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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).
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
1,2
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^(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
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 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:
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
Cf. A222598 (similar).
Sequence in context: A001649 A303220 A301894 * A353578 A324719 A170884
KEYWORD
nonn
AUTHOR
Michel Lagneau, Mar 02 2013
EXTENSIONS
Definition clarified, a(1) inserted, and a(21)-a(34) added by Kevin P. Thompson, Dec 15 2021
STATUS
approved