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a(n) = floor(T(n+1)!*T(n-1)!/(T(n)!)^2), where T(n) = n(n+1)/2 = the n-th triangular number.
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%I #8 Jul 14 2021 01:57:52

%S 6,20,42,71,108,152,204,263,330,403,485,573,669,773,884,1002,1128,

%T 1261,1401,1549,1704,1866,2036,2214,2398,2591,2790,2997,3211,3433,

%U 3662,3898,4142,4394,4652,4918,5192,5472,5761,6056,6359,6669,6987

%N a(n) = floor(T(n+1)!*T(n-1)!/(T(n)!)^2), where T(n) = n(n+1)/2 = the n-th triangular number.

%C The ratio itself is an integer only for n = 1, 2 and 3.

%e a(1) = 6 = (2*3)/1;

%e a(2) = 20 = (4*5*6)/(2*3), etc.

%t Floor[(#[[1]]!*#[[3]]!)/(#[[2]]!)^2]&/@Partition[Accumulate[ Range[ 0,50]],3,1] (* _Harvey P. Dale_, Aug 21 2015 *)

%K nonn

%O 1,1

%A _Amarnath Murthy_, Nov 09 2002

%E More terms from Mark Hudson (mrmarkhudson(AT)hotmail.com), Feb 11 2003

%E Definition clarified by _Harvey P. Dale_, Aug 21 2015