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Maximum length of arithmetic progression starting at n such that each term k has tau(k) = tau(n).
2

%I #12 Sep 24 2018 02:40:25

%S 1,2,3,2,5,3,7,4,2,5,11,3,13,7,6,2,17,3,19,5,7,11

%N Maximum length of arithmetic progression starting at n such that each term k has tau(k) = tau(n).

%C Implicitly, we require the difference d of the arithmetic progression to be positive.

%C a(n) <= n for all n.

%e 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.

%Y Cf. A165498, A165501, A088430.

%K hard,nonn,more

%O 1,2

%A _Hugo van der Sanden_, Sep 21 2009, Oct 09 2009

%E Extended to n=22 (taking advantage of A088430 for n=19) by _Hugo van der Sanden_, Jun 02 2015