A073359 Nested floor product of n and fractions (2k+2)/(2k+1) for all k>=0, divided by 2.

1, 3, 6, 9, 13, 19, 24, 31, 39, 45, 54, 66, 73, 90, 103, 111, 126, 144, 153, 174, 193, 199, 229, 240, 264, 283, 306, 324, 354, 381, 403, 421, 463, 474, 504, 546, 555, 594, 630, 660, 679, 735, 741, 789, 846, 859, 903, 949, 966, 1011

Offset: 1

Author

Paul D. Hanna, Jul 29 2002

Keywords

• easy
• nonn

Examples

			a(3) = 6 since (1/2)[[[[[[3(2/1)](4/3)](6/5)](8/7)](10/9)](12/11)] = (1/2)[[[[[6(4/3)](6/5)](8/7)](10/9)](12/11)] = (1/2)[[[[8(6/5)](8/7)](10/9)](12/11)] = (1/2)[[[9(8/7)](10/9)](12/11)] = (1/2)[[10(10/9)](12/11)] = (1/2)[11(12/11)] = 6. [Minor correction by _M. F. Hasler_, Nov 23 2016]
		

Links

• D. Wilson et al., Interesting sequence, SeqFan list, Nov. 2016

Crossrefs

Cf. A000960.

Programs

PARI

apply( A073359(n)=forstep(k=2,9e9,2,n==(n=floor(n*k/(k-1)))&&return(n\2)), [1..100]) \\ M. F. Hasler, Nov 23 2016

Formula

By definition, a(n)=(1/2)[...[[[[n(2/1)](4/3)](6/5)]...(2k+2)/(2k+1)]..., where [x] = floor of x; this infinite nested floor product will eventually level-off at a(n).

a(n) = (A000960(n+1)-1)/2, cf. link to posts on the SeqFan list. - M. F. Hasler, Nov 23 2016 [This may only be a conjecture? - N. J. A. Sloane, Nov 23 2016]