A351239 Numbers M such that 101 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

11, 10989011, 10989010989011, 10989010989010989011, 10989010989010989010989011, 10989010989010989010989010989011, 10989010989010989010989010989010989011, 10989010989010989010989010989010989010989011, 10989010989010989010989010989010989010989010989011

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

Table of n, a(n) for n=1..9.

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

Subsequence of A116436.

Cf. A329914, A329915.

Similar for: A095372 \ {1} (k = 21), A331630 (k = 23), A351237 (k = 83), A351238 (k = 87), this sequence (k = 101).

Sequence in context: A358155 A083443 A174089 * A162861 A022011 A175889

Adjacent sequences: A351236 A351237 A351238 * A351240 A351241 A351242

Keyword

nonn,base,easy

Author

Bernard Schott, Feb 05 2022

Status

approved