A083964 Smallest palindromic multiple of 2n-1 beginning with the digit string of 2n-1; or 0 if no such number exists.

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

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=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

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, Jun-Jul (1998), pp. 259-262.

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