|
|
A090315
|
|
Least k such that k and digit reversal of k both have n divisors, or 0 if no such number exists.
|
|
1
|
|
|
1, 2, 4, 6, 14641, 44, 0, 24, 484, 272, 0, 294, 0, 291008, 44944, 264, 0, 252, 0, 2992, 0, 2532352, 0, 2508, 10004000600040001, 2977792, 1002001, 2112, 0, 63536, 0, 4224, 0, 44356665344, 0, 2772, 0, 2380651036672, 0, 42224, 0, 6336, 0, 2937856, 698896, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
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.
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
|
|
LINKS
|
|
|
EXAMPLE
|
a(8) =24, tau(24) = tau(42) = 8.
|
|
CROSSREFS
|
|
|
KEYWORD
|
base,nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|