%I #16 Jan 22 2019 06:31:39

%S 1,3,5,7,9,11,131131,0,174471,1911191,219912,238832,0,27972,2930392,

%T 31713,33,0,3701073,393393,412214,4357534,0,476674,49294,513315,

%U 5305035,55,579975,59295,6188816,6317136,0,6735376,6933396,7109017,7305037,0,77,793397

%N Smallest palindromic multiple of 2n-1 beginning with the digit string of 2n-1; or 0 if no such number exists.

%C Conjecture: If n is not of the type (5k+1)/2 then a(n) is not zero.

%C 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=2n-1 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

%H Lars Blomberg, <a href="/A083964/b083964.txt">Table of n, a(n) for n = 1..4090</a>

%H M. Harminc and R. Sotak, <a href="https://www.fq.math.ca/Scanned/36-3/harminc.pdf">Palindromic numbers in arithmetic progressions</a>, Fibonacci Quarterly Journal, Jun-Jul (1998), pp. 259-262.

%Y Cf. A083965.

%K base,nonn

%O 1,2

%A _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 20 2003

%E Name corrected and a(11)-a(40) from _Lars Blomberg_, Sep 22 2013