cp's OEIS Frontend

A343269 a(n) is the smallest integer whose orbit length is n under iteration of the map r -> A061602(r).

%I A343269 #23 Apr 13 2021 18:33:05
%S A343269 1,0,169,78,69,26,24,4,22,5,122,25,14,127,6,3,12,33,136,256,57,247,
%T A343269 148,38,1478,368,79,1458,48,44,29,7,13,34,9,8,23,234,37,337,58,46,139,
%U A343269 138,369,239,267,36,334,289,3555,49,144,45,229,2569,22888,136789,334479,1479,1233466
%N A343269 a(n) is the smallest integer whose orbit length is n under iteration of the map r -> A061602(r).
%C A343269 A303935 provides the orbit's lengths, i.e., the number of needed steps, starting from a given number, to reach a value that already exists in trajectory.
%C A343269 This sequence is infinite. Actually, given a number x whose orbit's length is k, one can always build a number y whose orbit's length is (k+1).
%C A343269 For instance, just consider either the number 10^(x-1), or Rx (the repunit of length x), or any other x-digit binary string, all of them leading to the number x after application of the mapping function: A061602(y) = x.
%C A343269 Indeed, none of them will correspond to the smallest integer m such that A303935(m) = k + 1.
%C A343269 In fact, it becomes computationally hard to determine further terms since, as in the Collatz mapping function and other similar problems, there is no predictable way to define the exact complete path without calculating all intermediary orbit's components until one reaches a previously calculated or encountered number.
%C A343269 a(59) = 334479, a(60) = 1479, a(61) = 1233466, next terms = ?
%e A343269 a(6) = 26 because A303935(26) = 6, and 26 is the smallest nonnegative integer m such that A303935(m) = 6.
%Y A343269 Cf. A303935 (orbit's length), A061602 (sum of factorials of digits), A014080 (factorions).
%Y A343269 Cf. A193163, A214285, A254499, A188283, A244090.
%K A343269 nonn,base
%O A343269 1,3
%A A343269 _Lamine Ngom_, Apr 10 2021