cp's OEIS Frontend

A268253 a(n) begins the first chain of 5 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.

%I A268253 #25 Jun 04 2017 02:44:06
%S A268253 98,130,290,354,386,387,418,507,514,610,628,802,840,841,866,943,944,
%T A268253 945,1003,1121,1122,1154,1172,1186,1272,1314,1378,1442,1494,1495,1496,
%U A268253 1497,1538,1634,1680,1681,1682,1683,1684,1698,1699,1826,1890,1922,1923,1991,1992
%N A268253 a(n) begins the first chain of 5 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 A268253 a(1) = A078441(5).
%C A268253 Or numbers k such that h(k) + h(k+4) = h(k+1) + h(k+3) and h(k+2) = (h(k) + h(k+4))/2, where h(k) is the length of k, f(k), f(f(k)), ..., 1 in the Collatz (or 3x + 1) problem.
%C A268253 The 5-tuple of consecutive h(k) are symmetric about the central value h(k+2) which are averages of both their immediate neighbors and their second neighbors.
%C A268253 A majority of numbers generate trivial 5-tuples {m, m, m, m, m}.
%C A268253 The 5-tuples {h(k)} of the form {m, p, p, p, q} are generated by the numbers of the sequence 507, 1003, ...
%C A268253 The 5-tuples {h(k)} of the form {m, p, m, q, m} are generated by the numbers of the sequence 1272, 3672, ...
%H A268253 Vincenzo Librandi, <a href="/A268253/b268253.txt">Table of n, a(n) for n = 1..1560</a>
%e A268253 In 5-tuple of consecutive {h(k)}: {h(1272),h(1273),h(1274),h(1275),h(1276)} = {57,31,57,83,57}, the central value is 57, and 57+57 = 31+83 = 2*57. Hence, 1272 is in the sequence.
%e A268253 Alternatively, the symmetry can be seen from the differences between consecutive {h(k)}. For {57,31,57,83,57}, the differences {h(k+1)-h(k)} are {-26,26,26,-26}.
%t A268253 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+4]==f[m+1]+f[m+3]&&f[m+2]==(f[m]+f[m+4])/2,AppendTo[lst,m]],{m,1,4000}];lst
%Y A268253 Cf. A006577, A078441, A268268.
%K A268253 nonn
%O A268253 1,1
%A A268253 _Michel Lagneau_, Jan 29 2016