%I #7 May 13 2013 01:48:41
%S 0,0,1,0,1,0,1,1,0,1,0,0,0,1,1,0,1,0,0,1,0,1,1,0,0,0,0,0,2,0,1,0,1,0,
%T 1,1,0,1,1,0,1,0,0,0,2,2,0,0,0,1,0,1,1,1,1,0,1,0,0,1,2,0,0,0,2,0,1,0,
%U 0,1,1,0,0,0,0,1,0,1,1,0,1,0,0,0,0,1,0,0,0,1,1,0,1,0,0,1,0,2,0,1,0,0,0,0,1
%N a(n) = floor(p(n)*((log of p(n+2) to base p(n+1))-1)), where p(n) is n-th prime.
%C For the first 10000 terms: 1) values of terms are from 0..6. 2) numbers of occurrences of k = 0..6: {6010, 2959, 791, 197, 29, 13, 1} 3) positions of k=0..6 in the sequence: k=0: {1, 2, 4, 6, 9, 11, 12, 13, 16, 18, 19, 21, 24, ...}, k=1: {3, 5, 7, 8, 10, 14, 15, 17, 20, 22, 23, 31, 33, ...}, k=2: {29, 45, 46, 61, 65, 98, 136, 145, 149, 153, 161, 179, 204, ...}, k=3: {188, 262, 326, 366, 428, 445, 461, 589, 649, 684, 707, 731, 737, ..}, k=4: {216, 1830, 1878, 2190, 2343, 3076, 3426, 3731, 3794, 3860, 4008, 4563, 4753, ..}, k=5: {2224, 2809, 3643, 3792, 4230, 4259, 4521, 4611, 5948, 7809, 8359, 8687, 9833, ..}, k=6: {3384, ..}. Is the sequence bounded?
%C By Rankin's theorem the sequence is unbounded: in particular there are infinitely many n for which a(n) >> log log n log log log log n / (log log log n)^2. [_Charles R Greathouse IV_, Dec 20 2011]
%t a[n_]:=Floor[Prime[n]*(Log[Prime[n+1],Prime[n+2]]-1];
%o (PARI) p=2;q=3;forprime(r=5,1e3,print1(p*log(r/q)\log(q)", ");p=q;q=r) \\ _Charles R Greathouse IV_, Dec 20 2011
%K nonn
%O 1,29
%A _Zak Seidov_, Apr 01 2007