A073360 Nested floor product of n and fractions (k+1)/k for all k>0 (mod 3), divided by 3.

1, 4, 9, 20, 29, 44, 69, 104, 121, 180, 241, 284, 349, 420, 521, 664, 701, 860, 1009, 1184, 1301, 1540, 1789, 1964, 2181, 2380, 2701, 3124, 3301, 3704, 4029, 4444, 4809, 5144, 5789, 6340, 6729, 7244, 7981, 8420, 9101

Offset: 1

Author

Paul D. Hanna, Jul 29 2002

Keywords

• easy
• nonn

Examples

			a(2) = 4 since (1/3)[[[[[[2(2/1)](3/2)](5/4)](6/5)](8/7)](9/8)](11/10)](12/11)]
= (1/3)[[[[[4(3/2)](5/4)](6/5)](8/7)](9/8)](11/10)](12/11)]
= (1/3)[[[[6(5/4)](6/5)](8/7)](9/8)](11/10)](12/11)]
= (1/3)[[[[7(6/5)](8/7)](9/8)](11/10)](12/11)]
= (1/3)[[[[8(8/7)](9/8)](11/10)](12/11)]
= (1/3)[[[[9(9/8)](11/10)](12/11)]
= (1/3)[[[[10(11/10)](12/11)]
= 4.
Note that the denominators consist of positive integers not == 0 mod 3.
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 1..300

Crossrefs

Cf. A073359.

Formula

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).