A112726 -id:A112726 - cp's OEIS frontend

A062567 First multiple of n whose reverse is also divisible by n, or 0 if no such multiple exists.

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

Offset: 1

Erich Friedman, Jul 03 2001

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

			48 and 84 are both divisible by 12.

Cf. A112725, A112726.

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 *)

Offset corrected by Sean A. Irvine, Apr 03 2023

A112725 Smallest positive palindromic multiple of 3^n.

Original entry on oeis.org

1, 3, 9, 999, 999999999, 29799999792, 39789998793, 39989598993, 68899199886, 68899199886, 68899199886, 68899199886, 68899199886, 2699657569962, 146189959981641, 191388777883191, 191388777883191, 18641845754814681

Offset: 0

Farideh Firoozbakht, Nov 12 2005

a(0)=1; a(1)=3 and it is easily shown that for n>1, 10^3^(n-2)-1 is a palindromic multiple of 3^n(see comments line of A062567). So for each n, a(n) exists and for n>1, a(n)<=10^3^(n-2)-1. This sequence is a subsequence of A020485(a(n)=A020485(3^n)) and for all n, A062567(3^n)<=a(n) because for all n, A062567(n)<= A020485(n). Jud McCranie conjectures that for n>1 A062567(3^n) =10^3^(n-2)-1, if his conjecture were true then from the above facts we conclude that for n>1 a(n)=10^3^(n-2)-1, but we see that for 4

			a(18)=18771463736417781 because 18771463736417781=3^18*48452429 is the smallest positive palindromic multiple of 3^18.

Cf. A020485, A062567, A112726.

b[n_]:=(For[m=1, !FromDigits[Reverse[IntegerDigits[m*n]]]==m*n, m++ ]; m*n);Do[Print[b[3^n]], {n, 0, 18}]

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cp's OEIS Frontend

A062567 First multiple of n whose reverse is also divisible by n, or 0 if no such multiple exists.

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