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 A062567 First multiple of n whose reverse is also divisible by n, or 0 if no such multiple exists. 3
 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; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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, 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, Nov 13 2005 LINKS EXAMPLE 48 and 84 are both divisible by 12. MATHEMATICA 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}] (* 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}] (* Farideh Firoozbakht *) CROSSREFS Cf. A112725, A112726. Sequence in context: A321243 A067079 A080434 * A069554 A347346 A020485 Adjacent sequences: A062564 A062565 A062566 * A062568 A062569 A062570 KEYWORD base,nonn AUTHOR Erich Friedman, Jul 03 2001 STATUS approved

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Last modified November 27 09:16 EST 2022. Contains 358367 sequences. (Running on oeis4.)