OFFSET
1,2
COMMENTS
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.
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.
The area is not unique for different values of n. Examples:
a(63) = a(2342) = 101699;
a(174) = a(450) = 118514;
a(241) = a(332) = 17225;
a(563) = a(867) = 13787;
a(697) = a(1131) = 14991;
a(1178) = a(1426) = 44427.
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:
The primes of the sequence are 47, 137, 509, 2161, 3967, 4007, ...
The squares of the sequence are 0, 4900, 8281, 61009, 176400, 181476,…
LINKS
Michel Lagneau, Table of n, a(n) for n = 1..10000
EXAMPLE
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.
MAPLE
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:
MATHEMATICA
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Dec 13 2015
STATUS
approved