OFFSET
1,1
COMMENTS
There are only 15 numbers k such that there exist numbers M_k which, when 1 is placed at both ends of M_k, the number M_k is multiplied by k; 101 is the fourteenth such integer, so 101 = A329914(14), and a(1) = A329915(14) = 11; hence, the terms of this sequence form the infinite set {M_101}.
Every term M = a(n) has q = 6*n-4 digits, and 10^(q+1)+1 that has q = 6*n-4 zeros in its decimal expansion is equal to 91 * M, so a(n) = M is a divisor of 10^(6*n-3)+1. Example: a(2) = 10989011 has 8 digits and 91 * 10989011 = 1000000001 that has 8 zeros in its decimal expansion.
REFERENCES
D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.
LINKS
Index entries for linear recurrences with constant coefficients, signature (1000001,-1000000).
FORMULA
a(n) = (10^(6*n-3)+1)/91 for n >= 1.
EXAMPLE
101 * 11 = 1[11]1, hence 11 is a term.
101 * 10989011 = 1[10989011]1 and 10989011 is another term.
MAPLE
seq((10^(6*n-3)+1)/91, n=1..15);
MATHEMATICA
Table[(10^(6*n - 3) + 1)/91, {n, 1, 9}] (* Amiram Eldar, Feb 06 2022 *)
LinearRecurrence[{1000001, -1000000}, {11, 10989011}, 10] (* Harvey P. Dale, Sep 12 2022 *)
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Bernard Schott, Feb 05 2022
STATUS
approved