A351238 - OEIS

A351238

Numbers M such that 87 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

13, 12987013, 12987012987013, 12987012987012987013, 12987012987012987012987013, 12987012987012987012987012987013, 12987012987012987012987012987012987013, 12987012987012987012987012987012987012987013, 12987012987012987012987012987012987012987012987013, 12987012987012987012987012987012987012987012987012987013

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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; 87 is the twelfth such integer, so 87 = A329914(12), and a(1) = A329915(12) = 13; hence, the terms of this sequence form the infinite set {M_87}.

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 77 * M, so a(n) = M is a divisor of 10^(6*n-3)+1. Example: a(2) = 12987013 has 8 digits and 77 * 12987013 = 1000000001 that has 8 zeros in its decimal expansion.

REFERENCES

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

LINKS

Table of n, a(n) for n=1..10.

Index entries for linear recurrences with constant coefficients, signature (1000001,-1000000).

FORMULA

a(n) = (10^(6*n-3)+1)/77 for n >= 1.

EXAMPLE

87 * 13 = 1[13]1, hence 13 is a term.

87 * 12987013 = 1[12987013]1, and 12987013 is a term.

MAPLE

seq((10^(6*n-3)+1)/77, n=1..15);

MATHEMATICA

Table[(10^(6*n - 3) + 1)/77, {n, 1, 10}] (* Amiram Eldar, Feb 06 2022 *)