no to leading zeros

[Edwin Clark]

On Sat, 24 May 2003, N. J. A. Sloane wrote:

> here's the revised version of that sequence:
> 
> 
> %I A083960
> %S A083960 1,2,3,4,5,6,7,8,9,0,11,2112,131131,4441444,5115,61616,
> %T A083960 7111117,8118,99199,0,1121211,22,32223,4224,525,262262,722277772227,
> %U A083960 828828,922229,0
> %N A083960 Smallest palindromic multiple of n using only and all digits of n; or 0 if no such number exists.
> %Y A083960 Cf. A083961.
> %K A083960 base,more,nonn,new
> %O A083960 1,2
> %A A083960 Amarnath Murthy and Meenakshi Srikanth (amarnath_murthy(AT)yahoo.com), May 20 2003
> %E A083960 More terms from Edwin Clark, May 24, 2003.
> 

Assuming that a(n) = 0 when n mod 10 = 0 here's more of this
sequence. (You may not want to add so many terms. Note that if one
generates all palindromes using all the digits of n with k
digits where k starts with the number of digits of n and increases k till
one finds a multiple of n, one finds such palindromic multiples
much faster than looking for palindromes among the multiples of n. The
following were generated in about 1 min. Even finding the 22 digit
multiple 1888888888888888888881 of 81 took only about 1.6 sec.)
This begins where the above left off:

  133331, 23232, 33, 4333334, 535535, 6336, 333777333, 83888838,

        393393, 0, 11441114411, 2242422, 34443, 44, 5445, 64446,

        7477747, 8448, 9949499, 0, 1555551, 252252, 53335553335,

        444555555444, 55, 656656, 5755575, 855558, 9995995999, 0,

        11166111, 266662, 3363633, 46464, 565565, 66, 7667667667,

        8666668, 96669, 0, 1711171, 27277272, 37733773, 444777444,

        5775, 67666676, 77, 877887788778, 77977977, 0,

        1888888888888888888881, 22882228822, 333888333, 4484844,

        5888885, 68886, 8777778, 88, 8888998888, 0, 191191, 29992,

        399993, 4944494, 59555595, 69696, 779977, 8898988, 99, 0, 101,

        2101221012, 31000013, 40144104, 510000015, 610161016,

        701171107, 80111888811108, 911010119, 0, 111, 2112222112,

        13133333131, 411141114, 555111555, 61161116116, 171171,

        818111818, 111999111, 0, 121, 2112222112, 1233321, 421414124,

        521125, 22166266122, 2177777712, 211818112, 2211991122, 0,

        131, 213312, 311311113113, 4133333314, 513333315, 631363136,

        731717137, 813111318, 9313139, 0, 141, 2414142, 13144131,

        441414144, 541111145, 4616116164, 174414471, 4811184,

        19944991, 0, 151, 2155512, 513315, 41155114, 5115, 6156666516,

        1715555171, 8555115558, 59911995, 0, 161, 2116666661666666112,

        631136, 46166164, 516615, 666616616666, 6167117616, 861168,

        199696991, 0, 171, 271717172, 173333371, 474141474, 57155175,

        61677616, 777171777, 81178187118, 7199917, 0, 181, 2118228112,

        133838331, 441888144, 51518581515, 661868166, 18711781,

        88181118188, 18999981, 0, 191, 219222912, 19933133991,

        491111194, 59911995, 6961111696, 199797991, 819918,

        9911991199, 0


I continued the program up to n = 500 just to see how well the conjecture
that for n such that n mod 10 >0 there is always a palindromic multiple
with the same set of digits as n. I found no problem numbers. The record
size up to 500 is a(411) = 111111111441144111111111 which has 24 digits.

Edwin