A060572 Tower of Hanoi: the optimal way to move an even number of disks from peg 0 to peg 2 or an odd number from peg 0 to peg 1 is on move n to move disk A001511 from peg A060571 to peg A060572 (here).
1, 2, 2, 1, 0, 1, 1, 2, 2, 0, 0, 2, 1, 2, 2, 1, 0, 1, 1, 0, 2, 0, 0, 1, 1, 2, 2, 1, 0, 1, 1, 2, 2, 0, 0, 2, 1, 2, 2, 0, 0, 1, 1, 0, 2, 0, 0, 2, 1, 2, 2, 1, 0, 1, 1, 2, 2, 0, 0, 2, 1, 2, 2, 1, 0, 1, 1, 0, 2, 0, 0, 1, 1, 2, 2, 1, 0, 1, 1, 0, 2, 0, 0, 2, 1, 2, 2, 0, 0, 1, 1, 0, 2, 0, 0, 1, 1, 2, 2, 1, 0, 1, 1, 2, 2
Offset: 1
Examples
Start by moving first disk (from peg 0) to peg 1, second disk (from peg 0) to peg 2, first disk (from peg 1) to peg 2, etc., so sequence starts 1,2,2,...
Links
- Gary W. Adamson, Comments on A060572
- J.-P. Allouche, D. Astoorian, J. Randall, and J. Shallit, Morphisms, squarefree strings, and the Tower of Hanoi puzzle, Amer. Math. Monthly 101 (1994), 651-658.
- Index entries for sequences related to Towers of Hanoi
Programs
-
PARI
a(n) = (- (-1)^valuation(n,2) - n)%3; \\ Kevin Ryde, Aug 07 2021
Formula
If n > 2^A001511(n) then a(n) = a(n-2^A001511(n)) - (-1)^A001511(n) mod 3, otherwise a(k) = -(-1)^A001511(n) mod 3.
If a(n)=0 then a(2n)=0, If a(n)=1 then a(2n)=2, If a(n)=2 then a(2n)=1, Thus a(n)=a(4n). - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003
a(5n) = A060571(n) with the 1's and 2s reversed. - Donald Sampson (marsquo(AT)hotmail.com), Dec 08 2003
Comments