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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.
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%I #22 Jan 02 2023 12:30:48

%S 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,

%T 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,

%U 2,1,1,2,2,2,1,1,1,3,1,1,1,1,1,1,3,1

%N 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.

%C Question. Is the sequence bounded?

%C 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.

%H V. Shevelev, <a href="http://list.seqfan.eu/oldermail/seqfan/2011-October/015755.html">Collatz-like problem with prime iterations</a>

%e 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.

%t 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 *)

%o (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

%Y Cf. A074473, A126241.

%K nonn

%O 3,1

%A _Vladimir Shevelev_, Oct 29 2011