A268486 Least k starting a chain of (2n+1) consecutive integers {h(k+i)}, i=0,1,...,2n, where h(k) is the length of the finite set {k, f(k), f(f(k)), ..., 1} in the Collatz (or 3x + 1) problem, with the property that h(k) = h(k+2n), h(k+1) = h(k+2n-1), ..., h(k+n-1) = h(k+n+1).

24, 48, 230, 229, 228, 2987, 7083, 7083, 14168, 15959, 57346, 57346, 119388, 182852, 365740, 365739, 365738, 596310, 596310, 1088124, 1088123, 2901713, 2901712, 3264428, 3264428

Offset: 1

Author

Michel Lagneau, Feb 06 2016

Keywords

• nonn
• more

Comments

It is interesting to look for symmetries in the sequence A006577 (number of halving and tripling steps to reach 1 in '3x+1' problem). The symmetrical architecture is different from A268468 with h(k) + h(k+2n) = h(k+1)+ h(k+2n-1) = ... = h(k+n-1)+ h(k+n+1) = 2*h(k+n) where h(k+n) is the symmetrical center. Here the property is h(k)= h(k+2n), h(k+1) = h(k+2n-1), ..., h(k+n-1) = h(k+n+1), and the corresponding symmetrical centers h(k+n) have no special properties (for instance, a symmetrical center is not calculable according to the others terms of the chain).

We observe two essential families of chains containing symmetries:

(i) The majority of trivial chains are obtained when a(n) begins the first chain of 2n+1 consecutive positive integers where h(k) = h(k+1) = ... = h(k+2n).

The numbers of the sequence having this property are 2987, 7083, 57346, 596310, 3264428, ...

(ii) Chains having several distinct values. This case is more interesting, with an important question: how many maximum distinct values can contain such a set {h(k)}?

The numbers of the sequence having this property are 24, 48, 230, 229, 228, 14168, 15959, 119388, 182852, 365740, 365739, 365738, 1088124, 2901713, ... For each corresponding chain of the sequence, the number of distinct elements is 2, 2, 4, 4, 4, 1, 1, 1, 2, 3, 1, 1, 2, 2, 2, 2, 2, 1, 1, 3, 3, 2, 2, 1, ...

The symmetrical centers h(k+n) are 23, 24, 83, 83, 83, 48, 57, 57, 151, 190, 78, 78, 242, 85, 197, ... with the corresponding k = 25, 50, 233, 233, 233, 2993, 7090, 7091, 14177, 15969, 57357, ...

Examples

			a(3) = 230 because in the first 7-tuple {h(230), h(231), h(232), h(233), h(234), h(235), h(236)} = {34, 127, 21, 83, 21, 127, 34}, the numbers are symmetric w.r.t. its central h(233) = 83. Hence 230 belongs to the sequence.
The first elements k of the other 7-tuples {h(k+i)}, i=0..6, are 362, 810, 836, 943, 1222, 1256, 1322, 1410, ...
		

Crossrefs

Cf. A006577, A268468.

Programs

Maple

nn:=10^7:T:=array(1..nn):
for j from 1 to 5*10^6 do:
  k:=0:m:=j:it:=0:
    for i from 1 to nn while(m<>1) do:
     if irem(m,2)=0
      then
       m:=m/2:
      else
      m:=3*m+1:
     fi:
    it:=it+1:
    od:
    k:=j:T[j]:=it:
   od:
   for n from 43 by 2 to 60 do:
   ii:=0:
   for j from 1 to nn while(ii=0)do:
     itr:=0:lst:={}:
     for jj from 1 to (n-1)/2 do:
      lst:=lst union {T[j+jj-1]} union {T[j+n-jj]}:
      if T[j+jj-1]= T[j+n-jj]
       then
       itr:=itr+1:
       else fi:
      od:
      if itr=(n-1)/2 then ii:=1:
      printf("%d %d \n",n,j):
      else
      fi:
     od:
     od: