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Let c = Sum_{k>=0} 1/2^(k!), sequence gives values of terms congruent to 5 of the continued fraction for c.
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%I #14 Aug 05 2024 14:58:10

%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

%O 1,1

%A _Benoit Cloitre_, Nov 02 2002