

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



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
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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]=k1
....return A235452


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



