OFFSET
1,2
COMMENTS
Initial conditions: a(1)=1; a(2)=1; a(3)=4;
With a(3)=3, the recursion gives the positive integers.
MATHEMATICA
a[1]:= 1; a[2]:= 2; a[3]:= 4;
a[n_Integer]:=a[n]=a[n-1]+ Times@@Table[(a[n-2i+1]-a[n-2i]), {i, 1, (n-1)/2}];
Table[a[n], {n, 1, 15}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Olivier Gérard, Aug 24 2013
STATUS
approved