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Search: id:A062567
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| A062567 |
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First multiple of n whose reverse is also divisible by n, or 0 if no such multiple exists. |
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+0
4
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| 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 11, 48, 494, 252, 510, 272, 272, 216, 171, 0, 168, 22, 161, 696, 525, 494, 999, 252, 232, 0, 434, 2112, 33, 272, 525, 216, 111, 494, 585, 0, 656, 252, 989, 44, 540, 414, 141, 2112, 343, 0, 969, 676, 212, 4698, 55, 616, 171, 232, 767
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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a(81) = 999999999. 10^27-1 is a solution for a(3^5), but it may not be the smallest one. However, it seems likely (and perhaps easy to prove) that a(3^i) is 3^(i-2) "9"s, for i > 1. - Jud McCranie (j.mccranie(AT)comcast.net), Aug 07 2001
a(3^5)=4899999987<10^27-1 so Jud McCranie's conjecture "for n>1, a(3^n)=10^3^(n-2)-1 " is incorrect. I found a(3^n) for n<21; A112726 gives this subsequence. From the terms of A112726 we see that for n>4, a(3^n) is much smaller than 10^3^(n-2)-1. It seems that only for n=2,3 & 4 we have a(3^n)=10^3^(n-2)-1. - Farideh Firoozbakht (f.firoozbakht(AT)math.ui.ac.ir), Nov 13 2005
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EXAMPLE
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48 and 84 are both divisible by 12
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MATHEMATICA
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Block[{k = 1}, While[ !IntegerQ[k/n] || !IntegerQ[ FromDigits[ Reverse[ IntegerDigits[k]]]/n] && k < 10^5, k++ ]; If[k != 10^5, k, 0]]; Table[ a[n], {n, 1, 60}] (from Robert G. Wilson v)
a[n_]:=(For[m=1, !IntegerQ[FromDigits[Reverse[IntegerDigits[m*n]]]/n], m++ ]; m*n); Do[Print[a[n]], {n, 60}] (Firoozbakht)
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CROSSREFS
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Cf. A112725, A112726.
Adjacent sequences: A062564 A062565 A062566 this_sequence A062568 A062569 A062570
Sequence in context: A056967 A067079 A080434 this_sequence A069554 A020485 A083116
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KEYWORD
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base,nonn
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AUTHOR
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Erich Friedman (efriedma(AT)stetson.edu), Jul 03 2001
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