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

1,2

Comments

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

Links

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

Example

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.

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 *)

Prog

(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=", ")

Crossrefs

Cf. A061641, A177729.

Sequence in context: A240947 A023398 A186501 * A171438 A200683 A286541

Adjacent sequences: A235449 A235450 A235451 * A235453 A235454 A235455

Keyword

nonn

Author

Martin Y. Champel, Jan 10 2014

Status

approved