A198586 a(n) = (4^A001651(n+1) - 1)/3: numbers (4^k-1)/3 for k > 1, not multiples of 3.

5, 85, 341, 5461, 21845, 349525, 1398101, 22369621, 89478485, 1431655765, 5726623061, 91625968981, 366503875925, 5864062014805, 23456248059221, 375299968947541, 1501199875790165, 24019198012642645, 96076792050570581, 1537228672809129301

Offset: 1

Author

T. D. Noe, Oct 30 2011

Keywords

• nonn
• easy

Comments

Numbers coprime to 6 producing 2 odd numbers in the Collatz iteration.

Numbers appearing in A198585 (sorted and duplicates removed). These numbers occur in A002450, numbers of the form (4^k-1)/3, for k = 2, 4, 5, 7, 8, 10, ... (note that k a multiple of 3 does not appear).

A124477 \ {0,1} is a subset: for these n, 3n+1 = 2^(p-3) with p > 3 prime, whence also n !== 0 (mod 3). - M. F. Hasler, Oct 16 2018

These are exactly the odd non-multiples of 3 such 3n+1 = 2^m for some m, i.e., n = (2^m-1)/3. This is possible iff m = 2k, so we get n = (4^k-1)/3. Then n == 0 (mod 3) <=> 4^k == 1 (mod 9) <=> k == 0 (mod 3) <=> k not in A001651. This yields the FORMULA. (Multiples of 3 are excluded because the original definition implied that the terms are in the Collatz-orbit of another odd number, i.e., of the form n = (3x+1)/2^r, which is impossible for x a multiple of 3.) - M. F. Hasler, Oct 16 2018

From Wolfdieter Lang, Jan 14 2022: (Start)

a(n) mod 8 = 5. As subsequence of A002450 for n >= 1.

{a(n) mod 6} == repeat{5, 1}. See the first comment, and the periodicity modulo 6 of A002450 for n >= 1.

{a(n) mod 72} == repeat{5, 13, 53, 61, 29, 37}. Proof by induction: First with the bisection formulas, a(1+2*k) = (4^(2+3*k) - 1)/3 and a(2+2*k) = (4^(3*k+4) - 1)/3, for k >= 0, then trisection, using (4^9 - 1)/3 = 873819 = 9*9709. (End)

Links

• T. D. Noe, Table of n, a(n) for n = 1..100
• Index entries for linear recurrences with constant coefficients, signature (1,64,-64).

Crossrefs

Cf. A001651, A002450, A124477, A198584.

Programs

Magma

[4^(3*n  div 2 + 1) div 3: n in [1..25]]; // Vincenzo Librandi, Oct 20 2018

Mathematica

e = 19; ex = Complement[Range[2,3*e], 3*Range[e]]; (4^ex - 1)/3
(* Second program: *)
Rest@ Map[(4^# - 1)/3 &, LinearRecurrence[{1, 1, -1}, {1, 2, 4}, 21]] (* Michael De Vlieger, Oct 17 2018 *)

PARI

is(n)=gcd(n,6)==1&&(n=3*n+1)>>valuation(n,2)==1 \\ M. F. Hasler, Oct 16 2018

PARI

A198586(n)=4^(3*n\2+1)\3 \\ M. F. Hasler, Oct 16 2018

PARI

Vec(x*(5 + 80*x - 64*x^2) / ((1 - x)*(1 - 8*x)*(1 + 8*x)) + O(x^20)) \\ Colin Barker, Jan 17 2020

Formula

a(n) = (4^A001651(n+1) - 1)/3. - M. F. Hasler, Oct 16 2018

From Colin Barker, Jan 17 2020: (Start)

G.f.: x*(5 + 80*x - 64*x^2) / ((1 - x)*(1 - 8*x)*(1 + 8*x)).

a(n) = a(n-1) + 64*a(n-2) - 64*a(n-3) for n>3.

a(n) = (-1 + (-8)^n + 3*8^n) / 3.

(End)

Extensions

Definition corrected by M. F. Hasler, Oct 16 2018