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%I #63 Mar 18 2015 08:49:56
%S 0,0,0,0,1,0,1,0,1,1,0,0,1,0,1,1,2,0,0,1,2,0,1,0,1,2,1,1,1,0,2,0,1,1,
%T 0,1,1,1,0,1,1,1,1,0,1,2,0,0,1,1,2,2,0,0,2,1,2,0,1,0,1,1,2,1,3,0,2,1,
%U 0,0,1,1,2,1,0,0,0,2,0,2,1,2,1,0,3,1,0,1,1,0,2,2,2,0,0,0,1,2,1,3,1,0,0,2,2
%N Number of bases less than sqrt(n) in which n is a palindrome.
%C Is there a number m such that a(n) > 0 for all n > m? I call the set of numbers for which a(n)=0 "unkempt" for refusing to use a mirror in any base. Is there an infinite number of unkempt numbers? a(n) can be arbitrarily large.
%C The sequence A123586 gives the values of n where a(n)=0. - _Robert G. Wilson v_, Nov 01 2014
%C Is there a closed-form formula for this function? - _Robert G. Wilson v_, Nov 01 2014
%C From _Robert G. Wilson v_, Nov 26 2014: (Start)
%C The first occurrence, beginning at 0, of n is: 1, 5, 17, 65, 121, 562, 1432, 1477, 4369, 36582, 35101, 86677, 83161, 360361, 291721, 720721, 887041, 1496881, 1670761, 3931201, 3341521, 5654881, 7207201, 7761601,...
%C Positions where a(n)=k:
%C k = 0: A123586;
%C k = 1: 5, 7, 9, 10, 13, 15, 16, 20, 23, 25, 27, 28, 29, 33, 34, 36, 37, 38, 40, ...;
%C k = 2: 17, 21, 26, 31, 46, 51, 52, 55, 57, 63, 67, 73, 78, 80, 82, 91, 92, 93, 98, ...;
%C k = 3: 65, 85, 100, 130, 154, 164, 170, 178, 191, 195, 203, 209, 242, 282, 292, ...;
%C k = 4: 121, 235, 255, 257, 273, 300, 325, 341, 343, 373, 400, 495, 601, 610, 626, 666, ...;
%C k = 5: 562, 676, 771, 819, 1009, 1111, 1220, 1333, 1365, 1441, 1543, 1978, 1981, 2000, ...;
%C k = 6: 1432, 2380, 2666, 2925, 3280, 4035, 4095, 4161, 4225, 4401, 4525, 4561, 4681, ...;
%C k = 7: 1477, 4097, 4591, 7141, 7993, 8191, 9640, 10081, 10297, 10626, 10858, 11761, ...; etc.
%C (End)
%H Robert G. Wilson v, <a href="/A100563/b100563.txt">Table of n, a(n) for n = 1..1000</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Unary_numeral_system">Base 1</a>
%F a(n) = A135551(n) - A033831(n). - _Robert G. Wilson v_, Nov 01 2014
%e 100 is a palindrome in bases 3, 7 and 9, so a(100) = 3.
%t f[n_] := Module[{p}, Table[ p = IntegerDigits[n, b]; If[p == Reverse@ p, {b, p}, Sequence @@ {}], {b, 2, Sqrt@ n}]]; Array[ Length@ f@# &, 105] (* _Robert G. Wilson v_, Nov 01 2014 *)
%o (PARI) a(n) = {my(nb = 0); for (b=2, sqrt(n), d = digits(n, b); nb+= (Vecrev(d) == d);); nb;} \\ _Michel Marcus_, Nov 05 2014
%Y Cf. A060873, A060874, A060875, A060876, A060877, A060878, A060879, A060947, A060948, A060949, A123586.
%Y Cf. A000042.
%K easy,base,nonn
%O 1,17
%A _Gordon Hamilton_, Nov 29 2004
%E a(58) from _Robert G. Wilson v_, Nov 05 2014
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