Search: id:A069943
Results 1-1 of 1 results found.

%I A069943
%S A069943 1,1,1,2,5,13,19,29,191,131,1187,2231,17519,71063,29881,323423,2887921,
%T A069943 13237457,2397389,15030317,742458253,3748521653,9670072483,25451905333,
%U A069943 10932619111,78684575461,4163946939067,11799518538967,136025604432743
%N A069943 Let b(1)=b(2)=1, b(n+2)=(1/(n+1))*(b(n+1)+b(n)); then a(n)=numerator(b(n)).
%C A069943 Sum(k=1,infinity,b(k))=e^(3/2) where e=2,718... More generally if b(1)=b(2)=...=b(m)=1 
               and b(n+m+1)=1/(n+m)*(b(n+m)+b(n+m-1)+...+b(n)) then Sum(k=1,infinity,
               b(k))=e^H(m) where H(m)=1+1/2+1/3+...+1/m is the m-th harmonic number 
               (Benoit Cloitre and Boris Gourevitch).
%F A069943 Numerators in the power series of exp(x+x^2/2) (EGF for involutions, 
               cf. A000085). exp(x+x^2/2) = 1 + x + x^2 + 2/3*x^3 + 5/12*x^4 + 13/
               60*x^5 + 19/180*x^6 + 29/630*x^7 + 191/10080*x^8 + ... - Joerg Arndt 
               (arndt(AT)jjj.de), May 10 2008
%Y A069943 Cf. A069944.
%Y A069943 a(n)/A069944(n) = A000085(n-1)/A000142(n-1) in lowest terms. (Christian 
               G. Bower, Jan 14 2006).
%Y A069943 Sequence in context: A019390 A073770 A077545 this_sequence A094158 A068374 
               A068371
%Y A069943 Adjacent sequences: A069940 A069941 A069942 this_sequence A069944 A069945 
               A069946
%K A069943 easy,frac,nonn
%O A069943 1,4
%A A069943 Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 27 2002

Search completed in 0.001 seconds