A198724 Let P(n) be the maximal prime divisor of 3*n+1. Then a(n) is the smallest number of iterations of P(n) such that the a(n)-th iteration < n, and a(n) = 0, if such number does not exist.

2, 3, 1, 6, 4, 1, 1, 6, 3, 2, 1, 2, 2, 1, 1, 1, 2, 3, 1, 3, 1, 2, 1, 2, 6, 1, 1, 1, 4, 3, 1, 2, 2, 3, 1, 1, 5, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 3, 1, 6, 1, 2, 2, 1, 1, 1, 4, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 3, 2, 1, 1, 2, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 3, 1

Offset: 3

Author

Vladimir Shevelev, Oct 29 2011

Keywords

• nonn

Comments

Question. Is the sequence bounded?

By private communication from Alois P. Heinz, the places of records are 3, 4, 6, 286, 29866 with values 2, 3, 6, 8, 10. No more up to 46000000.

Examples

			For n=52 we have iterations: P^(1)=157, P^(2)=59, P^(3)=89, P^(4)=67, P^(5)=101, P^(6)=19<52. Thus a(52)=6.
		

Links

• V. Shevelev, Collatz-like problem with prime iterations

Crossrefs

Cf. A074473, A126241.

Programs

Mathematica

P[n_] := FactorInteger[3*n + 1][[-1, 1]]; Table[k = 1; m = n; While[m = P[m]; m >= n, k++]; k, {n, 3, 100}] (* T. D. Noe, Oct 30 2011 *)

PARI

a(n) = {nb = 1; na = n; while((nna=vecmax(factor(3*na+1)[,1])) >= n,na = nna; nb++); nb;} \\ Michel Marcus, Feb 06 2016