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A061906
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Obtain m by omitting trailing zeros from n; a(n) = smallest k such that k*m is a palindrome.
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4
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1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 21, 38, 18, 35, 17, 16, 14, 9, 1, 12, 1, 7, 29, 21, 19, 37, 9, 8, 1, 14, 66, 1, 8, 15, 7, 3, 13, 15, 1, 16, 6, 23, 1, 13, 9, 3, 44, 7, 1, 19, 13, 4, 518, 1, 11, 3, 4, 13, 1, 442, 7, 4, 33, 9, 1, 11, 4, 6, 1, 845, 88, 4, 3, 7, 287, 1, 11, 6, 1, 12345679, 8
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,13
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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.
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LINKS
| P. De Geest, Smallest multipliers to make a number palindromic.
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EXAMPLE
| For n = 30 we have m = 3, 1*m = 3 is a palindrome, so a(30) = 1. For n = m = 12 the smallest palindromic multiple is 21*m = 252, so a(12) = 21.
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PROG
| (ARIBAS): stop := 20000000; 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(k, " "); else write(-1, " "); end; end;
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CROSSREFS
| Cf. A050782, A062293, A061915, A061916, A061816. Values of k*m are given in A061906.
Sequence in context: A083567 A109211 A050782 * A139768 A176071 A072708
Adjacent sequences: A061903 A061904 A061905 * A061907 A061908 A061909
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KEYWORD
| base,easy,nonn
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AUTHOR
| Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Jun 25 2001
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