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%I #17 Jun 04 2017 02:42:52
%S 943,1377,1494,1495,1680,1681,1682,1991,1992,1993,2358,2359,2987,2988,
%T 2989,2990,2991,2992,2993,2994,3288,3289,3360,3542,3543,3982,3983,
%U 3984,3985,3986,3987,3988,4193,4481,4482,4722,4723,4724,4725,4897,4936,4937,5313,5314
%N a(n) begins the first chain of 7 consecutive positive integers of h-values with symmetrical gaps about the center, where h(k) is the length of the finite sequence k, f(k), f(f(k)), ..., 1 in the Collatz (or 3x + 1) problem.
%C Or numbers k such that h(k) + h(k+6) = h(k+1) + h(k+5) and h(k+3) = (h(k) + h(k+6))/2, where h(k) is the length of k, f(k), f(f(k)), ..., 1 in the Collatz (or 3x + 1) problem.
%C a(1) = A078441(7).
%C The symmetry can be seen from the differences between consecutive h(k) (see the example).
%C The 7-tuple of consecutive h(k) are symmetric about the central value h(k+3) which are averages of both their immediate neighbors, their second neighbors and their third neighbors.
%C A majority of numbers of the sequence generate trivial 7-tuples {m, m, m, m, m, m, m}.
%C The 7-tuples {h(k)} of the form {m, p, p, p, p, p, q} are generated by the numbers of the sequence 1377, 4897, ...
%C The 7-tuples {h(k)} of the form {m, m, p, m, q, m, m} are generated by the numbers of the sequence 5511, 58757, ...
%C The 7-tuples {h(k)} of the form {m, p, m, m, m, q, p} are generated by the numbers of the sequence 9514, ...
%C The 7-tuples {h(k)} of the form {m, m, p, p, p, q, q} are generated by the numbers of the sequence 21442, 25666, ...
%C The 7-tuples {h(k)} of the form {m, m, m, p, q, q, q} are generated by the numbers of the sequence 55108, 55293, ...
%e In 7-tuple of consecutive {h(k)}: {h(9514),h(9515),h(9516),h(9517),h(9518),h(9519),h(9520)} = {78,52,78,78,78,104,78}, the central value is 78, and 78+78 = 52+104 = 2*78. Hence, 9514 is in the sequence.
%e Alternatively, the symmetry can be seen from the differences between consecutive {h(k)}. For {78,52,78,78,78,104,78}, the differences {h(k+1)-h(k)} are {-26,26,0,0,26,-26}.
%t lst={};f[n_]:=Module[{a=n,k=0},While[a!=1,k++;If[EvenQ[a],a=a/2,a=a*3+1]];k];Do[If[f[m]+f[m+6]==f[m+1]+f[m+5]&&f[m+2]+f[m+4]==f[m]+f[m+6]&& f[m]+f[m+6]==f[m+2]+f[m+4]&&f[m+3]==(f[m]+f[m+6])/2,AppendTo[lst,m]],{m,1,6000}];lst
%Y Cf. A006577, A078441, A268253.
%K nonn
%O 1,1
%A _Michel Lagneau_, Jan 29 2016
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