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%I #4 Mar 30 2012 18:36:32
%S 1,7,28,84,175,421,847,1288,1939,3780,5656,9247,15148,22099,25375,
%T 39676,54607,75208,90559,129360,166321,209832,240268,320719,399595,
%U 536956,672672,816733,906444,1115275,1321741,1595832,1908088,2323944
%N Nested floor product of n and fractions (k+1)/k for all k>0 (mod 6), divided by 6.
%C When is a(n) not divisible by 7?
%F a(n)=(1/6)[...[[[[[[n(2/1)](3/2)](4/3)](5/4)](6/5)](8/7]...(k+1)/k]..., k>0 (mod 6), where [x] = floor of x; this infinite nested floor product will eventually level-off at a(n).
%e a(1)=1 since (1/6)[[[[1(2/1)](3/2)](4/3)](5/4)](6/5)]=1
%Y Cf. A073359, A073360, A073361, A073362.
%K easy,nonn
%O 1,2
%A _Paul D. Hanna_, Jul 29 2002