A055662 -id:A055662 - cp's OEIS frontend

A055661 Tower of Hanoi positions (A055662) converted from base 3 to base 10.

0, 1, 7, 8, 17, 15, 12, 13, 67, 68, 65, 63, 72, 73, 79, 80, 161, 159, 156, 157, 139, 140, 137, 135, 108, 109, 115, 116, 125, 123, 120, 121, 607, 608, 605, 603, 612, 613, 619, 620, 593, 591, 588, 589, 571, 572, 569, 567, 648, 649, 655, 656, 665, 663, 660, 661

Offset: 0

Henry Bottomley, Jun 06 2000

Zoran Šunić, Tree morphisms, transducers and integer sequences, arXiv:math/0612080 [math.CO], 2006, see after proposition 5.4.
Index entries for sequences related to Towers of Hanoi

Cf. A055662, A128202.

a[n_]:=Sum[3^j*(Mod[(-1)^j*Floor[(n/2^j + 1)/2],3]), {j,0,Floor[Log2[n]]}]; Array[a, 80, 0] (* Stefano Spezia, Oct 27 2018 *)

a(n) = Sum_{j=0..floor(log_2(n))} 3^j*((-1)^j*floor((n/2^j + 1)/2) mod 3). - Corrected by Stefano Spezia, Oct 27 2018

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).

Original entry on oeis.org

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

Henry Bottomley, Apr 03 2001

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

			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,...

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

Cf. A001511, A055662, A060571, A060573, A060574, A060575.

a(n) = (- (-1)^valuation(n,2) - n)%3; \\ Kevin Ryde, Aug 07 2021

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.
a(n) = A001511(n)-th digit from right of A055662(n).
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

A060571 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 (here) to peg A060572.

Original entry on oeis.org

0, 0, 1, 0, 2, 2, 0, 0, 1, 1, 2, 1, 0, 0, 1, 0, 2, 2, 0, 2, 1, 1, 2, 2, 0, 0, 1, 0, 2, 2, 0, 0, 1, 1, 2, 1, 0, 0, 1, 1, 2, 2, 0, 2, 1, 1, 2, 1, 0, 0, 1, 0, 2, 2, 0, 0, 1, 1, 2, 1, 0, 0, 1, 0, 2, 2, 0, 2, 1, 1, 2, 2, 0, 0, 1, 0, 2, 2, 0, 2, 1, 1, 2, 1, 0, 0, 1, 1, 2, 2, 0, 2, 1, 1, 2, 2, 0, 0, 1, 0, 2, 2, 0, 0, 1

Offset: 1

Henry Bottomley, Apr 03 2001

a(n) is equal to a(2n) with the 1's and 2s reversed, thus a(n) = a(4n). - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003

			Start by moving first disk from peg 0 (to peg 1), second disk from peg 0 (to peg 2), first disk form peg 1 (to peg 2), etc. so sequence starts 0,0,1,...

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

Cf. A001511, A055662, A060572, A060573, A060574, A060575.

If n>2^A001511(n) then a(n)=a(n-2^A001511(n))-(-1)^A001511(n) mod 3 =A060572(n-2^A001511(n)), otherwise a(k)=0. Also A001511(n)-th digit from right of A055662(n-1).

A060574 Tower of Hanoi: using 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, a(n) is the smallest disk on peg 1 after n moves (or 0 if there are no disks on peg 1).

Original entry on oeis.org

0, 1, 1, 0, 3, 3, 2, 1, 1, 2, 3, 3, 0, 1, 1, 0, 5, 5, 2, 1, 1, 2, 5, 5, 4, 1, 1, 4, 3, 3, 2, 1, 1, 2, 3, 3, 4, 1, 1, 4, 5, 5, 2, 1, 1, 2, 5, 5, 0, 1, 1, 0, 3, 3, 2, 1, 1, 2, 3, 3, 0, 1, 1, 0, 7, 7, 2, 1, 1, 2, 7, 7, 4, 1, 1, 4, 3, 3, 2, 1, 1, 2, 3, 3, 4, 1, 1, 4, 7, 7, 2, 1, 1, 2, 7, 7, 6, 1, 1, 6, 3, 3, 2, 1, 1

Offset: 0

Henry Bottomley, Apr 03 2001

			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 0, 1, 1, 0, ...

Index entries for sequences related to Towers of Hanoi

Cf. A060573 (smallest on peg 0), A060575 (smallest on peg 2), A055662 (whole configuration).

Cf. A001511, A060571, A060572.

A060574_upto(n, p=[[]|n<-[1..3]])={p[1]=[1..n]; vector(2^n-1, n, my(a = n\2%3+1, b = a%3+1); bittest(n,0) || if( !p[a = b%3+1] || (#p[b] && p[b][1] < p[a][1]), [a, b] = [b, a]); p[b] = concat(p[a][1], p[b]); p[a] = p[a][^1]; if(p[2], p[2][1]))} \\ This produces the sequence for n pegs, i.e., of length 2^n-1 ! - M. F. Hasler, Mar 28 2022
apply( {A060574(n,t=n%3>0)=n\=3<A060574(n)= if ( n%3, 1+2*valuation(1+bitor(n\3, 4^exponent(n)\6),4), n && n = bitand(n\6,4^exponent(n)\3), 2*valuation(n,4)+2, 0)}, [0..99])
\\ M. F. Hasler, Mar 29 2022

From M. F. Hasler, Mar 29 2022: (Start)
a(3k+1) = a(3k+2) = 2*b(k) + 1, where b(k) = valuation_4((k OR 2*(4^m-1)/3)+1) is the number of times k must be divided by 4 (discarding a remainder) until the result is even, i.e., the position of the rightmost even base-4 digit of k.
(Here and below, m is any integer > log_4(k).)
a(6k) = a(6k+3) = 2*c(k), where c(k) = 1 + valuation_4(k AND (4^m-1)/3) is the number of times 2*k must be divided by 4 (discarding a remainder) until the result is odd (i.e., the position of the rightmost odd base-4 digit in 2*k), or c(k) = 0 if this never happens <=> n is in 12*A000695 + {0, 3} <=> a(n) = 0. (End)

A060573 Tower of Hanoi: using 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, a(n) is the smallest disk on peg 0 after n moves.

Original entry on oeis.org

1, 2, 3, 3, 4, 1, 1, 4, 5, 5, 2, 1, 1, 2, 5, 5, 6, 1, 1, 6, 3, 3, 2, 1, 1, 2, 3, 3, 6, 1, 1, 6, 7, 7, 2, 1, 1, 2, 7, 7, 4, 1, 1, 4, 3, 3, 2, 1, 1, 2, 3, 3, 4, 1, 1, 4, 7, 7, 2, 1, 1, 2, 7, 7, 8, 1, 1, 8, 3, 3, 2, 1, 1, 2, 3, 3, 8, 1, 1, 8, 5, 5, 2, 1, 1, 2, 5, 5, 4, 1, 1, 4, 3, 3, 2, 1, 1, 2, 3, 3, 4, 1, 1, 4, 5

Offset: 0

Henry Bottomley, Apr 03 2001

			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 1,2,3,3,...

Index entries for sequences related to Towers of Hanoi

Cf. A060574 (smallest on peg 1), A060575 (smallest on peg 2), A055662 (whole configuration).

Cf. A001511, A060571, A060572.

A060575 Tower of Hanoi: using 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, a(n) is the smallest disk on peg 2 after n moves (or 0 if there are no disks on peg 2).

Original entry on oeis.org

0, 0, 2, 1, 1, 2, 0, 0, 4, 1, 1, 4, 3, 3, 2, 1, 1, 2, 3, 3, 4, 1, 1, 4, 0, 0, 2, 1, 1, 2, 0, 0, 6, 1, 1, 6, 3, 3, 2, 1, 1, 2, 3, 3, 6, 1, 1, 6, 5, 5, 2, 1, 1, 2, 5, 5, 4, 1, 1, 4, 3, 3, 2, 1, 1, 2, 3, 3, 4, 1, 1, 4, 5, 5, 2, 1, 1, 2, 5, 5, 6, 1, 1, 6, 3, 3, 2, 1, 1, 2, 3, 3, 6, 1, 1, 6, 0, 0, 2, 1, 1, 2, 0, 0, 4

Offset: 0

Henry Bottomley, Apr 03 2001

			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 0,0,2,1,...

Index entries for sequences related to Towers of Hanoi

Cf. A060573 (smallest on peg 0), A060574 (smallest on peg 1), A055662 (whole configuration).

Cf. A001511, A060571, A060572.

A060586 Minimum number of moves required to reach position n starting from 0 in Tower of Hanoi (with three pegs: 0,1,2), where with position n written in base 3, xyz means smallest disk is on peg z, second smallest is on peg y, third smallest on peg x, etc. and leading zeros indicate largest disks are all on peg 0.

Original entry on oeis.org

0, 1, 1, 3, 3, 2, 3, 2, 3, 7, 6, 7, 6, 7, 7, 5, 5, 4, 7, 7, 6, 5, 4, 5, 6, 7, 7, 15, 15, 14, 13, 12, 13, 14, 15, 15, 12, 13, 13, 15, 15, 14, 15, 14, 15, 11, 10, 11, 10, 11, 11, 9, 9, 8, 15, 14, 15, 14, 15, 15, 13, 13, 12, 11, 11, 10, 9, 8, 9, 10, 11, 11, 12, 13, 13, 15, 15, 14, 15, 14

Offset: 0

Henry Bottomley, Apr 04 2001

			a(46) = 10 since 46 written in base 3 is 1201 (i.e. with the smallest and fourth smallest disks on the first peg, the third smallest disk on the second peg and the second smallest and any other disks on the zeroth peg) and the optimal moves starting from position 0 go through positions 2, 12, 11, 211, 210, 220, 222, 1222, 1220, 1210 taking ten moves.

Index entries for sequences related to Towers of Hanoi

k appears A001316(k) times in the sequence.

Cf. A001511, A055661, A055662.

a(n) = A060585(A060583(n)).

A060592 Square table by antidiagonals of minimum number of moves between two positions in the Tower of Hanoi (with three pegs: 0,1,2), where with position n written in base 3, xyz means smallest disk is on peg z, second smallest is on peg y, third smallest on peg x, etc. and leading zeros indicate largest disks are all on peg 0.

Original entry on oeis.org

0, 1, 1, 1, 0, 1, 3, 1, 1, 3, 3, 3, 0, 3, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 0, 2, 2, 3, 2, 2, 1, 1, 1, 1, 2, 2, 3, 1, 3, 1, 0, 1, 3, 1, 3, 7, 2, 2, 1, 1, 1, 1, 2, 2, 7, 6, 6, 3, 2, 2, 0, 2, 2, 3, 6, 6, 7, 5, 7, 2, 3, 2, 2, 3, 2, 7, 5, 7, 6, 6, 6, 6, 3, 3, 0, 3, 3, 6, 6, 6, 6, 7, 5, 7, 5, 5, 3, 1, 1, 3, 5, 5, 7, 5, 7

Offset: 0

Henry Bottomley, Apr 06 2001

			T(4,9)=5 since 4 and 9 written in base 3 are 11 and 100, i.e. the starting position has the first and second disks on peg 1 and the others on peg 0, while the end position has the third disk on peg 1 and the others on peg 0; the five optimal moves between these positions are: move the third disk to peg 2, then the first to peg 2, the second to peg 0, the first to peg 0 and finally the third to peg 1.

Cf. A001511, A007798, A055661, A055662.

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A055661 Tower of Hanoi positions (A055662) converted from base 3 to base 10.

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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).

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A060571 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 (here) to peg A060572.

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A060574 Tower of Hanoi: using 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, a(n) is the smallest disk on peg 1 after n moves (or 0 if there are no disks on peg 1).

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A060573 Tower of Hanoi: using 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, a(n) is the smallest disk on peg 0 after n moves.

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A060575 Tower of Hanoi: using 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, a(n) is the smallest disk on peg 2 after n moves (or 0 if there are no disks on peg 2).

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