login
Numbers M such that 83 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.
5

%I #22 Nov 01 2022 19:38:25

%S 137,13698630137,1369863013698630137,136986301369863013698630137,

%T 13698630136986301369863013698630137,

%U 1369863013698630136986301369863013698630137,136986301369863013698630136986301369863013698630137

%N Numbers M such that 83 * 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; 83 is the eleventh such integer, so 83 = A329914(11), and a(1) = A329915(11) = 137; hence, the terms of this sequence form the infinite set {M_83}.

%C Every term M = a(n) has q = 8*n-5 digits, and 10^(q+1)+1 that has q = 8*n-5 zeros in its decimal expansion is equal to 73 * M, so a(n) = M is a divisor of 10^(8*n-4)+1. Example: a(2) = 13698630137 has 11 digits and 73 * 13698630137 = 1000000000001 that has 11 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 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (100000001,-100000000).

%F a(n) = (10^(8*n-4)+1)/73 for n >= 1.

%e 83 * 137 = 1[137]1, hence 137 is a term.

%e 83 * 13698630137 = 1[13698630137]1, and 13698630137 is another term.

%p seq((10^(8*n-4)+1)/73, n=1..15);

%t Table[(10^(8*n-4)+1)/73, {n, 1, 7}] (* _Amiram Eldar_, Feb 06 2022 *)

%t LinearRecurrence[{100000001,-100000000},{137,13698630137},20] (* _Harvey P. Dale_, Nov 01 2022 *)

%Y Subsequence of A116436.

%Y Cf. A329914, A329915.

%Y Similar for: A095372 \ {1} (k=21), A331630 (k=23), this sequence (k=83), A351238 (k=87), A351239 (k=101).

%K nonn,base,easy

%O 1,1

%A _Bernard Schott_, Feb 05 2022