OFFSET
1,2
COMMENTS
This sequence is infinite because 1/(10^k-1) has a period of k for all k, so the period can be arbitrarily large.
Are 1, 3, 289 and 361 the only terms that are not in A001913? - Robert Israel, Feb 10 2019
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
Project Euler, Reciprocal cycles: Problem 26
Eric Weisstein's World of Mathematics, Repeating Decimal
FORMULA
RECORDS transform of A051626.
EXAMPLE
7 is a term because 1/7 has a period of 6, which is greater than the periods of 1/m for m < 7.
MAPLE
count:= 1: A[1]:= 1: m:= 0:
for k from 0 to 100 do
for d in [3, 7, 9, 11] do
x:= 10*k+d;
p:= numtheory:-order(10, x);
if p > m then
m := p;
count:= count+1;
A[count]:= x
fi
od od:
seq(A[i], i=1..count); # Robert Israel, Feb 10 2019
MATHEMATICA
ResourceFunction["ProgressiveMaxPositions"]@
Map[n |->
First[RealDigits[n]] /. {{___, list_?ListQ} :> Length[list],
list_?ListQ -> 0}][
1/Range[1050]] (* Peter Cullen Burbery, Aug 05 2023 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Matthew Schulz, Feb 09 2019
STATUS
approved
