The following fractions are written on the board 1/n, 2/(n-1), 3/(n-2), ... , n/1 where n is a natural number. Alice calculated the differences of the neighboring fractions in this row and found among them 10000 fractions of type 1/k (with natural k). Prove that he can find even 5000 more of such these differences.
I proved none of the speculations below. It's just what I was able to uncover as plausible.
Finding the number of unit fractions for small n gives (starting with 1)
0,0,0,0,2,0,0,0,2,0,2,0,2,2,0,0,2,0,2,2,2,0,2,0,2,0,2,0,6,...
Which leads to http://oeis.org/A087893
Number of numbers m satisfying 1 < m < n such that m^2 == m (mod n)
Which seems to fit the bill (with offset 1)
The first 2 with with n=5 for this problem:
3/3-2/4=1/2 and 4/2-3/3=1
The first 2 in the OEIS sequence is for n=6 (m=3 and m=4)
3^2=9==3mod6
4^2=16==4mod6
Looking at the 5000 terms given in the OEIS sequence, the first appearance of other values besides 2 and 6 are
14 at n=210
30 at n=2730
The sequence 0,2,6,14,30 appears to be 2^n - 2
If this is the case then it continues to the 13th term 8190 and 14th term 16382. In other words any term above 10000 is also above 15000.
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Since I'm sleuthing, how about the numbers for n that give records in the sequence:
6,30,210,2730,...
This gives 3 hits in OEIS. The most likely is http://oeis.org/A093449
Least number with n distinct prime divisors arising as the product of two or more consecutive integers
If so, the first n in the original problem with over 10000 fraction of type 1/k is the 14th term minus one:
660393717163700520-1=660393717163700519
Which would require a really, really big board!
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Posted by Jer
on 2021-02-20 12:09:44 |
Just some guesses
