A268288 -id:A268288 - cp's OEIS frontend

A268468 Least k starting a chain or (2n+1)-tuple of consecutive integers {h(k+i)}, i=0,1,...,2n (excluding the trivial chain when h(k) = h(k+1) = ... = h(k+2n)) with symmetrical gaps about the center, where h(k) is the length of the finite set {k, f(k), f(f(k)),...,1} in the Collatz (or 3x + 1) problem.

4, 507, 1377, 12608, 55291, 55290, 55289, 145645, 104455, 104454, 336734, 336733, 336732, 525907, 1960873, 1836239, 2176265, 2176264, 2176263, 2176262, 2176261, 2176260, 2176259, 2176258

Offset: 1

Michel Lagneau, Feb 05 2016

It is interesting to search for symmetries in the sequence A006577 (number of halving and tripling steps to reach 1 in '3x+1' problem).  The symmetrical architecture is given by the following property: 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.
We observe two essential families of chains containing symmetries:
(i) A 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). This case is not considered here, but is mentioned in A078441. If this case were to be considered, it would be found that a(n) = A078441(2n+1) = 28, 98, 943, 1680, 2987, 2987, 7083, 7083, ...
(ii) Chains having several distinct values. This case is more interesting, with an important question: how many distinct values can contain such a set {h(k)}? It appears that this value is equal to 3, and the sequence of the reduced corresponding sets is {4, 5, 6}, {35, 48, 61}, {96, 127, 158}, {32, 63, 94}, {91, 122, 153}, {91, 122, 153}, {91, 122, 153}, {126, 188, 250}, ... The corresponding symmetrical centers are 5, 48, 127, 63, 122, 122, 122, 188, 172, 172, 184, 184, 184, 164, 184, 202, 210, 210, 210, 210, 210, 210, 210, 210, ...

			a(1) = 4 because in the first 3-tuple {h(4),h(5),h(6)} = {2, 5, 8}, the numbers are symmetric w.r.t. the central h(5)= 5 since 2+8 = 2*5. Hence 4 belongs to the sequence.
Alternatively, the symmetry can be seen from the differences between consecutive h(k). For {2,5,8}, the set of the differences is {3,3}.
a(3) = 10136 because in 7-tuple of consecutive {h(k)} = {34, 34, 34, 60, 86, 86, 86}, the numbers are symmetric w.r.t. its central h(k+3) = 60, since 34+86 = 2*60, and this is the smallest such 7-tuple. Hence 10136 belongs to the sequence.
Alternatively, the symmetry can be seen from the differences between consecutive h(k). From the set {34, 34, 34, 60, 86, 86, 86}, the set of the differences is {0,0,26,26,0,0}.

Cf. A006577, A078441, A268253, A268268, A268288.

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 3 by 2 to 50 do:
   ii:=0:
   for j from 1 to nn while(ii=0)do:
     q:=T[j]+T[j+n-1]:
     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]=q and T[j+(n-1)/2]=q/2
       then
       itr:=itr+1:
       else fi:
      od:
      if itr=(n-1)/2 and nops(lst)>1 then ii:=1:
      printf("%d %d \n",n,j):
      else
      fi:
     od:
     od:

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