A351237 Numbers M such that 83 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

137, 13698630137, 1369863013698630137, 136986301369863013698630137, 13698630136986301369863013698630137, 1369863013698630136986301369863013698630137, 136986301369863013698630136986301369863013698630137

Offset: 1

Author

Bernard Schott, Feb 05 2022

Keywords

• nonn
• base
• easy

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; 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}.

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.

Examples

			83 * 137 = 1[137]1, hence 137 is a term.
83 * 13698630137 = 1[13698630137]1, and 13698630137 is another term.
		

References

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

Links

• Index entries for linear recurrences with constant coefficients, signature (100000001,-100000000).

Crossrefs

Subsequence of A116436.

Cf. A329914, A329915.

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

Programs

Maple

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

Mathematica

Table[(10^(8*n-4)+1)/73, {n, 1, 7}] (* Amiram Eldar, Feb 06 2022 *)
LinearRecurrence[{100000001,-100000000},{137,13698630137},20] (* Harvey P. Dale, Nov 01 2022 *)

Formula

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