|
%I #9 Jun 07 2013 03:40:16
%S 4095,4722366482869645213695,4095,
%T 3121748550315992231381597229793166305748598142664971150859156959625371738819765620120306103063491971159826931121406622895447975679288285306290175,
%U 4095,4722366482869645213695,4095
%N Let c=sum_{k>=0} 1/2^(k!), sequence gives values of terms congruent to 5 of the continued fraction for c.
%C Observation: if b(k) denotes the sequence of all elements of the continued fraction for c, b(k)=4095 if k==6 or 19 (mod 24); b(k)=4722366482869645213695 if k==12 or 37 (mod 48) ...
%F It seems that for n>=1, a(2n-1)=4095; a(4n-2)=4722366482869645213695 etc.
%e The continued fraction for c is shown in A076157. The "big terms" are all congruent to 5.
%Y Cf. A076152, A076157, A076187.
%K nonn,cofr
%O 1,1
%A _Benoit Cloitre_, Nov 02 2002
| 