A213214 -id:A213214 - cp's OEIS frontend

A213215 For the Collatz (3x+1) iterations starting with the odd numbers k, a(n) is the smallest k such that the trajectory contains at least n successive odd numbers == 3 (mod 4).

1, 3, 7, 15, 27, 27, 127, 255, 511, 1023, 1819, 4095, 4255, 16383, 32767, 65535, 77671, 262143, 459759, 1048575, 2097151, 4194303, 7456539, 16777215, 33554431, 67108863, 125687199, 125687199, 125687199, 1073741823, 2147483647, 4294967295, 8589934591, 17179869183

Offset: 1

Michel Lagneau, Mar 02 2013

nonn

The count of odd numbers includes the starting number n if it is part of the longest chain of odd numbers in the sequence.
The sequence is infinite because the Collatz trajectory starting at k = 2^n - 1 contains at least n consecutive odd numbers == 3 (mod 4) such that 3*2^n - 1 -> 3^2*2^(n-1)-1 -> ... -> 2*3^(n-1)-1 and then -> 3^n-1 -> ...  but the numbers of this sequence are not always of this form, for example 27, 1819, 4255, 77671, 459759, ...
Equivalently, a(n) is the smallest k such that the Collatz sequence for k suffers at least n consecutive (3x+1)/2 operations (i.e., no consecutive divisions by 2). - Kevin P. Thompson, Dec 15 2021

			a(4)=15 because the Collatz sequence for 15 (15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1) is the first Collatz sequence to contain 4 consecutive odd numbers congruent to 3 (mod 4): 15, 23, 35, and 53.

Kevin P. Thompson, Table of n, a(n) for n = 1..36

Cf. A006370, A006577, A000225, A024023, A213214.

Cf. A222598 (similar).

nn:=200:T:=array(1..nn):
for n from 1 to 20 do:jj:=0:
         for m from 3 by 2 to 10^8 while(jj=0) do:
               for i from 1 to nn while(jj=0) do:
               T[i]:=0:od:a:=1:T[1]:=m:x:=m:
                     for it from 1 to 100 while (x>1) do:
                         if irem(x,2)=0 then
                         x := x/2:a:=a+1:T[a]:=x:
                         else
                         x := 3*x+1: a := a+1: T[a]:=x:
                        fi:
                     od:
                     jj:=0:aa:=a:
                       for j from 1 to aa while(jj=0) do:
                         if irem(T[j],4)=3 then
                         T[j]:=1:
                         else
                         T[j]:=0:
                       fi:
                      od:
                         for p from 0 to aa-1 while (jj=0) do:
                         s:=sum(T[p+k],k=1..2*n):
                         if s=n then
                         jj:=1: printf ( "%d %d \n",n,m):
                         else
                         fi:
                  od:
              od:
           od:

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; countThrees[t_] := Module[{mx = 0, cnt = 0, i = 0}, While[i < Length[t], i++; If[t[[i]] == 3, cnt++; i++, If[cnt > mx, mx = cnt]; cnt = 0]]; mx]; nn = 15; t = Table[0, {nn}]; n = 1; While[Min[t] == 0, n = n + 2; c = countThrees[Mod[Collatz[n], 4]]; If[c <= nn && t[[c]] == 0, t[[c]] = n; Do[If[t[[i]] == 0, t[[i]] = n], {i, c}]]]; t (* T. D. Noe, Mar 02 2013 *)

Definition clarified, a(1) inserted, and a(21)-a(34) added by Kevin P. Thompson, Dec 15 2021

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A213215 For the Collatz (3x+1) iterations starting with the odd numbers k, a(n) is the smallest k such that the trajectory contains at least n successive odd numbers == 3 (mod 4).

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