Displaying 1-10 of 16 results found.
1, 2, 4, 8, 16, 20, 24, 32, 40, 48, 52, 56, 64, 68, 72, 80, 84, 88, 96, 100, 104, 112, 116
a(n) = (1 + number of halving and tripling steps to reach 1 in the Collatz (3x+1) problem), or -1 if 1 is never reached.
+10
45
1, 2, 8, 3, 6, 9, 17, 4, 20, 7, 15, 10, 10, 18, 18, 5, 13, 21, 21, 8, 8, 16, 16, 11, 24, 11, 112, 19, 19, 19, 107, 6, 27, 14, 14, 22, 22, 22, 35, 9, 110, 9, 30, 17, 17, 17, 105, 12, 25, 25, 25, 12, 12, 113, 113, 20, 33, 20, 33, 20, 20, 108, 108, 7, 28, 28, 28, 15, 15, 15, 103
COMMENTS
The number of steps (iterations of the map A006370) to reach 1 is given by A006577, this sequence counts 1 more. - M. F. Hasler, Nov 05 2017 When Collatz 3N+1 function is seen as an isometry over the dyadics, the halving step necessarily following each tripling is not counted, hence N -> N/2, if even, but N -> (3N+1)/2, if odd. Counting iterations of this map until reaching 1 leads to sequence A064433. [Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009]
REFERENCES
R. K. Guy, Unsolved Problems in Number Theory, E16.
FORMULA
a(n) = f(n,1) with f(n,x) = if n=1 then x else f( A006370(n),x+1).
MAPLE
a:= proc(n) option remember; 1+`if`(n=1, 0,
a(`if`(n::even, n/2, 3*n+1)))
end:
MATHEMATICA
Table[Length[NestWhileList[If[EvenQ[ # ], #/2, 3 # + 1] &, i, # != 1 &]], {i, 75}]
PROG
(Haskell)
a008908 = length . a070165_row
(Python)
def a(n):
if n==1: return 1
x=1
while True:
if n%2==0: n//=2
else: n = 3*n + 1
x+=1
if n<2: break
return x
EXTENSIONS
More terms from Larry Reeves (larryr(AT)acm.org), Apr 27 2001
Number of numbers not greater than n occurring in Collatz (3x+1) trajectory starting with n.
+10
8
1, 2, 3, 3, 4, 6, 5, 4, 7, 6, 7, 9, 7, 10, 7, 5, 9, 14, 11, 8, 6, 12, 9, 11, 14, 10, 10, 16, 14, 11, 10, 6, 17, 12, 9, 20, 18, 17, 18, 9, 13, 8, 20, 16, 14, 12, 13, 12, 20, 21, 18, 12, 10, 18, 15, 20, 24, 19, 22, 16, 14, 17, 15, 7, 23, 25, 22, 15, 13, 12, 16, 23
COMMENTS
If the Collatz conjecture is true, there are no cycles in the 3x+1 trajectory and the difference between the counts here and those of A076228 is that the start value is counted here but not there; then a(n) = 1+ A076228(n) [discovered by sequencedb.net]. - R. J. Mathar, Jun 24 2021
FORMULA
a(n) < n for n>6;
a(n) = f(n,n,1) with f(n,m,x) = if m=1 then x else f(n, A006370(m), if A006370(m)<n then x+1 else x).
EXAMPLE
a(9) = #{1,2,4,5,7,8,9} = 7, as
9-28-14-7-22-11-34-17-52-26-13-40-20-10-5-16-8-[4-2-1]*
9-..-..-7-..-..-..-..-..-..-..-..-..-..-5-..-8-[4-2-1]*.
MATHEMATICA
Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; f[n_] := Module[{c = Collatz[n]}, Length[Select[c, # <= n &]]]; Table[ f[n], {n, 100}] (* T. D. Noe, Mar 07 2013 *)
PROG
(Haskell)
a159999 n = length $ takeWhile (<= n) $ sort $ a070165_row n
Number of different initial values for 3x+1 trajectories in which the largest term appearing in the iteration is 2^n.
+10
7
1, 1, 1, 6, 1, 3, 1, 3, 1, 12, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 13, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 9, 1, 3, 1, 3, 1, 11, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 21, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 78, 1, 3, 1, 3, 1, 8, 1, 3, 1, 3, 1, 6, 1, 3, 1, 3, 1, 9, 1, 3, 1
COMMENTS
It would be interesting to know whether the ...1,3,1,3,1,x,1,3,1,3,1,... pattern persists. - John W. Layman, Jun 09 2004 The observed pattern should persist. Proof: [1] a(odd)=1 because -1+2^odd is not divisible by 3, so in Collatz-algorithm 2^odd is preceded by increasing inverse step. Thus 2^odd is the only suitable initial value; [2] a[2k]>=3 for k>1 because 2^(2k)-1=-1+4^k=3A so {b=2^2k, (b-1)/3 and (2a-2)/3} are three relevant initial values. No more case arises unless condition-[3] (see below) was satisfied; [3] a[6k+4]>=5 for k>=1, ..iv=c=2^(6k+4); here {c, (c-1)/3, 2(c-1)/3, (2c-5)/9, (4c-10)/9} is 5 suitable initial values, iff (2c-5)/9 is integer; e.g. at 6k+4=10, {1024<-341<-682<-227<-454} back-tracking the iteration. - Labos Elemer, Jun 17 2004 Except for a(2)=1 the sequence has the 6-element quasiperiod 1, 3, 1, x, 1, 3 where x>=6, but unequal to 7 and 10 (see links below and in A033496). Observe that for n=2^(6k+4)=16*2^(6k), n mod 9 = 7 so that (2n-5)/9 is an integer and a(n)>=6. Conjecture: All numbers m > 10 occur as values in A087256 (see A233293). The conjecture has been verified for all 10 < k < 133 for Collatz trajectories with maximum value through 2^(36000*6 + 4). The largest fan of initial values in this range, F(6*1993+4), has maximum 2^11962 and size 3958.
(End)
EXAMPLE
n = 10: 2^10 = 1024 = peak for trajectories started with initial value taken from the list: {151, 201, 227, 302, 341, 402, 454, 604, 682, 804, 908, 1024};
a trajectory with peak=1024: {201, 604, 302, 151, 454, 227, 682, 341, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1}
MATHEMATICA
c[x_]:=c[x]=(1-Mod[x, 2])*(x/2)+Mod[x, 2]*(3*x+1); c[1]=1; fpl[x_]:=FixedPointList[c, x]; {$RecursionLimit=1000; m=0}; Table[Print[{xm-1, m}]; m=0; Do[If[Equal[Max[fpl[n]], 2^xm], m=m+1], {n, 1, 2^xm}], {xm, 1, 30}]
PROG
(PARI) f(n, m) = 1 + if(2*n <= m, f(2*n, m), 0) + if (n%6 == 4, f(n\3, m), 0);
Numbers that are highest in their respective Collatz (3x+1) trajectories only.
+10
5
1, 2, 4, 8, 20, 24, 32, 48, 56, 68, 72, 80, 84, 96, 104, 116, 128, 132, 144, 152, 168, 176, 180, 192, 200, 212, 224, 228, 240, 260, 264, 272, 276, 288, 296, 308, 312, 320, 324, 336, 344, 356, 360, 368, 372, 384, 392, 404, 408, 416, 452, 456, 464, 468, 480, 488
COMMENTS
This is effectively the complement of A176869 in A033496, excluding numbers which are also highest in trajectories less than the number itself.
MATHEMATICA
Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; oldMax = {}; t = {}; Do[c = Collatz[n]; If[! MemberQ[oldMax, n] && Max[c] == n, AppendTo[t, n]]; oldMax = Union[oldMax, {Max[c]}], {n, 416}]; t (* T. D. Noe, Feb 28 2013 *)
Number of different initial values for 3x+1 trajectories in which the largest term appearing in the iteration is 2^(6n+4).
+10
4
6, 12, 8, 6, 13, 8, 6, 9, 11, 6, 21, 8, 6, 78, 8, 6, 9, 13, 6, 15, 8, 6, 16, 8, 6, 9, 20, 6, 12, 8, 6, 13, 8, 6, 9, 11, 6, 14, 8, 6, 32, 8, 6, 9, 32, 6, 23, 8, 6, 24, 8, 6, 9, 14, 6, 12, 8, 6, 13, 8, 6, 9, 11, 6, 14, 8, 6, 19, 8, 6, 9, 13, 6, 80, 8, 6, 29, 8, 6, 9, 18, 6, 12, 8, 6, 13, 8, 6, 9, 11
COMMENTS
The sequence has the quasiperiod 6, x, 8, 6, y, 8, 6, 9, z of length 9 starting at index 0 where x, y, z > 10; in addition, a(3*9*n+1) = 12, a(3*9*n+4) = 13 and a(3*9*n+8) = 11 for all n>=0; proof by induction (see this link) as in the link in A087256 based on the modular identities in the link in A033496. Conjecture: All numbers greater than 10 appear in the sequence (see also A033496 and A233293). (End)
EXAMPLE
a(1) = 12, i.e. the number of initial values for 2^10, since 804 -> 402 -> 201 -> 604 -> 302 -> 151 -> 454 -> 227 -> 682 -> 341 -> 1024 and 908 -> (454 -> ... -> 1024) are the two maximal trajectories containing all 12 initial values. a(8) = 11 since 2^(6*8+4) has 11 different initial values for Collatz trajectories leading to it. - Hartmut F. W. Hoft, Jun 24 2016
MATHEMATICA
trajectory[start_] := NestWhileList[If[OddQ[#], 3#+1, #/2]&, start, #!=1&]
fanSize[max_] := Module[{active={max}, fan={}, current}, While[active!={}, current=First[active]; active=Rest[active]; AppendTo[fan, current]; If[2*current<=max, AppendTo[active, 2*current]]; If[Mod[current, 3]==1 && OddQ[(current-1)/3] && current>4, AppendTo[active, (current-1)/3]]]; Length[fan]]/; max==Max[trajectory[max]]
a105730[low_, high_] := Map[fanSize[2^(6#+4)]&, Range[low, high]]
Numbers that are the maximum value attained by the Collatz (3x+1) iteration of some odd number.
+10
4
1, 16, 40, 52, 64, 88, 100, 112, 136, 148, 160, 184, 196, 208, 232, 244, 256, 280, 304, 340, 352, 400, 424, 448, 472, 520, 532, 544, 592, 616, 628, 640, 688, 712, 724, 736, 784, 808, 820, 832, 868, 904, 916, 928, 952, 964, 976, 1024, 1048, 1072, 1108, 1120
COMMENTS
Here the 3x+1 steps and the halving steps are applied separately. We use odd numbers because then the Collatz iteration always increases to a maximum value before (hopefully) going to 1. This is a subsequence of A033496. Except for the first term, all these numbers appear to equal 4 (mod 12). Some terms are the maximum value for the Collatz iteration of many numbers. For example, 9232 is the maximum value of the Collatz iteration of 408 odd numbers, the smallest of which is 27.
CROSSREFS
Cf. A025586 (maximum value in the Collatz iteration of n)
Numbers k whose Collatz (3x+1) trajectory includes at least one term that is greater than k.
+10
4
3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 69, 70, 71, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83
EXAMPLE
a(1) = 3 because 3 is the first number k whose trajectory goes above k.
MATHEMATICA
Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Select[Range[100], Max[Collatz[#]] > # &] (* T. D. Noe, Mar 01 2013 *)
Number of terms of the Collatz (3x+1) trajectory of n that are greater than n.
+10
4
0, 0, 5, 0, 2, 3, 12, 0, 13, 1, 8, 1, 3, 8, 11, 0, 4, 7, 10, 0, 2, 4, 7, 0, 10, 1, 102, 3, 5, 8, 97, 0, 10, 2, 5, 2, 4, 5, 17, 0, 97, 1, 10, 1, 3, 5, 92, 0, 5, 4, 7, 0, 2, 95, 98, 0, 9, 1, 11, 4, 6, 91, 93, 0, 5, 3, 6, 0, 2, 3, 87, 0, 97, 1, 7, 1, 3, 10, 13, 0, 6
MATHEMATICA
Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Table[Length[Select[Collatz[n], # > n &]], {n, 100}]
a(n) = gcd(4n, A025586(4n)), greatest common divisor of 4n and largest value in 3x+1 iteration list started at 4n.
+10
2
4, 8, 4, 16, 20, 24, 4, 32, 4, 40, 4, 48, 52, 56, 20, 64, 68, 72, 4, 80, 84, 88, 4, 96, 100, 104, 4, 112, 116, 40, 4, 128, 132, 136, 20, 144, 148, 152, 4, 160, 4, 168, 4, 176, 180, 184, 4, 192, 196, 200, 4, 208, 212, 8, 4, 224, 228, 232, 4, 240, 244, 8, 4, 256, 260, 264
MATHEMATICA
c[x_] := (1-Mod[x, 2])*(x/2)+Mod[x, 2]*(3*x+1)c[1]=1; fpl[x_] := Delete[FixedPointList[c, x], -1] Table[GCD[4*w, Max[fpl[4*w]]], {w, 1, 256}]
Search completed in 0.012 seconds
| 