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%I #9 Jul 10 2017 10:47:51
%S 1,2,4,6,14641,44,0,24,484,272,0,294,0,291008,44944,264,0,252,0,2992,
%T 0,2532352,0,2508,10004000600040001,2977792,1002001,2112,0,63536,0,
%U 4224,0,44356665344,0,2772,0,2380651036672,0,42224,0,6336,0,2937856,698896,0
%N Least k such that k and digit reversal of k both have n divisors, or 0 if no such number exists.
%C For a(7) one needs a number of the form p^6 whose digit reversal is q^6, p, q are primes. Hence a(7) perhaps is zero (not sure). Conjecture: There are infinitely many nonzero terms as well as zeros in this sequence.
%C Zeros are unproved. I have checked for a(21) up to 10^13, a(46) up to 10^14, a(33) up to 10^18, a(39) up to 10^20, a(35) up to 10^30 and the rest (7, 11, 13, 17, 19, 23, 29, 31, 37, 41 and 43) up to at least 10^48. - _David Wasserman_, Nov 01 2005
%e a(8) =24, tau(24) = tau(42) = 8.
%Y Cf. A083753.
%K base,nonn
%O 1,2
%A _Amarnath Murthy_, Dec 01 2003
%E More terms from _David Wasserman_, Nov 01 2005