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%I #4 Mar 30 2012 18:36:32
%S 1,5,15,31,65,105,151,275,420,631,695,1050,1411,1685,2385,2941,3425,
%T 4410,5326,6995,7350,9316,10880
%N Nested floor product of n and fractions (k+1)/k for all k>0 (mod 4), divided by 4.
%F a(n)=(1/3)[...[[[[n(2/1)](3/2)](4/3)](6/5)]...(k+1)/k]..., k>0 (mod 4), where [x] = floor of x; this infinite nested floor product will eventually level-off at a(n).
%e a(1)=1 since (1/4)[[[[1(2/1)](3/2)](4/3)](6/5)]=(1/4)[4(6/5)]=1
%Y Cf. A073359, A073360.
%K easy,nonn
%O 1,2
%A _Paul D. Hanna_, Jul 29 2002
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