A120686 Integers m such that the sequence defined by f(0)=m and f(n+1)=2+gpf(f(n)), with gpf(n) being the greatest prime factor of n (A006530), ends up in the period 3 cycle 5 -> 7 -> 9 -> 5 -> ...

5, 10, 15, 19, 20, 25, 30, 31, 38, 40, 45, 47, 50, 53, 57, 60, 61, 62, 75, 76, 80, 90, 93, 94, 95, 97, 100, 103, 106, 109, 114, 120, 122, 124, 125, 133, 135, 141, 149, 150, 152, 155, 159, 160, 163, 171, 173, 180, 183, 186, 188, 190, 191, 194, 199, 200

Offset: 0

Author

Carlos Alves, Jun 25 2006

Keywords

• nonn

Comments

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.

Examples

			Oscillation between 5,7,9: 2+gpf(5)=2+5=7; 2+gpf(7)=2+7=9; 2+gpf(9)=2+3=5.
Fixed point is 4: 2+gpf(4)=2+2=4.
		

Crossrefs

Cf. A120685, A120684, A072268, A006530.

Programs

Mathematica

fi = Function[n, FactorInteger[n][[ -1, 1]] + 2]; mn = Map[(NestList[fi, #, 6][[ -1]]) &, Range[2, 200]]; Cc4 = Flatten[Position[mn, 4]] + 1;Cc5 = Flatten[Position[mn, 5]] + 1; Cc7 = Flatten[Position[mn, 7]] + 1;Cc9 = Flatten[Position[mn, 9]] + 1; Cc5