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

1,2

Links

Table of n, a(n) for n=1..50.

D. Wilson et al., Interesting sequence, SeqFan list, Nov. 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]

Example

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]

Prog

(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

Crossrefs

Cf. A000960.

Sequence in context: A307270 A310160 A117469 * A310161 A137041 A153006

Adjacent sequences: A073356 A073357 A073358 * A073360 A073361 A073362

Keyword

easy,nonn

Author

Paul D. Hanna, Jul 29 2002

Status

approved