A112725 -id:A112725 - 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

A112726 First positive multiple of 3^n whose reverse is also a multiple of 3^n.

Original entry on oeis.org

1, 3, 9, 999, 999999999, 4899999987, 19899999972, 28999899936, 49989892689, 49999917897, 68899199886, 68899199886, 68899199886, 2678052898989, 17902896898419, 137530987695297, 189281899170567, 368055404997498

Offset: 0

Farideh Firoozbakht, Nov 13 2005

a(0)=1; a(1)=3 and it is easily shown that for n>1, 10^3^(n-2)-1 is a multiple of 3^n whose reverse is also a 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 A062567, a(n)=A062567(3^n). Jud McCranie conjectures that for n>1, a(n)=10^3^(n-2)-1 (see comments line of A062567), but we see that for n>4, a(n) is much smaller than 10^3^(n-2)-1, so his conjecture is rejected. It seems that only for n=2,3 & 4 we have, a(n)=10^3^(n-2)-1.

			a(20)=218264275944702783 because 218264275944702783=3^20*62597583
387207449572462812=3^20*111050012 & 218264275944702783 is the
smallest positive multiple of 3^20 whose reverse is also amultiple
of 3^20. I found a(n) for n<21, a(18) & a(19) are respectively
14048104419899757 & 171101619858478932.

Cf. A062567, A112725.

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

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions