login
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A265673 Area A under the trajectory of each odd number in the "3x+1" problem. 1

%I

%S 0,47,33,284,334,253,112,686,205,509,137,621,645,101426,427,101088,

%T 824,522,516,2324,101310,1316,474,100955,781,1104,372,102288,1624,

%U 2157,985,105066,1132,1425,543,100754,102682,1482,886,4344,1422,102053,553,2672,1905

%N Area A under the trajectory of each odd number in the "3x+1" problem.

%C For each odd number, we consider the 3x+1 function iterates (x, T(x), T(T(x)),...,4,2,1) plotted on standard vertical and horizontal scale. Each point (0,x), (1,T(x)), (2, T(T(x)),...,(r,1), where r is the number of iterations needed to reach 1, is connected by a straight line. The area A is the sum of trapezes of base b = 1, such that A =(x + T(x))/2 + (T(x) + T(T(x)))/2 + ... + 3/2.

%C The area A is an integer number if x is odd because A = ((x+1) + 2s)/2 = (x+1)/2 + s with s = T(x) + T(T(x) + T(T(T(x))) + ... + 4 + 2.

%C The area is not unique for different values of n. Examples:

%C a(63) = a(2342) = 101699;

%C a(174) = a(450) = 118514;

%C a(241) = a(332) = 17225;

%C a(563) = a(867) = 13787;

%C a(697) = a(1131) = 14991;

%C a(1178) = a(1426) = 44427.

%C There exists index i, j such that a(i) = a(j) with trajectories having the same length, this implies that A006577(2i-1) = A006577(2j-1) where A006577 is the number of halving and tripling steps to reach 1 in "3x+1" problem. Examples:

%C a(2696) = a(17195) = 231417 and A006577(5391) = A006577(34389) = 28;

%C a(5485) = a(6025) = 183649 and A006577(10969) = A006577(12049) = 42;

%C a(9137) = a(9297) = 950016 and A006577(18273) = A006577(18593) = 154.

%C The primes of the sequence are 47, 137, 509, 2161, 3967, 4007, ...

%C The squares of the sequence are 0, 4900, 8281, 61009, 176400, 181476,…

%H Michel Lagneau, <a href="/A265673/b265673.txt">Table of n, a(n) for n = 1..10000</a>

%e a(2) = 47 because the trajectory of the second odd number 3 is 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1, and the area A = (3+10)/2 + (10+5)/2 + (5+16)/2 +(16+8)/2 + (8+4)/2 +(4+2)/2 +(2+1)/2 = 94/2 = 47.

%p nn:=2000:for k from 1 to 100 do:n:=2*k-1:s:=0:m:=n:for i from 1 to nn while(m<>1) do:if irem(m,2)=0 then s:=s+(m+m/2)/2:m:=m/2:else s:=s+(m+3*m+1)/2:m:=3*m+1:fi:od:printf(`%d, `,s):od:

%t lst={};f[n_]:=Module[{s=0,a=2*n-1,k=0},While[a!=1,k++;If[EvenQ[a],s=s+(a+a/2)/2;a=a/2,s=s+(a+a*3+1)/2;a=a*3+1]];k;AppendTo[lst,s]];Table[f[n],{n,100}];lst

%Y Cf. A006577.

%K nonn

%O 1,2

%A _Michel Lagneau_, Dec 13 2015

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified April 14 12:11 EDT 2021. Contains 342949 sequences. (Running on oeis4.)