%I #21 Mar 05 2023 05:52:29
%S 1,1,3,7,22,78,315,1417,7087,38981,233887,1520268,10641881,79814109,
%T 638512875,5427359437,48846234937,464039231906,4640392319062,
%U 48724119350156,535965312851718,6163601097794765,73963213173537187,924540164669214843,12019022140699792968
%N a(n) = floor(Pt(n)/n!), where Pt(n) is product of first n positive triangular numbers (A000217).
%H Harvey P. Dale, <a href="/A233005/b233005.txt">Table of n, a(n) for n = 1..505</a>
%F a(n) = floor((n+1)!/2^n). - _Yifan Xie_, Mar 05 2023
%e a(4) = 7, because, the first four triangular numbers being 1, 3, 6, 10, their product is 180, which divided by 4! is 15/2 = 7.5.
%e a(5) = 22, because, the first five triangular numbers being 1, 3, 6, 10, 15, their product is 2700, which divided by 5! is 45/2 = 22.5.
%t With[{nn=30},Floor[#[[1]]/#[[2]]]&/@Thread[{FoldList[Times,Accumulate[ Range[ nn]]],Range[nn]!}]] (* _Harvey P. Dale_, Apr 02 2017 *)
%o (Python)
%o f=t=1
%o for n in range(1,33):
%o t*=n*(n+1)/2
%o f*=n
%o print str(t/f)+',',
%Y Cf. A000142, A000217.
%Y Cf. A006472 (triangular factorial, essentially equal to Pt(n)).
%Y Cf. A067667 (Pt(n)/n! for n's of the form 2^k-1).
%K nonn
%O 1,3
%A _Alex Ratushnyak_, Dec 03 2013