(n+1,k)*C(n,k-1)), where C(n,k) = binomial(n,k).
%C In Pascal's triangle, the product of the six entries surrounding C(n,k) is a perfect square.
%C .............................................
%C ..............C(n-1,k-1)____C(n-1,k).........
%C .............../.................\...........
%C ............C(n,k-1)...C(n,k)....C(n,k+1)....
%C ...............\................./...........
%C ..............C(n+1,k)______C(n+1,k+1).......
%C .............................................
%C In fact, C(n-1,k-1)*C(n,k+1)*C(n+1,k) = C(n-1,k)*C(n+1,k+1)*C(n,k-1).
%H Seiichi Manyama, <a href="/A197208/b197208.txt">Rows n = 2..141, flattened</a>
%F T(n,k) = sqrt(C(n-1,k-1)*C(n-1,k)*C(n,k+1)*C(n+1,k+1)*C(n+1,k)* C(n,k-1)).
%F T(n,k) = C(n-1,k-1)*C(n,k+1)*C(n+1,k) = C(n-1,k)*C(n+1,k+1)*C(n,k-1).
%F T(n,k) = 1/2*(n^3-n)*A056939(n-2,k-1), for n >= 2 and 1 <= k <= n-1.
%F Row sums are A197209.
%e .n\k.|....1......2......3......4......5......6
%e = = = = = = = = = = = = = = = = = = = = = = = =
%e ..2..|....3...
%e ..3..|...12.....12
%e ..4..|...30....120.....30
%e ..5..|...60....600....600.....60
%e ..6..|..105...2100...5250...2100....105
%e ..7..|..168...5880..29400..29400...5880....168
%e ...
%e T(4,3) = sqrt(1*3*6*10*5*1) = sqrt(900) = 30
%e ..............1..............
%e ............1...1............
%e ..........1...2...1..........
%e ........1...3...3____1.......
%e .............../......\......
%e ......1...4...6...4....1.....
%e ...............\....../......
%e ...1...5...10...10___5.....1.
%Y Cf. A007318, A056939, A197209 (row sums).
%K nonn,easy,tabl
%O 2,1
%A _Peter Bala_, Oct 12 2011