cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A080434 Duplicate of A062567.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 11, 48, 494, 252, 510, 272, 272, 216, 171, 0, 168, 22, 161, 696
Offset: 1

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Author

Keywords

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

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Author

Farideh Firoozbakht, Nov 12 2005

Keywords

Comments

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

Examples

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

Crossrefs

Programs

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

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

Author

Farideh Firoozbakht, Nov 13 2005

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

Programs

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