OFFSET
1,2
FORMULA
a(n)=(1/5)[...[[[[n(2/1)](3/2)](4/3)](5/4)](7/6)]...(k+1)/k]..., k>0 (mod 5), where [x] = floor of x; this infinite nested floor product will eventually level-off at a(n).
EXAMPLE
a(1)=1 since (1/5)[[[[1(2/1)](3/2)](4/3)](5/4)]=1
MATHEMATICA
f[n_] := Block[{k = 1, p = n}, While[q = Floor[p*(k + 1)/k]; q != p, p = q; k++; If[ Mod[k, 5] == 0, k++ ]]; p/5]; Table[ f[n], {n, 1, 37}] (* Robert G. Wilson v *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul D. Hanna, Jul 29 2002
EXTENSIONS
More terms from Robert G. Wilson v, Dec 27 2003
STATUS
approved