This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A268468 #11 Jan 05 2020 07:17:58
%S A268468 4,507,1377,12608,55291,55290,55289,145645,104455,104454,336734,
%T A268468 336733,336732,525907,1960873,1836239,2176265,2176264,2176263,2176262,
%U A268468 2176261,2176260,2176259,2176258
%N 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.
%C A268468 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.
%C A268468 We observe two essential families of chains containing symmetries:
%C A268468 (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, ...
%C A268468 (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, ...
%e A268468 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.
%e A268468 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}.
%e A268468 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.
%e A268468 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}.
%p A268468 nn:=10^7:T:=array(1..nn):
%p A268468 for j from 1 to 5*10^6 do:
%p A268468 k:=0:m:=j:it:=0:
%p A268468 for i from 1 to nn while(m<>1) do:
%p A268468 if irem(m,2)=0
%p A268468 then
%p A268468 m:=m/2:
%p A268468 else
%p A268468 m:=3*m+1:
%p A268468 fi:
%p A268468 it:=it+1:
%p A268468 od:
%p A268468 k:=j:T[j]:=it:
%p A268468 od:
%p A268468 for n from 3 by 2 to 50 do:
%p A268468 ii:=0:
%p A268468 for j from 1 to nn while(ii=0)do:
%p A268468 q:=T[j]+T[j+n-1]:
%p A268468 itr:=0:lst:={}:
%p A268468 for jj from 1 to (n-1)/2 do:
%p A268468 lst:=lst union {T[j+jj-1]} union {T[j+n-jj]}:
%p A268468 if T[j+jj-1]+T[j+n-jj]=q and T[j+(n-1)/2]=q/2
%p A268468 then
%p A268468 itr:=itr+1:
%p A268468 else fi:
%p A268468 od:
%p A268468 if itr=(n-1)/2 and nops(lst)>1 then ii:=1:
%p A268468 printf("%d %d \n",n,j):
%p A268468 else
%p A268468 fi:
%p A268468 od:
%p A268468 od:
%Y A268468 Cf. A006577, A078441, A268253, A268268, A268288.
%K A268468 nonn,more
%O A268468 1,1
%A A268468 _Michel Lagneau_, Feb 05 2016