[go: up one dir, main page]

login
Positive integers k such that the number of steps it takes to reach 1 in the '3x+1' problem is different for all j < k.
2

%I #26 Jan 31 2021 16:36:55

%S 1,2,3,4,5,6,7,8,9,10,11,12,14,16,17,18,22,24,25,27,28,31,33,34,36,39,

%T 41,43,47,48,49,54,57,62,65,71,72,73,78,82,86,91,94,97,98,103,105,107,

%U 108,111,114,121,123,124,129,130,135,137,142,145,153,155,159

%N Positive integers k such that the number of steps it takes to reach 1 in the '3x+1' problem is different for all j < k.

%C Positive integers k such that A337144(k) = 1.

%C Or positive integers k such that A006577(k) != A006577(j) for all j = 1..k-1.

%C Different from A129304.

%H Alois P. Heinz, <a href="/A337149/b337149.txt">Table of n, a(n) for n = 1..1000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Collatz_conjecture">Collatz Conjecture</a>

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>

%F A006577(a(n)) = A337150(n).

%p collatz:= proc(n) option remember; `if`(n=1, 0,

%p 1 + collatz(`if`(n::even, n/2, 3*n+1)))

%p end:

%p b:= proc() 0 end:

%p g:= proc(n) option remember; local t;

%p `if`(n=1, 0, g(n-1));

%p t:= collatz(n); b(t):= b(t)+1

%p end:

%p a:= proc(n) option remember; local k; for k

%p from 1+a(n-1) while g(k)>1 do od; k

%p end: a(0):=0:

%p seq(a(n), n=1..100);

%t collatz[n_] := collatz[n] = If[n==1, 0,

%t 1+collatz[If[EvenQ[n], n/2, 3n+1]]];

%t b[_] = 0;

%t g[n_] := g[n] = Module[{t}, If[n==1, 0, g[n-1]];

%t t = collatz[n]; b[t] = b[t]+1];

%t a[n_] := a[n] = Module[{k}, For[k = 1+a[n-1],

%t g[k] > 1, k++]; k]; a[0] = 0;

%t Array[a, 100] (* _Jean-François Alcover_, Jan 30 2021, after _Alois P. Heinz_ *)

%Y Cf. A006577, A129304, A337144, A337150.

%K nonn,look

%O 1,2

%A _Alois P. Heinz_, Jan 27 2021