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A060572
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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).
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5
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| If written in a fractal pattern of 4 X 4 squares, skipping the first square, going right then down then right then down, etc.:
X122 1011 ...
1011 0200
2200 1122
2122 1011
a number of patterns become apparent. Most notably the central diagonal going from the X down and to the right, when the 1's and 2's are reversed, gives the sequence A060571. When the same process is applied to A060571, this sequence emerges. - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003
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
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FORMULA
| a(n)=A060571(n)-(-1)^A001511(n) mod 3. 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. Also A001511(n)-th digit from right of A055662(n).
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EXAMPLE
| 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,...
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CROSSREFS
| Cf. A001511, A055662, A060571, A060572, A060573, A060574, A060575.
Sequence in context: A189996 A016390 A055800 * A163543 A180009 A180010
Adjacent sequences: A060569 A060570 A060571 * A060573 A060574 A060575
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KEYWORD
| easy,nonn
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AUTHOR
| Henry Bottomley (se16(AT)btinternet.com), Apr 03 2001
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