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%I #33 May 29 2022 13:33:06
%S 0,3,1,8,6,4,4,11,2,36,9,9,34,16,7,7,32,5,14,5,5,39,30,12,12,12,3,46,
%T 37,37,28,19,10,10,10,10,44,35,35,26,26,17,17,17,8,8,8,8,42,33,33,33,
%U 24,6,24,15,15,15,15,6,6,49,6,40,40,40,31,31,31,31,22
%N Number of repeated applications of the '4x+-1' operation needed to reach 1 from n, or -1 if 1 is never reached.
%C Start with any positive number n. Define f(n) as 'if n mod 3 = 1, f(n) = 4n-1; if n mod 3 = 2, f(n) = 4n+1; if n is divisible by 3, f(n) = n/3'. Apply this procedure repeatedly to reach 1. This sequence gives the number of steps required to reach 1, or -1 if 1 is never reached. The procedure is a modification of the famous Collatz procedure.
%C We can make a conjecture similar to that of Collatz to claim that the process will always reach 1. However, this is not proven yet.
%C It seems that more than 65 percent of the terms in this sequence satisfy a(i) = a(i+1). For example, 6655 of the first 10000 terms satisfy this condition.
%H Md. Towhidul Islam, <a href="/A340672/b340672.txt">Table of n, a(n) for n = 1..10001</a>
%e a(5)=6 because the trajectory of 5 is (5,21,7,27,9,3,1). That needs 6 repeated applications of the procedure to reach 1.
%p a:= proc(n) option remember; `if`(n=1, 0, 1 + (r->
%p a(`if`(r=0, q, 4*n+2*r-3)))(irem(n, 3, 'q')))
%p end:
%p seq(a(n), n=1..75); # _Alois P. Heinz_, Jan 20 2021
%t a[n_] := a[n] = If[n == 1, 0, {q, r} = QuotientRemainder[n, 3]; 1 +
%t a[If[r == 0, q, 4n + 2r - 3]]];
%t Table[a[n], {n, 1, 75}] (* _Jean-François Alcover_, May 29 2022, after _Alois P. Heinz_ *)
%o (PARI) f(n) = my(m=n%3); if (m==1, 4*n-1, if (m==2, 4*n+1, n/3));
%o a(n) = my(s=n, c=0); while(s>1, s=f(s); c++); c; \\ _Michel Marcus_, Jan 18 2021
%Y Cf. A006577.
%K nonn,easy,look
%O 1,2
%A _Md. Towhidul Islam_, Jan 15 2021
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