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%I #11 Apr 29 2024 14:23:24
%S 1,4,9,20,29,44,69,104,121,180,241,284,349,420,521,664,701,860,1009,
%T 1184,1301,1540,1789,1964,2181,2380,2701,3124,3301,3704,4029,4444,
%U 4809,5144,5789,6340,6729,7244,7981,8420,9101
%N Nested floor product of n and fractions (k+1)/k for all k>0 (mod 3), divided by 3.
%H Paul D. Hanna, <a href="/A073360/b073360.txt">Table of n, a(n) for n = 1..300</a>
%F a(n)=(1/3)[...[[[[n(2/1)](3/2)](5/4)](6/5)]...(k+1)/k]..., k>0 (mod 3), where [x] = floor of x; this infinite nested floor product will eventually level-off at a(n).
%e a(2) = 4 since (1/3)[[[[[[2(2/1)](3/2)](5/4)](6/5)](8/7)](9/8)](11/10)](12/11)]
%e = (1/3)[[[[[4(3/2)](5/4)](6/5)](8/7)](9/8)](11/10)](12/11)]
%e = (1/3)[[[[6(5/4)](6/5)](8/7)](9/8)](11/10)](12/11)]
%e = (1/3)[[[[7(6/5)](8/7)](9/8)](11/10)](12/11)]
%e = (1/3)[[[[8(8/7)](9/8)](11/10)](12/11)]
%e = (1/3)[[[[9(9/8)](11/10)](12/11)]
%e = (1/3)[[[[10(11/10)](12/11)]
%e = 4.
%e Note that the denominators consist of positive integers not == 0 mod 3.
%Y Cf. A073359.
%K easy,nonn
%O 1,2
%A _Paul D. Hanna_, Jul 29 2002
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