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.

98, 130, 290, 354, 386, 387, 418, 507, 514, 610, 628, 802, 840, 841, 866, 943, 944, 945, 1003, 1121, 1122, 1154, 1172, 1186, 1272, 1314, 1378, 1442, 1494, 1495, 1496, 1497, 1538, 1634, 1680, 1681, 1682, 1683, 1684, 1698, 1699, 1826, 1890, 1922, 1923, 1991, 1992

Offset

1,1

Comments

a(1) = A078441(5).

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.

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.

A majority of numbers generate trivial 5-tuples {m, m, m, m, m}.

The 5-tuples {h(k)} of the form {m, p, p, p, q} are generated by the numbers of the sequence 507, 1003, ...

The 5-tuples {h(k)} of the form {m, p, m, q, m} are generated by the numbers of the sequence 1272, 3672, ...

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1560

Example

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.
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}.

Mathematica

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

Crossrefs

Cf. A006577, A078441, A268268.

Sequence in context: A244207 A344843 A109553 * A105405 A341180 A044430

Adjacent sequences: A268250 A268251 A268252 * A268254 A268255 A268256

Keyword

nonn

Author

Michel Lagneau, Jan 29 2016

Status

approved