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G.f.: A(x) = 1/[1-x - Sum_{n>=1} A001147(n)*x^(2n) ] where A001147(n) = (2n)!/(n!*2^n) is the double factorials.
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%I #3 Mar 30 2012 18:37:01

%S 1,1,2,3,8,14,43,81,283,556,2243,4512,21374,43469,243817,497217,

%T 3289606,6697795,51583952,104698998,922789643,1867079621,18522929815,

%U 37380015420,411572179999,828925168492,10014624164666,20140445929353

%N G.f.: A(x) = 1/[1-x - Sum_{n>=1} A001147(n)*x^(2n) ] where A001147(n) = (2n)!/(n!*2^n) is the double factorials.

%C Is this sequence equal to A076876 (meandric numbers for a river crossing two parallel roads at n points)?

%e G.f.: A(x) = 1/(1-x - x^2 - 3*x^4 - 15*x^6 - 105*x^8 - 945*x^10 -...).

%o (PARI) a(n)=polcoeff(1/(1-x-sum(k=1,n\2,(2*k)!/k!/2^k*x^(2*k))+x*O(x^n)),n)

%Y Cf. A001147.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Nov 04 2006