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%I #21 Mar 07 2023 08:53:06
%S 1,3,9,4,2,2,8,1,1,9,2,3,1,2,1,1,1,1,5,2,1,5,1,1,3,5,3,9,1,1,6,7,7,6,
%T 3,4,1,9,9,3,2,1,2,6,6,1,2,3,5,1,5,2,5,1,1,5,8,2,7,3,4,1,1,5,5,2,3,1,
%U 8,1,8,9,1,6,3,1,4,6,4,1,8,1,1,9,1,4,8,8,8,9,1,3,3,2,1,5,4,2,3,3,1,1,4,6,6
%N The leading digit of (10^n)!.
%C I employed R. Wm. Gosper's approximation (A090583).
%H Google-tm Answers, <a href="http://answers.google.com/answers/threadview/id/509662.html"> mathtalk-ga on Apr 16 2005 19:20 PDT</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StirlingsApproximation.html">Stirling's Approximation</a>.
%e (10^1)! = 3628800 begins with 3.
%e (10^6)! begins with 8 and (10^100)! begins with 1.
%t f[n_] := IntegerPart[ 10^FractionalPart[ N[(n*Log[n] - n + (1/2) Log[2 Pi*n + 1/3])/Log[10], 150]]]; f[1] = 1; Table[ f[10^n], {n, 0, 104}]
%o (PARI) a(n)=my(g=lngamma(10^n+1)/log(10));g-=g\1;10^g\1 \\ _Charles R Greathouse IV_, Jan 09 2013
%Y Cf. A008905, A132826.
%K nonn,base
%O 0,2
%A _Robert G. Wilson v_, Jan 09 2013
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