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A351238 Numbers M such that 87 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1. 5
13, 12987013, 12987012987013, 12987012987012987013, 12987012987012987012987013, 12987012987012987012987012987013, 12987012987012987012987012987012987013, 12987012987012987012987012987012987012987013, 12987012987012987012987012987012987012987012987013, 12987012987012987012987012987012987012987012987012987013 (list; graph; refs; listen; history; text; internal format)
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
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 *)
CROSSREFS
Subsequence of A116436.
Similar for: A095372 \ {1} (k=21), A331630 (k=23), A351237 (k=83), this sequence (k=87), A351239 (k=101).
Sequence in context: A185213 A034248 A324270 * A177027 A158750 A145744
KEYWORD
nonn,base,easy
AUTHOR
Bernard Schott, Feb 05 2022
STATUS
approved

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Last modified June 9 16:35 EDT 2024. Contains 373248 sequences. (Running on oeis4.)