Back, Greg; Caragiu, Mihai The greatest prime factor and recurrent sequences. (English) Zbl 1221.11032 Fibonacci Q. 48, No. 4, 358-362 (2010). The authors consider sequences \(p_j = \text{gpf}(a_1p_j - 1+a_2p_{j - 2}+\cdots +a_dp_{j - d}+a_0)\), where for any integer \(x\geq 2\), gpf(\(x\)) denotes the greatest prime factor of \(x\). In the simple case of the ‘GPF-Fibonacci’ sequences corresponding to \(d = 2, a_0 = 0\), and \(a_1 = a_2 = 1\), they find that regardless of the initial conditions \(p_0\) and \(p_1\), all such sequences ultimately enter the cycle 7, 3, 5, 2. A computational exploration of the ‘GPF-Tribonacci’ analogue \(d = 3, a_0 = 0\), and \(a_1 = a_2 = a_3 = 1\) reveals four cycles of lengths, listed in the decreasing order of frequencies, 100, 212, 28 and 6, with the two larger cycles collecting more than 98% of the sequences as defined by the initial conditions \(p_0, p_1\), and \(p_2\). The paper concludes with a general ultimate periodicity conjecture and discusses its plausibility. Reviewer: Florin Nicolae (Berlin) MSC: 11B37 Recurrences 11A41 Primes Keywords:recurrent sequence; greatest prime factor PDFBibTeX XMLCite \textit{G. Back} and \textit{M. Caragiu}, Fibonacci Q. 48, No. 4, 358--362 (2010; Zbl 1221.11032) Full Text: Link Online Encyclopedia of Integer Sequences: Gpf(n): greatest prime dividing n, for n >= 2; a(1)=1. a(1)=a(2)=1; thereafter a(n) = gpf(a(n-1)+a(n-2)), where gpf = ”greatest prime factor”. a(1)=a(2)=a(3)=1; thereafter a(n) = gpf(a(n-1)+a(n-2)+a(n-3)), where gpf = ”greatest prime factor”. a(1)=a(2)=a(3)=a(4)=1; thereafter a(n) = gpf(a(n-1)+a(n-2)+a(n-3)+a(n-4)), where gpf is the greatest prime factor. Like A177904, but start with a(1)=0, a(2)=a(3)=1. a(0)=0, a(1)=1; thereafter a(n) = gpf(2*a(n-1)+a(n-2)), where gpf = ”greatest prime factor” (A006530).