a(n) is the last occurrence of n in A037968.
a(n) >= A049363(n+1)  1 for all n.  Derek Orr, Aug 31 2014
From Derek Orr, Aug 31 2014 (Start):
At least for 1 <= n <= 5, a(n)+1 fails when written in base n^2+1. Examples:
a(1) = 1 written in base 2 is 1 (1 distinct digit). 2 written in base (21)^2+1 = 2 is 10. Thus 2 fails.
a(2) = 43 written in base 3 is 1121 (2 distinct digits). 44 written in base 2^2+1 = 5 is 134. Thus 44 fails.
a(3) = 2462 written in base 4 is 212132 (3 distinct digits). 2463 written in base 3^2+1 = 10 is 2463. Thus 2463 fails.
Generalizing... (Conjecture)
a(n) written in base n+1 has n distinct digits. a(n)+1 written in base n^2+1 will always have n+1 distinct digits.
Further, for 1 < n <= 5, a(n)1 fails when written in base n^2+1.
(End)
a(1)a(6) are confirmed for all n <= 10^11.  Hiroaki Yamanouchi, Sep 21 2014
a(6) = 2610787117 written in base 7 is 121461216151 (5 distinct digits), and 2610787118 written in base 6^2+1 = 37 is (1)(0)(24)(1)(22)(2)(0) (5 distinct digits). Therefore, Derek Orr's conjecture seems to be wrong.
a(7) >= 314941024802.  Hiroaki Yamanouchi, Sep 21 2014
