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%I #42 Oct 02 2022 23:04:53
%S 2,0,0,8,5,5,3,6,9,2,3,1,8,7,6,6,7,7,4,0,9,2,8,5,2,9,6,5,4,5,8,1,7,1,
%T 7,8,9,6,9,8,7,9,0,7,8,3,8,5,5,4,1,5,0,1,4,4,3,7,8,9,3,4,2,2,9,6,9,8,
%U 8,4,5,8,7,8,0,9,1,9,7,3,7,3,1,2,0,4,4,9,7,1,6,0,2,5,3,0,1,7,7,0
%N Decimal expansion of e^3.
%C Also where x^(1/x^(1/3)) is a maximum. - _Robert G. Wilson v_, Oct 22 2014
%H Harry J. Smith, <a href="/A091933/b091933.txt">Table of n, a(n) for n = 2..20000</a>
%H <a href="/index/Tra#transcendental">Index entries for transcendental numbers</a>
%F From _Peter Bala_, Jan 12 2022: (Start)
%F e^3 = Sum_{n >= 0} 3^n/n!. Faster converging series include
%F e^3 = 18*Sum_{n >= 0} 3^n/(p(n-1)*p(n)*n!), where p(n) = n^2 - 3*n + 5 and
%F e^3 = -162*Sum_{n >= 0} 3^n/(q(n-2)*q(n-1)*n!), where q(n) = n^3 + 8*n - 3.
%F e^3 = 22 - Sum_{n >= 0} 3^(n+4)/((n+3)^2*(n+4)^2*n!) and
%F 22/e^3 = 1 - 2*Sum_{n >= 0} (-3)^(n+2)*n^2/(n+3)!.
%F e^3 = lim_{n -> oo} f(n+2)*f(n)/(n^2*f(n+1)^2), where f(n) = n^(n^2). Compare with e = lim_{n -> oo} g(n+1)/(n*g(n)), where g(n) = n^n. (End)
%e exp(3) = e^3 = 20.0855369231876677409285296545817178969879... - _Harry J. Smith_, Apr 30 2009
%p Digits:=100: evalf(exp(3)); # _Wesley Ivan Hurt_, Jul 07 2014
%t RealDigits[E^3, 10, 100][[1]] (* _Alonso del Arte_, Jul 07 2014 *)
%o (PARI) default(realprecision, 20080); x=exp(3)/10; for (n=2, 20000, d=floor(x); x=(x-d)*10; write("b091933.txt", n, " ", d)); \\ _Harry J. Smith_, Apr 30 2009
%Y Cf. A001113, A072334, A092426, A092511, A092512, A092513.
%K easy,nonn,cons
%O 2,1
%A _Mohammad K. Azarian_, Mar 16 2004
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