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

%I A073359 #13 Dec 23 2024 14:53:42
%S A073359 1,3,6,9,13,19,24,31,39,45,54,66,73,90,103,111,126,144,153,174,193,
%T A073359 199,229,240,264,283,306,324,354,381,403,421,463,474,504,546,555,594,
%U A073359 630,660,679,735,741,789,846,859,903,949,966,1011
%N A073359 Nested floor product of n and fractions (2k+2)/(2k+1) for all k>=0, divided by 2.
%H A073359 D. Wilson et al., <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2016-November/thread.html">Interesting sequence</a>, SeqFan list, Nov. 2016
%F A073359 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).
%F A073359 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]
%e A073359 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]
%o A073359 (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
%Y A073359 Cf. A000960.
%K A073359 easy,nonn
%O A073359 1,2
%A A073359 _Paul D. Hanna_, Jul 29 2002