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A076628 a(n) = 2^(2^(n-1))*b(n) where b(1) = 1/2 and b(n+1) = b(n) - b(n)^2.

%I A076628 #33 Sep 08 2022 08:45:07
%S A076628 1,1,3,39,8463,483008799,1841209495473815103,
%T A076628 30574267942866411492610617989187955839,
%U A076628 9469098402231365955845075782127013884605836706970528486326625808482327697663
%N A076628 a(n) = 2^(2^(n-1))*b(n) where b(1) = 1/2 and b(n+1) = b(n) - b(n)^2.
%C A076628 Michael Somos found that b(n) has some nice properties (see link titled "A Somos-Rusin recursion").
%C A076628 From Shai Covo (green355(AT)netvision.net.il), Mar 17 2010: (Start)
%C A076628 Define a sequence of fractions by b'(1) = 1/2 and b'(n+1) = b'(n) - b'(n)^2/2. Then, b'(n) = 2*b(n+1), n >= 1. The sequence b'(n) is the complement to 1 of the sequence f(n) defined in A167424 by f(1) = 1/2 and f(n+1) = (f(n)^2 + 1)/2, i.e., b'(n) + f(n) = 1. The numerator of b'(n) is given by a'(n) = 2^(2^n-1)*b'(n).
%C A076628 Thus a'(n) = a(n+1), n >= 1. The sequences a'(n) and A167424(n) are related by a'(n) + A167424(n) = 2^(2^n-1). (End)
%C A076628 From Shai Covo (green355(AT)netvision.net.il), Mar 25 2010: (Start)
%C A076628 The following is the counterpart of a comment in A167424. Suppose that U_1, U_2, ... is a sequence of independent uniform (0,1) random variables, and define random variables X_1, X_2, ... as follows: X_1 = U_1, and, for n >= 1, X_{n+1} = X_n or U_{n+1} according as U_{n+1} > E(X_n) or U_{n+1} < E(X_n), respectively, where E() denotes expectation. Then, the sequence E(X_n) is identical to the sequence b'(n) introduced in the comment dated Mar 17 2010. Sketch of proof. E(X_1) = 1/2; for n >= 1, by the law of total expectation, we have E(X_{n+1}) = (1 - E(X_n))*E(X_n) + E(X_n)*E(X_n)/2. Hence E(X_{n+1}) = E(X_n) - E(X_n)^2/2. (End)
%C A076628 b(n) = sqrt(p(n-1)/2) with p(n) as defined in A187131, so Sum_n b(n)^2 = 1/2 - _Henry Bottomley_, Mar 05 2011
%H A076628 Michael Somos, <a href="http://www.math.niu.edu/Papers/Rusin/known-math/99/somos">A Somos-Rusin recursion</a>
%F A076628 a(n) = 2^(2^(n-1))*[1/(n + log(n) + (1-e(oo)) + (log(n)/n)*(1+o(1)))], where 1 - e(oo) is a constant equal to about 1 - 0.232006 = 0.767994. The expression in the square brackets is Dave Rusin's "best estimate" for b(n), taken from the link "Michael Somos, A Somos-Rusin recursion" (where b(n) is denoted a(n)). - Shai Covo (green355(AT)netvision.net.il), Mar 15 2010
%F A076628 a(n+1) = a(n) * 2^(2^(n-1)) - a(n)^2 starting from a(1) = 1, also a(n) = sqrt(A187131(n-1)) - _Henry Bottomley_, Mar 05 2011
%F A076628 It appears that a(n) = Product_{k = 1..n} A100441(k). - _Peter Bala_, Feb 11 2015
%t A076628 a[ n_] := If[ n < 2, Boole[n == 1], a[n - 1] (2^(2^(n - 2)) - a[n - 1])]; (* _Michael Somos_, Jul 24 2018 *)
%o A076628 (PARI) b(n)=if(n<2,1/2,b(n-1)-b(n-1)^2); a(n)=numerator(b(n))
%o A076628 (Magma) [1] cat [n eq 1 select 1 else Self(n-1)*2^(2^(n-1))-Self(n-1)^2: n in [1..8]]; // _Vincenzo Librandi_, Jun 17 2015
%Y A076628 Cf. A100441, A167424, A187131.
%K A076628 nonn,easy
%O A076628 1,3
%A A076628 _Benoit Cloitre_, Oct 22 2002
%E A076628 Typo in definition corrected by Shai Covo (green355(AT)netvision.net.il), Mar 14 2010