%I #63 Sep 06 2023 16:01:46
%S 1,3,7,17,19,23,29,47,59,61,97,109,113,131,149,167,179,181,193,223,
%T 229,233,257,263,269,289,313,337,361,367,379,383,389,419,433,461,487,
%U 491,499,503,509,541,571,577,593,619,647,659,701,709,727,743,811,821,823
%N Numbers k such that the period of 1/k, or 0 if 1/k terminates, is strictly greater than the period of the decimal expansion of 1/m for all m < k.
%C This sequence is infinite because 1/(10^k-1) has a period of k for all k, so the period can be arbitrarily large.
%C Are 1, 3, 289 and 361 the only terms that are not in A001913? - _Robert Israel_, Feb 10 2019
%H Robert Israel, <a href="/A306355/b306355.txt">Table of n, a(n) for n = 1..10000</a>
%H Project Euler, <a href="https://projecteuler.net/problem=26">Reciprocal cycles: Problem 26</a>
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RepeatingDecimal.html">Repeating Decimal</a>
%F RECORDS transform of A051626.
%e 7 is a term because 1/7 has a period of 6, which is greater than the periods of 1/m for m < 7.
%p count:= 1: A[1]:= 1: m:= 0:
%p for k from 0 to 100 do
%p for d in [3,7,9,11] do
%p x:= 10*k+d;
%p p:= numtheory:-order(10,x);
%p if p > m then
%p m := p;
%p count:= count+1;
%p A[count]:= x
%p fi
%p od od:
%p seq(A[i],i=1..count); # _Robert Israel_, Feb 10 2019
%t ResourceFunction["ProgressiveMaxPositions"]@
%t Map[n |->
%t First[RealDigits[n]] /. {{___, list_?ListQ} :> Length[list],
%t list_?ListQ -> 0}][
%t 1/Range[1050]] (* _Peter Cullen Burbery_, Aug 05 2023 *)
%Y Cf. A051626, A007732.
%Y Contains A001913.
%K nonn,base
%O 1,2
%A _Matthew Schulz_, Feb 09 2019
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