

A277215


a(n) is the smallest even number not congruent to 1 modulo 3 that starts a (2n+1)element alternating sequence of x/2 and (3x+1) iterations ending in the maximum of its Collatz trajectory.


3



0, 26, 6, 14, 30, 1214, 1662, 254, 510, 1022, 2046, 28670, 40958, 180222, 32766, 65534, 131070, 1835006, 5767166, 1048574, 2097150, 4194302, 8388606, 16777214, 33554430, 469762046, 671088638, 268435454, 536870910, 7516192766, 2147483646, 4294967294, 8589934590, 17179869182, 34359738366, 755914244094
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OFFSET

0,2


COMMENTS

a(n) starts a maximal alternating Collatz sequence v_0, ..., v_2n of 2n+1 elements and must have the form v_0 = 2*(q*2^n  1) where q is the smallest odd number not a multiple of 3 such that v_(2n) = 2*(q*3^n  1) is the maximum of its Collatz trajectory.
The intermediate elements of the sequence for 1 <= j <= n are v_(2j1) = q * 2^(nj+1) * 3^(j1)  1, which is odd, and v_(2j) = 2 * (q * 2^(nj) * 3^j  1), which is congruent to 2 modulo 4 except for j=n.
A277875(n) is the odd multiplier q in the expression for a(n).
Subsequences of a(n) are related to subsequences of the following sequences depending on the value of q:
a(n) = 2*A000225(n) = A000918(n+1) when A277875(n) = 1;
a(n) = 2*A153894(n) = A131051(n+3) when A277875(n) = 5;
a(n) = 2*A086224(n) = A086224(n+1)1 = A176448(n+1) when A277875(n) = 7;
a(n) = A086225(n+1)1 when A277875(n) = 11;
a(n) = A198274(n+1)1 when A277875(n) = 13;
a(n) = A198276(n+1)1 when A277875(n) = 19;
For small q > 1, the positions of 2*(q*2^n  1) among the first 200 numbers in the sequence are:
q = 5: 12, 26, 36, 46, 58, 62, 174;
q = 7: 1, 11, 17, 25, 29, 45, 49, 53, 57, 61, 65, 77, 93, 103, 109, 113, 117, 125, 139, 141, 145, 157, 165, 173, 187, 189, 193;
q = 11: 13, 18, 35, 59, 69, 75, 83, 114, 133, 179;
q = 13: 6, 118;
q = 19: 5;
and among the first 400 numbers are:
q = 17: 222, 229, 230, 268;
(see A277875).
Conjecture: For every n there is an odd number q such that the alternating sequence ends in v_(2n), the maximum of the Collatz trajectory of a(n)=v_0.


LINKS

Table of n, a(n) for n=0..35.


EXAMPLE

a(0) = 0 = 2*(1*2^0  1) since it is the start and end of the first alternating sequence of 1 element and the maximum of its trajectory.
a(1) = 26 = 2*(7*2^1  1) since sequence 26, 13, 40 has 3 elements and ends in the maximum of its trajectory and since 2, 10 and 18 do not satisfy the conditions for a(1).
a(5) = 1214 = 2*(19*2^5  1) starts the alternating sequence of 11 elements  1214, 607, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232  that ends in the trajectory maximum 9232 while the 11element alternating sequences starting at 2*(q*2^5  1) with odd q<19 either do not end at the trajectory maximum or are congruent to 1 modulo 3 and therefore do not have maximal length.


MATHEMATICA

collatz[n_] := If[OddQ[n], 3n+1, n/2]
altdata[low_, high_] := Module[{n, q, notDone, v, a, m, list={}}, For[n=low, n<=high, n++, q=1; notDone=True; While[notDone, q+=2; v=2*(q*2^n1); If[Mod[v, 3]!=1, a=NestWhile[collatz, v, Mod[#, 4]!=0&]; m=Max[NestWhileList[collatz, a, #!=1&]]; notDone=(a!=m)]]; AppendTo[list, {n, q, v, a}]]; list]/; (low>1)
a277215[n_]:=Map[#[[3]]&, altdata[2, n]]
Join[{0, 26}, a277215[35]] (* sequence data *)


CROSSREFS

Cf. A000225, A000918, A086224, A086225, A131051, A153894, A176448, A198274, A198276, A277875.
Sequence in context: A073029 A040657 A309073 * A331481 A040656 A070661
Adjacent sequences: A277212 A277213 A277214 * A277216 A277217 A277218


KEYWORD

nonn


AUTHOR

Hartmut F. W. Hoft, Nov 03 2016


STATUS

approved



