OFFSET
1,1
COMMENTS
Start with a number x and construct a successor by the following iterative procedure: first remove all factors 2, 3, 5, ..., p(k) from x, where p(k) is the k-th prime number. When no further such factors remain then take the number ([p(k+1)*x]+1)/2 as the successor. The (3x+1) problem is the special case k=1 in the sequence that lists the p(k+1) leading to a one-cycle.
For other primes there is at least 1 supplementary cycle: e.g. when p(k+1)=11 there is also a cycle starting with 17; when p(k+1)=19 there is also a cycle starting with 46063; when p(k+1)=61 there are 3 supplementary cycles starting resp. with 97, 199, 26833; etc.
MATHEMATICA
v[n_, k_]:=Block[{m=n}, Do[While[Mod[m, Prime[i]]==0, m=m/Prime[i]], {i, k}]; If[m!=1, Prepend[v[m*Prime[k+1]+1, k], m], v[m, k]={1}]] b[r_, s_, t_]:=Table[v[n, r], {n, s, t}]
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Herman Roelants, Oct 11 2004
STATUS
approved