A347789 a(n) is the number of times that only 2 pegs have disks on them during the optimal solution to a Towers of Hanoi problem with n disks.

0, 2, 4, 8, 12, 20, 28, 44, 60, 92, 124, 188, 252, 380, 508, 764, 1020, 1532, 2044, 3068, 4092, 6140, 8188, 12284, 16380, 24572, 32764, 49148, 65532, 98300, 131068, 196604, 262140, 393212, 524284, 786428, 1048572, 1572860, 2097148, 3145724, 4194300, 6291452

Offset: 1

Author

John Bonomo, Sep 13 2021

Keywords

• nonn
• easy

Comments

Zero together with the partial sum of the even terms of A016116. - Omar E. Pol, Sep 14 2021

For n >= 2, a(n+1) is the number of n X n arrays of 0's and 1's with every 2 X 2 square having density exactly 1. - David desJardins, Oct 27 2022

Links

• J. Bonomo, Back to the Tower, The College Mathematics Journal 52(2021), 265-273.
• Index entries for sequences related to Towers of Hanoi
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2).

Crossrefs

Cf. A016116, A027383, A052955, A341579.

The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Programs

Maple

a:= proc(n) option remember;
      `if`(n<3, 2*n-2, 2*(a(n-2)+2))
    end:
seq(a(n), n=1..42);  # Alois P. Heinz, Sep 14 2021

Mathematica

LinearRecurrence[{1, 2, -2}, {0, 2, 4}, 42] (* Jean-François Alcover, May 14 2022 *)

PARI

a(n) = (3+(n % 2))*(2^(n\2)) - 4; \\ Michel Marcus, Sep 14 2021

Python

def a(n): return (3 + n%2) * 2**(n//2) - 4
print([a(n) for n in range(1, 43)]) # Michael S. Branicky, Sep 14 2021

Formula

a(n) = (3+(n mod 2))*(2^floor(n/2)) - 4.

a(n) = 4 * A052955(n-3) for n >= 3. - Joerg Arndt, Sep 14 2021

a(n) = A027383(n) - 2. - Omar E. Pol, Sep 14 2021

a(n) = 2 * A027383(n-2) for n >= 2. - Alois P. Heinz, Sep 14 2021

From Stefano Spezia, Sep 14 2021: (Start)

G.f.: 2*x^2*(1+x)/((1-x)*(1-2*x^2)).

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