

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).


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
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OFFSET

1,2


COMMENTS

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^(n1)1 > ... > 2*3^(n1)1 and then > 3^n1 > ... 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


LINKS



EXAMPLE

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.


MAPLE

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 aa1 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:


MATHEMATICA

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 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS

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


STATUS

approved



