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A165500
Maximum length of arithmetic progression starting at n such that each term k has tau(k) = tau(n).
2
1, 2, 3, 2, 5, 3, 7, 4, 2, 5, 11, 3, 13, 7, 6, 2, 17, 3, 19, 5, 7, 11
OFFSET
1,2
COMMENTS
Implicitly, we require the difference d of the arithmetic progression to be positive.
a(n) <= n for all n.
EXAMPLE
For n=4, tau(n)=3 so each term of the arithmetic progression must be the square of a prime. The difference d must be odd for n+d to qualify, in which case n+2d is even and does not qualify; so a(4)=2 is an upper bound.
CROSSREFS
KEYWORD
hard,nonn,more
AUTHOR
Hugo van der Sanden, Sep 21 2009, Oct 09 2009
EXTENSIONS
Extended to n=22 (taking advantage of A088430 for n=19) by Hugo van der Sanden, Jun 02 2015
STATUS
approved