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A346775 Starting from n!+1, the length of the longest sequence of consecutive numbers which all take the same number of steps to reach 1 in the Collatz (or '3x+1') problem.

%I A346775 #37 Aug 28 2024 11:16:41
%S A346775 1,1,1,1,1,1,3,5,1,7,1,1,3,1,3,1,1,1,30,1,30,3,7,1,3,3,7,1,1,7,15,3,1,
%T A346775 1,3,15,26,15,1,1,1,1,7,7,26,7,1,7,3,1,1,3,1,7,3,7,1,1,26,15,7,30,1,1,
%U A346775 1,1,3,15,3,1,1,31,648,26,26,30,90,1,1,3,15
%N A346775 Starting from n!+1, the length of the longest sequence of consecutive numbers which all take the same number of steps to reach 1 in the Collatz (or '3x+1') problem.
%C A346775 The largest value known in this sequence is a(219)=78553595.
%C A346775 2^32 < a(238) < 11442739136455298475. - _Martin Ehrenstein_, Aug 21 2021
%C A346775 Jeremy Sawicki found that a(238) = 107150589645. - _Dmitry Kamenetsky_, Aug 25 2024
%H A346775 Dmitry Kamenetsky, <a href="/A346775/b346775.txt">Table of n, a(n) for n = 0..237</a>
%H A346775 Dmitry Kamenetsky, <a href="https://math.stackexchange.com/questions/4222776/smallest-m1-such-that-the-number-of-collatz-steps-needed-for-238m-to-reac">Smallest m>1 such that the number of Collatz steps needed for 238!+m to reach 1 differs from that of 238!+1.</a>, Mathematics StackExchange, Aug 2021.
%H A346775 <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%e A346775 a(6) = 3, because 6!+1, 6!+2 and 6!+3 all take 46 steps to reach 1, while 6!+4 requires 20 steps to reach 1.
%t A346775 f[n_] := Length[NestWhileList[If[EvenQ@#, #/2, 3 # + 1] &, n, # != 1 &]] - 1; Table[k = 1; While[f[n! + k] == f[n! + k + 1], k++]; k, {n, 0, 100}] (* _Bence Bernáth_, Aug 14 2021 *)
%o A346775 (PARI) a6577(n0)={my(n=n0,k=0);while(n>1,k++;n=if(n%2,3*n+1,n/2));k};
%o A346775 for(n=0,80,my(n0=n!+1,nc=a6577(n0),k=1);while(a6577(n0++)==nc,k++);print1(k,", ")) \\ _Hugo Pfoertner_, Aug 04 2021
%Y A346775 Cf. A006577, A070974, A277109.
%K A346775 nonn
%O A346775 0,7
%A A346775 _Dmitry Kamenetsky_, Aug 03 2021