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 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 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified October 22 07:29 EDT 2019. Contains 328315 sequences. (Running on oeis4.)