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Second term of continued fraction for exp(n).
1

%I #16 Nov 15 2014 08:41:52

%S 1,2,11,1,2,2,1,1,11,2,7,1,2,3,2,1,1,7,1,2,2,7,4,1,2,1,1,2,23,2,2,1,

%T 15,1,1,4,3,4,1,2,4,2,2,1,28,7,3,2,10,2,6,1,1,2,1,6,1,1,3,3,9,26,3,1,

%U 2,16,2,2,1,8,1,2,1,1,69,1,1,2,2,1,5,3,1,2,2,1,1,1,1,1,68,1,1,1,1,1,2

%N Second term of continued fraction for exp(n).

%H Eric M. Schmidt, <a href="/A057213/b057213.txt">Table of n, a(n) for n = 1..10000</a>

%F floor(1/(exp(n)-floor(exp(n))))

%t Table[ContinuedFraction[Exp[n],3][[2]],{n,100}] (* _Harvey P. Dale_, Nov 15 2014 *)

%o (PARI) default(realprecision, 10000);

%o for(n=1, 10000, print(n, " ", contfrac(exp(n))[2]))

%o \\ _Eric M. Schmidt_, Mar 05 2014

%Y Cf. A138324.

%K easy,nonn

%O 1,2

%A _Leroy Quet_, Sep 29 2000