

A083964


Smallest palindromic multiple of 2n1 beginning with the digit string of 2n1; or 0 if no such number exists.


2



1, 3, 5, 7, 9, 11, 131131, 0, 174471, 1911191, 219912, 238832, 0, 27972, 2930392, 31713, 33, 0, 3701073, 393393, 412214, 4357534, 0, 476674, 49294, 513315, 5305035, 55, 579975, 59295, 6188816, 6317136, 0, 6735376, 6933396, 7109017, 7305037, 0, 77, 793397
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OFFSET

1,2


COMMENTS

Conjecture: If n is not of the type (5k+1)/2 then a(n) is not zero.
There are two requirements: A) the number must be a multiple of k and end with the digits of reverse(k), and B) it must be palindromic. A) Let k=2n1 and k has c digits. We seek an m such that m*k == reverse(k) (mod 10^c). This equation has solutions iff GCD(k,10^c)  reverse(k), or since k is odd, iff GCD(k,5^c)  reverse(k). B) When m exists, make an arithmetic progression a0 + i*d, i=0,1,2,... with a0=m*k and d=k*10^c. All terms end in reverse(k) and are divisible by k. While d is divisible by 10, a0 is not, since reverse(k) is never divisible by 10. According to Harminc and Sotaks theorem: "An arithmetic progression a0 + i*d contains infinitely many palindromic numbers unless both a0 and d are multiples of ten", we conclude that when m exists, there also exists a palindrome with the required properties. Example: k=5015 has a solution since GCD=5 which divides 5105, but k=5025 has not since GCD=25 which does not divide 5205.  Lars Blomberg, Sep 22 2013


LINKS

Lars Blomberg, Table of n, a(n) for n = 1..4090
M. Harminc and R. Sotak, Palindromic numbers in arithmetic progressions, Fibonacci Quarterly Journal, JunJul (1998), pp. 259262.


CROSSREFS

Cf. A083965.
Sequence in context: A046497 A061512 A356750 * A340439 A131628 A079091
Adjacent sequences: A083961 A083962 A083963 * A083965 A083966 A083967


KEYWORD

base,nonn


AUTHOR

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 20 2003


EXTENSIONS

Name corrected and a(11)a(40) from Lars Blomberg, Sep 22 2013


STATUS

approved



