cp's OEIS Frontend

A080095 Let sum(k>=0, k^n/(2*k+1)!) = (x(n)*e + y(n)/e)/z(n), where x(n) and z(n) are >0, then a(n)=z(n).

%I A080095 #10 Apr 09 2012 01:35:57
%S A080095 2,8,16,16,64,128,128,512,1024,1024,4096,8192,8192,32768,65536,65536,
%T A080095 262144,524288,524288,2097152,4194304,4194304,16777216,33554432,
%U A080095 33554432,134217728,268435456,268435456,1073741824,2147483648
%N A080095 Let sum(k>=0, k^n/(2*k+1)!) = (x(n)*e + y(n)/e)/z(n), where x(n) and z(n) are >0, then a(n)=z(n).
%H A080095 <a href="/index/Di#divseq">Index to divisibility sequences</a>
%F A080095 a(n) = 2^b(n) and {b(n)}={1, 3, 4, 4, 6, 7, 7, 9, 10, 10, 12, 13, 13, 15, ..} where b(3n-2)=3n-2, b(3n-1)=3n, b(3n)=b(3n+1)=3n+1, for n>0.
%o A080095 (PARI) apply(n->2^n,Vec((1+2*x+x^2-x^3)/(1-x-x^3+x^4)+O(x^30))) \\ _Charles R Greathouse IV_, Apr 09 2012
%Y A080095 Cf. A080093, A080094.
%K A080095 nonn,easy
%O A080095 1,1
%A A080095 _Benoit Cloitre_ and _Paul D. Hanna_, Jan 28 2003