

A061816


Obtain m by omitting trailing zeros from n (cf. A004151); a(n) = smallest multiple k*m which is a palindrome.


6



0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 252, 494, 252, 525, 272, 272, 252, 171, 2, 252, 22, 161, 696, 525, 494, 999, 252, 232, 3, 434, 2112, 33, 272, 525, 252, 111, 494, 585, 4, 656, 252, 989, 44, 585, 414, 141, 2112, 343, 5, 969, 676, 212, 27972, 55, 616, 171, 232
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OFFSET

0,3


COMMENTS

Every positive integer is a factor of a palindrome, unless it is a multiple of 10 (D. G. Radcliffe, see links).
Every integer n has a multiple of the form 99...9900...00. To see that n has a multiple that's a palindrome (allowing 0's on the left) with even digits, let 9n divide 99...9900...00; then n divides 22...2200...00.  Dean Hickerson, Jun 29 2001


LINKS

Table of n, a(n) for n=0..58.
P. De Geest, Smallest multipliers to make a number palindromic.


EXAMPLE

For n = 30 we have m = 3, 1*m = 3 is a palindrome, so a(30) = 3. For n = m = 12 the smallest palindromic multiple is 21*m = 252, so a(12) = 252.


PROG

(ARIBAS): stop := 200000; for n := 0 to maxarg do k := 1; test := true; while test and k < stop do mp := omit_trailzeros(n)*k; if test := mp <> int_reverse(mp) then inc(k); end; end; if k < stop then write(mp, " "); else write(1, " "); end; end;


CROSSREFS

Cf. A050782, A062293, A061915, A061916. Values of k are given in A061906.
Sequence in context: A084011 A004086 A121760 * A083960 A138795 A297233
Adjacent sequences: A061813 A061814 A061815 * A061817 A061818 A061819


KEYWORD

base,easy,nonn


AUTHOR

Klaus Brockhaus, Jun 25 2001


STATUS

approved



