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

%I A351239 #14 Sep 12 2022 14:56:29
%S A351239 11,10989011,10989010989011,10989010989010989011,
%T A351239 10989010989010989010989011,10989010989010989010989010989011,
%U A351239 10989010989010989010989010989010989011,10989010989010989010989010989010989010989011,10989010989010989010989010989010989010989010989011
%N A351239 Numbers M such that 101 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.
%C A351239 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}.
%C A351239 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.
%D A351239 D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.
%H A351239 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1000001,-1000000).
%F A351239 a(n) = (10^(6*n-3)+1)/91 for n >= 1.
%e A351239 101 * 11 = 1[11]1, hence 11 is a term.
%e A351239 101 * 10989011 = 1[10989011]1 and 10989011 is another term.
%p A351239 seq((10^(6*n-3)+1)/91, n=1..15);
%t A351239 Table[(10^(6*n - 3) + 1)/91, {n, 1, 9}] (* _Amiram Eldar_, Feb 06 2022 *)
%t A351239 LinearRecurrence[{1000001,-1000000},{11,10989011},10] (* _Harvey P. Dale_, Sep 12 2022 *)
%Y A351239 Subsequence of A116436.
%Y A351239 Cf. A329914, A329915.
%Y A351239 Similar for: A095372 \ {1} (k = 21), A331630 (k = 23), A351237 (k = 83), A351238 (k = 87), this sequence (k = 101).
%K A351239 nonn,base,easy
%O A351239 1,1
%A A351239 _Bernard Schott_, Feb 05 2022