A331630 Numbers M such that 23 * M = 1M1, where 1M1 denotes the concatenation of 1, M and 1.

77, 76923077, 76923076923077, 76923076923076923077, 76923076923076923076923077, 76923076923076923076923076923077, 76923076923076923076923076923076923077, 76923076923076923076923076923076923076923077

Offset: 1

Author

Bernard Schott, Jan 23 2020

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; 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.

Examples

			23 * 77 = 1771, hence 77 is a term.
23 * 76923076923077 = 1(76923076923077)1, and 76923076923077 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

• Colin Barker, Table of n, a(n) for n = 1..150
• Index entries for linear recurrences with constant coefficients, signature (1000001,-1000000).

Crossrefs

Subsequence of A116436.

Cf. A329914, A329915, A095372 \ {1} (similar for k = 21).

Programs

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 *)

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

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