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; 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}.
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.
REFERENCES
D. Wells, 112359550561797732809 entry, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1997, p. 196.
LINKS
Colin Barker, Table of n, a(n) for n = 1..150
Index entries for linear recurrences with constant coefficients, signature (1000001,-1000000).
FORMULA
a(n) = (10^(6*n-3)+1)/13 for n >= 1.
From Colin Barker, Jan 25 2020: (Start)
G.f.: 77*x*(1 - 1000*x) / ((1 - x)*(1 - 1000000*x)).
a(n) = 1000001*a(n-1) - 1000000*a(n-2) for n>2.
a(n) = (1000 + 1000^(2*n))/13000 for n>0.
(End)
E.g.f.: exp(x)*(1000 + exp(999999*x))/13000 - 77/1000. - Stefano Spezia, Jan 26 2020
EXAMPLE
23 * 77 = 1771, hence 77 is a term.
23 * 76923076923077 = 1(76923076923077)1, and 76923076923077 is another term.
MAPLE
seq((10^(6*m-3)+1)/13, m=1..15);
MATHEMATICA
Array[(10^(6 # - 3) + 1)/13 &, 9] (* Michael De Vlieger, Jan 24 2020 *)
LinearRecurrence[{1000001, -1000000}, {77, 76923077}, 10] (* Harvey P. Dale, Mar 03 2023 *)
PROG
(PARI) vector(9, n, (10^(6*n-3)+1)/13) \\ Michel Marcus, Jan 25 2020
(PARI) Vec(77*x*(1 - 1000*x) / ((1 - x)*(1 - 1000000*x)) + O(x^10)) \\ Colin Barker, Jan 25 2020
(PARI) apply( {A331630(n)=10^(6*n-3)\/13}, [1..9]) \\ M. F. Hasler, Jan 26 2020, following Michel Marcus
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Bernard Schott, Jan 23 2020
STATUS
approved