

A165498


Maximum length of arithmetic progression with difference n, such that each term k has tau(k) = n


3




OFFSET

1,2


COMMENTS

a(n) = 1 for all odd n.
9 <= a(6) <= 13; a(8) = 17; 5 <= a(10) <= 8; 10 <= a(12) <= 103. (Improved bounds.  Hugo van der Sanden, Nov 29 2016)
a(10) >= 5, as witnessed by 43920665884407841463671+10*j, for j=0..4.  Giovanni Resta, Jul 28 2013


LINKS

Table of n, a(n) for n=1..5.


EXAMPLE

When tau(k) = 4, k cannot be divisible by 9 unless k = 27. An arithmetic progression of 9 terms with difference 4 must have a term divisible by 9, and k=27 is not part of a progression of 9 terms with tau(k)=4, so a(4) must be less than 9. Since a progression of 8 terms is achievable (eg starting at 5989), a(4) = 8 is proved.


CROSSREFS

Cf. A064491, A165497, A165499, A165500
Sequence in context: A084246 A141252 A276168 * A195731 A154294 A059526
Adjacent sequences: A165495 A165496 A165497 * A165499 A165500 A165501


KEYWORD

hard,nonn,more


AUTHOR

Hugo van der Sanden, Sep 21 2009


STATUS

approved



