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Numbers M such that 23 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.
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%I #39 Mar 03 2023 10:28:45

%S 77,76923077,76923076923077,76923076923076923077,

%T 76923076923076923076923077,76923076923076923076923076923077,

%U 76923076923076923076923076923076923077,76923076923076923076923076923076923076923077

%N Numbers M such that 23 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

%C 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; 23 is the second such integer, so 23 = A329914(2), and a(1) = A329915(2) = 77; hence, the terms of this sequence form the infinite set {M_23}.

%C Every term M = a(n) has q = 6*n-4 digits, and 10^(q+1)+1 that has 6*n-4 zeros in its decimal expansion is equal to 13 * M, so M is a divisor of 10^(6*n-3)+1. Example: a(2) = 76923077 has 8 digits and 13 * 76923077 = 1000000001 that has 8 zeros in its decimal expansion.

%D D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.

%H Colin Barker, <a href="/A331630/b331630.txt">Table of n, a(n) for n = 1..150</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1000001,-1000000).

%F a(n) = (10^(6*n-3)+1)/13 for n >= 1.

%F From _Colin Barker_, Jan 25 2020: (Start)

%F G.f.: 77*x*(1 - 1000*x) / ((1 - x)*(1 - 1000000*x)).

%F a(n) = 1000001*a(n-1) - 1000000*a(n-2) for n>2.

%F a(n) = (1000 + 1000^(2*n))/13000 for n>0.

%F (End)

%F E.g.f.: exp(x)*(1000 + exp(999999*x))/13000 - 77/1000. - _Stefano Spezia_, Jan 26 2020

%e 23 * 77 = 1771, hence 77 is a term.

%e 23 * 76923076923077 = 1(76923076923077)1, and 76923076923077 is another term.

%p seq((10^(6*m-3)+1)/13, m=1..15);

%t Array[(10^(6 # - 3) + 1)/13 &, 9] (* _Michael De Vlieger_, Jan 24 2020 *)

%t LinearRecurrence[{1000001,-1000000},{77,76923077},10] (* _Harvey P. Dale_, Mar 03 2023 *)

%o (PARI) vector(9, n, (10^(6*n-3)+1)/13) \\ _Michel Marcus_, Jan 25 2020

%o (PARI) Vec(77*x*(1 - 1000*x) / ((1 - x)*(1 - 1000000*x)) + O(x^10)) \\ _Colin Barker_, Jan 25 2020

%o (PARI) apply( {A331630(n)=10^(6*n-3)\/13}, [1..9]) \\ _M. F. Hasler_, Jan 26 2020, following Michel Marcus

%Y Subsequence of A116436.

%Y Cf. A329914, A329915, A095372 \ {1} (similar for k = 21).

%K nonn,base,easy

%O 1,1

%A _Bernard Schott_, Jan 23 2020