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 Neighbors (Posted on 2014-04-18)
Find the smallest integer n such that both n-1 and n+1 have the same number of divisors as n.

 See The Solution Submitted by Ady TZIDON No Rating

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 a bit of cheating (spoiler) | Comment 3 of 8 |
Not being a programmer I figured I could still solve this without a lot of brute force on my part:

I figured semi-primes will have 4 distinct factors (unless they are squares) and are rather numerous.  So I used the OEIS:
http://oeis.org/A006881
and scanned for triples.  The central numbers are 34, 86, 94, 142, 202

http://oeis.org/A169834
Numbers n such that d(n-1) = d(n) = d(n+1).

Interestingly, the first number on this sequence that would not be in the semiprimes sequence is 231
230 = 2*5*23
231 = 3*7*11
232 = 2^3*29
each has 8 divisors.

 Posted by Jer on 2014-04-18 12:58:37

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