A072507
David Wilson
davidwwilson at comcast.net
Fri Jul 28 03:45:20 CEST 2006
a(10) = 3.
Proof:
The smallest exemplar showing a(10) >= 3 is 7939375.
To show a(10) <= 3, suppose there were 4 numbers n in a row with d(n) = 10.
Two of these numbers, a and b = a+2, would be even, with d(a) = d(b) = 10.
Since d(a) = d(b) = 10, a and b must each be of the form p^9 or pq^4 with p
and q prime.
If either a or b is of the form p^9, then, being even, it must be equal to
2^9 = 512. But then the other does not have 10 divisors.
So a and b are both of the form pq^4.
a and b are even, so they must be of the form 2p^4 or 16p, with p and q odd
primes.
This implies that a and b must each be congruent to 2 or 16 modulo 32.
This is inconsistent with a+2 = b.
Hence, no two consecutive even integers n have d(n) = 10.
Therefore, at most three consecutive integers n have d(n) = 10.
This gives a(10) <= 3, so a(10) = 3.
Maybe this argument could be generalized for a(2p)?
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