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A351239
Numbers M such that 101 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.
5
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
David Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.
FORMULA
a(n) = (10^(6*n-3)+1)/91 for n >= 1.
From Elmo R. Oliveira, Apr 18 2026: (Start)
G.f.: 11*x*(1 - 1000*x)/((1 - x)*(1 - 1000000*x)).
E.g.f.: (-1001 + 1000*exp(x) + exp(1000000*x))/91000.
a(n) = 1000001*a(n-1) - 1000000*a(n-2) for n > 2. (End)
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.
Similar for: A095372 \ {1} (k = 21), A331630 (k = 23), A351237 (k = 83), A351238 (k = 87), this sequence (k = 101).
Sequence in context: A083443 A374238 A174089 * A162861 A022011 A175889
KEYWORD
nonn,base,easy
AUTHOR
Bernard Schott, Feb 05 2022
STATUS
approved