A235452 Take the union of all the sequences Collatz(i) for i <= n. The number a(n) is the largest of consecutive numbers beginning with 1.

1, 2, 5, 5, 5, 6, 8, 8, 11, 11, 11, 14, 14, 14, 17, 17, 17, 18, 20, 20, 23, 23, 23, 24, 26, 26, 29, 29, 29, 32, 32, 32, 35, 35, 35, 36, 38, 38, 41, 41, 41, 42, 44, 44, 47, 47, 47, 50, 50, 50, 53, 53, 53, 54, 56, 56, 59, 59, 59, 62, 62, 62, 65, 65, 65

Offset: 1

Author

Martin Y. Champel, Jan 10 2014

Keywords

• nonn

Comments

The Collatz sequence is also called the 3x+1 sequence.

Examples

			Let the C(n) function compute the Collatz sequence starting at n.
For n = 1, C(1) = {1} then term 1 is 1.
For n = 2, C(2) = {1,2} then term 2 is 2.
For n = 3, C(3) = {3,10,5,16,8,4,2,1} = {1,2,3,4,5,8,10,16} then it contains {1,2,3,4,5} but not {1,2,3,4,5,6} then term 3 is 5.
For n = 4, C(4) = C(3) then term 4 is 5.
For n = 5, C(5) = C(4) = C(3) then term 5 is 5.
For n = 6, C(6) = {1,2,3,4,5,6,8,10,16} then term 6 is 6.
		

Links

• Martin Y. Champel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A061641, A177729.

Programs

Mathematica

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; countConsec[lst_] := Module[{cnt = 0, i = 1}, While[i <= Length[lst] && lst[[i]] == i, cnt++; i++]; cnt]; mx = 0; u = {}; Table[c = Collatz[n]; u = Union[u, c]; mx = Max[mx, countConsec[u]], {n, 65}] (* T. D. Noe, Feb 23 2014 *)

Python

def A235452(n=100):
    a = set([])
    A235452 = {1: 1}
    for i in range(2, n):
        c = i
        a.add(c)
        while c != 1:
            if c % 2 == 1:
                c = 3 * c + 1
                a.add(c)
            c = c / 2
            a.add(c)
        k = 1
        while k in a:
            k += 1
        A235452[i] = k - 1
    return A235452
seq_map = A235452()
for n in range(1, len(seq_map) + 1):
    print(seq_map[n], end=", ")