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%I #14 Jan 29 2022 02:59:42
%S 1,2,10,204,18326,7157436,11867138452,81971848887192,
%T 2329289249771718630,270079267572894401313900,
%U 127115660247624311548253487740,242023658005438716992830183038644712
%N a(n) = 2^(n^2+n) * C(n-1 + 1/2^n, n) = [x^n] 1/(1 - 2^(n+1)*x)^(1/2^n).
%H Seiichi Manyama, <a href="/A159558/b159558.txt">Table of n, a(n) for n = 0..57</a>
%F G.f.: A(x) = Sum_{n>=0} a(n)*x^n/2^(n^2+n) = Sum_{n>=0} (-1)^n*log(1 - x/2^n)^n/n!.
%F a(n) ~ 2^(n^2) / n. - _Vaclav Kotesovec_, Oct 20 2020
%e G.f.: A(x) = 1 + 2*x/2^2 + 10*x^2/2^6 + 204*x^3/2^12 + 18326*x^4/2^20 +...
%e A(x) = 1 - log(1-x/2) + log(1-x/4)^2/2! - log(1-x/8)^3/3! +...+ (-1)^n*log(1-x/2^n)^n/n! +...
%e Illustrate a(n) = [x^n] 1/(1 - 2^(n+1)*x)^(1/2^n):
%e (1-4*x)^(-1/2) = 1 + (2)*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 +...
%e (1-8*x)^(-1/4) = 1 + 2*x + (10)*x^2 + 60*x^3 + 390*x^4 + 2652*x^5 +...
%e (1-16*x)^(-1/8) = 1 + 2*x + 18*x^2 + (204)*x^3 + 2550*x^4 + 33660*x^5 +...
%e (1-32*x)^(-1/16) = 1 + 2*x + 34*x^2 + 748*x^3 + (18326)*x^4 + 476476*x^5 +...
%e (1-64*x)^(-1/32) = 1 + 2*x + 66*x^2 + 2860*x^3 + 138710*x^4 + (7157436)*x^5 +...
%e where the coefficients in parenthesis form the initial terms of this sequence.
%e Particular values.
%e A(1) = 1 + log(2) + log(4/3)^2/2! + log(8/7)^3/3! + log(16/15)^4/4! +...
%e A(1/2) = 1 + log(4/3) + log(8/7)^2/2! + log(16/15)^3/3! +...
%e A(1/4) = 1 + log(8/7) + log(16/15)^2/2! + log(32/31)^3/3! +...
%e A(3/2) = 1 + log(4) + log(8/5)^2/2! + log(16/13)^3/3! + log(32/29)^4/4! +...
%e Explicitly,
%e A(1) = 1.734925215983391138169827514899...
%e A(3/2) = 2.498242012620581570762548014070...
%e A(r) = 2 at r=1.2139293567161900826815...
%e A(r) = 3 at r=1.6849757886374480509741...
%e A(-1) = 0.6191596458119190547682348949108188...
%e A(-2) = 0.3872099757580366707782339498635620...
%e A(2) is indeterminate.
%t Table[2^(n^2+n) * Binomial[n-1+1/2^n, n], {n,0,15}] (* _Vaclav Kotesovec_, Oct 20 2020 *)
%o (PARI) a(n)=2^(n^2+n)*binomial(n-1+1/2^n,n)
%Y Cf. A159478, A158093, A183131, A224883.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Apr 21 2009
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