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a(n) = Bigomega(Catalan(n)) - round( 3 n /(2 log(n+2))) (= A081399 - A137687).
3

%I #9 Sep 26 2017 10:58:41

%S 0,-1,-1,-2,-1,-1,0,-2,-1,-2,-1,-1,-1,0,1,-1,0,-1,1,0,0,0,1,0,0,1,0,

%T -1,1,0,1,-2,-1,0,0,-2,-1,-1,1,-1,0,2,2,2,2,1,3,1,2,0,2,1,1,1,1,-1,-1,

%U -1,1,0,2,3,3,0,0,0,1,0,3,2,2,0,2,3,3,2,2,3,4,1,0,1,1,1,1,1,3,1,4,2,2,1,2,2,3,2,3,1,2,0,1,0,2,1,2,2,3,1,3,2,3,1,2,3,3,2,3

%N a(n) = Bigomega(Catalan(n)) - round( 3 n /(2 log(n+2))) (= A081399 - A137687).

%C It is easy to show that A081399(n) = bigomega(Catalan(n)) is between n/log(n) and 2n/log(n) (for n>n0). The sequence A137687 is roughly the middle of this interval, which turns out to be a fair approximation to A081399. The present sequence lists the (signed) difference.

%H M. F. Hasler, <a href="/A137686/b137686.txt">Table of n, a(n) for n = 0..3000</a>.

%H Douglas M. Campbell, <a href="http://www.jstor.org/stable/2689673">The Computation of Catalan Numbers</a>, Mathematics Magazine, Vol. 57, No. 4. (Sep., 1984), pp. 195-208.

%F a(n) = A001222(A000108(n)).

%o (PARI) A137686(n) = bigomega(prod(i=2,n,(n+i)/i)) - round(3*n/log(n+2)/2)

%Y Cf. A000108, A081399, A120626, A137687.

%K easy,sign

%O 0,4

%A _M. F. Hasler_, Feb 06 2008