OFFSET
1,1
COMMENTS
The question about the boundedness of the sequence is equivalent to the question about its eventually periodicity. For example, the sequence defined by the same recurrence relation and initial values b(1)=2 and b(2)=5 is 2, 5, 15, 21, 25, 25, 21, 25, 25, ... so is periodic for n>=4 with the period {21,25,25}.
Problem. (a) Do there exist initial terms a(1) and a(2) depending on a given N for which the sequence has the least period of length >= N? (b) Do there exist initial terms a(1) and a(2) for which the sequence has no period?
Conjecture. Problem (a) is answered in affirmative, while problem (b) is answered in the negative.
This sequence is eventually periodic and therefore bounded: a(61)=a(85)=85 [sic] and a(62)=a(86)=73. [D. S. McNeil, Dec 14 2011]
MATHEMATICA
a[1] := 2; a[2] := 3; a[n_] := a[n] = 2(FactorInteger[a[n - 1]][[-1, 1]] + FactorInteger[a[n - 2]][[-1, 1]]) + 1; Table[a[n], {n, 50}] (* Alonso del Arte, Dec 14 2011 *)
nxt[{a_, b_}]:={b, 2(FactorInteger[a][[-1, 1]]+FactorInteger[b] [[-1, 1]])+ 1}; Transpose[NestList[nxt, {2, 3}, 120]][[1]] (* Harvey P. Dale, Dec 15 2011 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Dec 14 2011
STATUS
approved