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%I #8 Apr 24 2018 22:44:05
%S 1,1,2,17,62,1382,21844,929569,6404582,443861162,18888466084,
%T 1936767361654,58870668456604,8374643517010684,689005380505609448,
%U 129848163681107301953,1736640792209901647222,418781231495293038913922
%N The left hand column of the triangle A160468.
%C Resembles A002430, the numerators of the Taylor series for tan(x). The first difference occurs at a(12). (Its resemblance to this sequence led to the conjecture A160469(n) = A002430(n)*A089170(n-1).)
%F a(n) = A002430(n)*A089170(n-1) with A002430 (n) = numer((-1)^(n-1)*2^(2*n)*(2^(2*n)-1)* bernoulli(2*n)/(2*n)!) and A089170 (n-1) = numer(2*bernoulli(2*n)* (4^n-1)/(2*n))/ numer((4^n-1)*bernoulli(2*n)/(2*n)!) for n = 1, 2, 3, ....
%Y Equals the first left hand column of A160468.
%Y Equals A002430(n)*A089170(n-1).
%Y Equals (A002430(n)/A036279(n))*(A117972(n)/A000265(n)).
%Y Equals A048896(n-1)*A002425(n).
%Y Cf. A156769 (which resembles the denominators of the Taylor series for tan(x)).
%K easy,nonn
%O 1,3
%A _Johannes W. Meijer_, May 24 2009
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