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%I #10 Aug 26 2019 11:18:52
%S 1,0,1,0,1,0,0,0,1,0,1,1,0,1,0,0,0,0,0,0,0,0,1,1,0,0,0,1,0,0,0,0,1,0,
%T 0,1,1,0,1,0,1,1,0,1,0,0,1,0,0,1,0,1,1,1,1,1,1,1,0,0,1,0,0,1,0,1,0,0,
%U 1,0,0,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,1
%N Digits of the multiplicative inverse of A309754.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F By definition, (Sum_{i=-1..n} a(i)*2^(i+1)) * (Sum_{i=0..floor((n+1)/2)} 2^(2*i)/(2*i+1)) == 1 (mod 2^(n+2)).
%e arctanh(2) = ...0111000110010001010010010111010001111010, so 1/arctanh(2) = 010110010000100011000000001011010001010.1.
%o (PARI) a(n) = lift(Mod(sum(i=0, (n+1)/2, 2^(2*i)/(2*i+1)), 2^(n+2))^(-1))\2^(n+1)
%Y Cf. A309754, A309847.
%K nonn,base
%O -1
%A _Jianing Song_, Aug 20 2019
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