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A321689 Approximation of the 2-adic integer exp(-4) up to 2^n.

%I A321689 #9 Nov 18 2018 10:03:52
%S A321689 0,1,1,5,5,5,5,5,133,389,901,1925,3973,8069,8069,24453,57221,57221,
%T A321689 188293,450437,974725,974725,974725,974725,974725,17751941,17751941,
%U A321689 84860805,84860805,84860805,621731717,621731717,621731717,4916699013,4916699013
%N A321689 Approximation of the 2-adic integer exp(-4) up to 2^n.
%C A321689 Let 4Q_2 = {x belongs to Q_2 : |x|_2 <= 1/4} and 4Q_2 + 1 = {x belongs to Q_2: |x - 1|_2 <= 1/4}. Define exp(x) = Sum_{k>=0} x^k/k! and log(x) = -Sum_{k>=0} (1 - x)^k/k over 2-adic field, then exp(x) is a one-to-one mapping from 4Q_2 to 4Q_2 + 1, and log(x) is the inverse of exp(x).
%C A321689 a(n) is the multiplicative inverse of A320814(n) modulo 2^n.
%H A321689 Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F A321689 If Sum_{i=0..A320840(n)-1} (-4)^i/i! = p/q, gcd(p, q) = 1, then a(n) = p*q^(-1) mod 2^n.
%F A321689 a(n) = Sum_{i=0..n-1} A321692(i)*2^i.
%e A321689 A320840(1) = 1, (-4)^0/0! = 1, so a(1) = 1.
%e A321689 A320840(3) = 2, Sum_{i=0..1} (-4)^i/i! = -3 == 5 (mod 8), so a(3) = 5.
%e A321689 A320840(8) = 6, Sum_{i=0..5} (-4)^i/i! = -53/15 == 133 (mod 256), so a(8) = 133.
%e A321689 A320840(9) = 7, Sum_{i=0..6} (-4)^i/i! = 97/45 == 389 (mod 512), so a(9) = 389.
%e A321689 A320840(10) = 9, Sum_{i=0..8} (-4)^i/i! = 167/315 == 901 (mod 1024), so a(10) = 901.
%o A321689 (PARI) a(n) = lift(sum(i=0, n-1-(n>=2), Mod((-4)^i/i!, 2^n)))
%o A321689 (PARI) a(n) = lift(exp(-4 + O(2^n)));
%Y A321689 Cf. A320814, A320840, A321692.
%K A321689 nonn
%O A321689 0,4
%A A321689 _Jianing Song_, Nov 17 2018