login
a(1)=1, a(n)=ceiling(n*(a(n-1)+3/a(n-1))).
0

%I #9 Sep 12 2013 15:28:52

%S 1,8,26,105,526,3157,22100,176801,1591210,15912101,175033112,

%T 2100397345,27305165486,382272316805,5734084752076,91745356033217,

%U 1559671052564690,28074078946164421,533407499977124000,10668149999542480001

%N a(1)=1, a(n)=ceiling(n*(a(n-1)+3/a(n-1))).

%C More generally if m is an integer >=3 and a(1)=1, a(n)=ceiling(n*(a(n-1)+m/a(n-1))) there is a closed formula for a(n) namely : a(n)=floor(n!*(e+m-4/3))

%F a(n)=floor(n!*(exp(1)+5/3))

%t nxt[{n_,a_}]:={n+1,Ceiling[(n+1)(a+3/a)]}; Transpose[NestList[nxt,{1,1},20]][[2]] (* _Harvey P. Dale_, Sep 12 2013 *)

%K nonn

%O 1,2

%A _Benoit Cloitre_, May 06 2003

%E Offset changed by _Harvey P. Dale_, Sep 12 2013