A265946 Sums of positive differences and negated sums of negative differences in the Collatz trajectory of n.

0, 0, 0, 1, 18, 20, 0, 3, 11, 15, 18, 23, 111, 117, 0, 7, 130, 138, 11, 20, 96, 106, 18, 29, 38, 50, 111, 124, 267, 281, 0, 15, 73, 89, 130, 147, 194, 212, 11, 30, 43, 63, 96, 117, 236, 258, 18, 41, 245, 269, 38, 63, 40563, 40589, 111, 138, 155, 183, 267, 296

Offset

1,5

Comments

For each number n, we consider the Collatz trajectory represented by the ordered sequence {x(i)}, i = 1, 2, ..., r starting with x(1) = n and ending with x(r) = 1. Let d(i) be the difference x(i+1) - x(i). This sequence lists the pairs of positive integers (u, -v) such that u is the sum of all the positive differences and v is the sum of all the negative differences.

I present an interesting conjecture because of the linearity of the result.

Conjecture: a(2n) - a(2n-1) = n - 1.

Another definition:

Consider F(x) the 3x + 1 function iterates (x, T(x), T(T(x)), ..., 4, 2, 1) plotted on standard vertical and horizontal scale where each point (0, x), (1, T(x)), (2, T(T(x))), ..., (r, 1) is connected to the next by a straight line with r being the number of iterations needed to reach 1. Let F'(x) be the derivative of F(x). The sequence lists the pairs of integers (u, -v) such that u is the sum of the positive values of F'(x) and v is the sum of the negative values of F'(x) at x = 1, 2, ..., r.

Property: a(2^n - 1) = 0. In the general case, if m is an odd integer, a(m) = q => a(m*2^r + 2^r - 1) = q, r = 0,1,2,..

Example: for m = 5, a(5)=18 => a(11) = a(23) = a(47) = a(95) = ... = 18.

Links

Michel Lagneau, Table of n, a(n) for n = 1..10000

Example

n=3 => (a(2n-1), a(2n)) = (a(5),a(6)) = (18,20) because the Collatz trajectory of 3 is 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1. The sum of the positive differences is u = (10 - 3) + (16 - 5) = 18 and the sum of the negative differences is u = (5 - 10) + (8 - 16) + (4 - 8) + (2 - 4) + (1 - 2)= -20 with the property a(6) - a(5) = 3 - 1 = 2.

Maple

T:=array(1..1000):nn:=1000:
for n from 1 to 100 do:
  kk:=1:m:=n:T[kk]:=n:it:=0:
    for i from 1 to nn while(m<>1) do:
     if irem(m, 2)=0
      then
      m:=m/2:kk:=kk+1:T[kk]:=m:it:=it+1:
      else
      m:=3*m+1:kk:=kk+1:T[kk]:=m:it:=it+1:
     fi:
    od:
   jj:=0:s1:=0:s2:=0:
    for j from 1 to it do:
     if T[j]<T[j+1] then s1:=s1+T[j+1]-T[j]:
      else
      s2:=s2+T[j]-T[j+1]:fi:
    od:
      printf(`%d, `, s1):printf(`%d, `, s2):
       od:

Crossrefs

Cf. A006577.

Sequence in context: A041680 A041681 A041682 * A027888 A332925 A333973

Adjacent sequences: A265943 A265944 A265945 * A265947 A265948 A265949

Keyword

nonn

Author

Michel Lagneau, Dec 23 2015

Status

approved