cp's OEIS Frontend

Let f(0)=m; f(n+1)=1+gpf(f(n)), where gpf(n) is the greatest prime factor of n (A006530). For any m, for sufficiently large n the sequence f(n) oscillates between 3 and 4. Given a sufficiently large n, this allows us to divide integers in two classes: C3 (m such that the sequence f(n) enters the cycle 3, 4, 3, ...) and C4 (m such that the sequence f(n) enters the cycle 4, 3, 4, ...). We present here C4 as the one that begin with 4. In A120684 we present C3 as the one that begin with 3.

Let f(0)=m; f(n+1)= c + d gpf(f(n)), where gpf(n) is the largest prime factor of n (A006530). For any m, for sufficiently large n the sequence f(n) oscillates. In A120684, A120685 the values d=c=1 were considered. Here we consider d=1, c=2 and this allows us to divide integers in 4 classes: C4 (m such that f(n)=4, which is a fixed point); C5 (m such that f(n)=5, then oscillates between 5,7,9); C7 (m such that f(n)=7, then oscillates between 7,9,5); C9 (m such that f(n)=9, then oscillates between 9,5,7); In A120686 (here) we present C5 as the one that includes 5. In A120687 we present C7 as the one that includes 7. In A120688 we present C9 as the one that includes 9.

Note that if f(n) is not prime then f(n+1)= 2 + gpf(f(n)) <= 2 + f(n)/2 and the sequence decreases. If f(n) is prime and 2+f(n) is prime, the sequence will decrease when 2k+f(n) is not prime, which must occur for k>2. The bottom limit case is the cycle (5 7 9). The only other possibility occurs for 2^k numbers that go to the fixed point 4 because 2+gpf(2^k)=2+2=4.

Let f(0)=m; f(n+1)= c + d gpf(f(n)), where gpf(n) is the largest prime factor of n (A006530). For any m, for sufficiently large n the sequence f(n) oscillates. In A120684, A120685 the values d=c=1 were considered. Here we consider d=1, c=2 and this allows us to divide integers in 4 classes: C4 (m such that f(n)=4, which is a fixed point); C5 (m such that f(n)=5, then oscillates between 5,7,9); C7 (m such that f(n)=7, then oscillates between 7,9,5); C9 (m such that f(n)=9, then oscillates between 9,5,7); In A120686 we present C5 as the one that includes 5. In A120687 (here) we present C7 as the one that includes 7. In A120688 we present C9 as the one that includes 9.

Note that if f(n) is not prime then f(n+1)= 2 + gpf(f(n)) <= 2 + f(n)/2 and the sequence decreases. If f(n) is prime and 2+f(n) is prime, the sequence will decrease when 2k+f(n) is not prime, which must occur for k>2. The bottom limit case is the cycle (5 7 9). The only other possibility occurs for 2^k numbers that go to the fixed point 4 because 2+gpf(2^k)=2+2=4.

Let f(0)=m; f(n+1)= c + d gpf(f(n)), where gpf(n) is the largest prime factor of n (A006530). For any m, for sufficiently large n the sequence f(n) oscillates. In A120684, A120685 the values d=c=1 were considered. Here we consider d=1, c=2 and this allows us to divide integers in 4 classes: C4 (m such that f(n)=4, which is a fixed point); C5 (m such that f(n)=5, then oscillates between 5,7,9); C7 (m such that f(n)=7, then oscillates between 7,9,5); C9 (m such that f(n)=9, then oscillates between 9,5,7); In A120686 we present C5 as the one that includes 5. In A120687 we present C7 as the one that includes 7. In A120688 (here) we present C9 as the one that includes 9.

Note that if f(n) is not prime then f(n+1)= 2 + gpf(f(n)) <= 2 + f(n)/2 and the sequence decreases. If f(n) is prime and 2+f(n) is prime, the sequence will decrease when 2k+f(n) is not prime, which must occur for k>2. The bottom limit case is the cycle (5 7 9). The only other possibility occurs for 2^k numbers that go to the fixed point 4 because 2+gpf(2^k)=2+2=4.