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A007664 Reve's puzzle: number of moves needed to solve the Towers of Hanoi puzzle with 4 pegs and n disks, according to the Frame-Stewart algorithm.
(Formerly M2449)
+0
8
0, 1, 3, 5, 9, 13, 17, 25, 33, 41, 49, 65, 81, 97, 113, 129, 161, 193, 225, 257, 289, 321, 385, 449, 513, 577, 641, 705, 769, 897, 1025, 1153, 1281, 1409, 1537, 1665, 1793, 2049, 2305, 2561, 2817, 3073, 3329, 3585, 3841, 4097, 4609, 5121, 5633 (list; graph; listen)
OFFSET

0,3

COMMENT

The Frame-Stewart algorithm minimizes the number of moves a(n) needed to first move k disks to an intermediate peg (requiring a(k) moves), then moving the remaining n-k disks to the destination peg without touching the k smallest disks (requiring 2^(n-k)-1 moves) and finally moving the k smaller disks to the destination.

This leads to the given recursive formula a(n) = min{...}. It follows that the sequence of first differences is A137688 = (1,2,2,4,4,4,...) = 2^A003056(n), which in turn gives the explicit formulae for a(n) as partial sums of A137688.

It is conjectured that the algorithm always gives the optimal solution; for n<=30 this is confirmed by exhaustive search, but no proof is known for the general case.

"Numerous others have rediscovered this algorithm over the years [several references omitted]; many of these failed to derive the correct value for the parameter i, most mistakenly thought that they had actually proved optimality and almost none contributed anything new to what was done by Frame and Stewart". [Stockmeyer]

Numbers of the form 2^k+1 appear for n = 2, 3, 4, 6, 8, 11, 15, 15+4 = 19, 19+5 = 24, 24+6 = 30, 30+7 = 37, 37+8 = 45... - Max Alekseyev, Feb 06 2008

REFERENCES

J.-P. Allouche, Note on the cyclic towers of Hanoi, Theoret. Comput. Sci., 123 (1994), 3-7.

A. Brousseau, Tower of Hanoi with more pegs, J. Recreational Math., 8 (1972), 169-176.

J. S. Frame, Solution to Problem 3918, Amer. Math. Monthly, 48 (1941), 216-217.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

B. M. Stewart, Advanced Problem 3918, Amer. Math. Monthly, 46 (1939), 363.

B. M. Stewart, Solution to Problem 3918, Amer. Math. Monthly, 48 (1941), 217-219.

D. Wood, Towers of Brahma and Hanoi revisited, J. Recreational Math., 14 (1981), 17-24.

LINKS

M. F. Hasler, Table of n, a(n) for n=0,...,1000.

S. Alejandre, Legend of Towers of Hanoi

A. M. Hinz, An iterative algorithm for the Tower of Hanoi with four pegs

B. Houston and H. Masum, Explorations in 4-peg Tower of Hanoi [Ppaper]

B. Houston and H. Masum, Explorations in 4-peg Tower of Hanoi [Web site]

S. Klavzar et al., Hanoi graphs and some classical numbers

S. Klavzar and U. Milutinovic, Simple explicit formulas for the Frame-Stewart's numbers

S. Klavzar, U. Milutinovic and C. Petr, On the Frame-Stewart algorithm for the multi-peg Tower of Hanoi problem, Discrete Appl. Math. 120, 1-3 (2002), 141 - 157.

Mathnet at U. Toronto, Generalizing the Towers of Hanoi Problem

Richard E. Korf and Ariel Felner. Recent Progress in Heuristic Search: a Case Study of the Four-Peg Towers of Hanoi Problem. IJCAI 2007: 2324-2329.

P. Stockmeyer, Variations on the Four-Post Tower of Hanoi Puzzle, CONGRESSUS NUMERANTIUM 102 (1994), pp. 3-12. [Has extensive bibliography]

Eric Weisstein's World of Mathematics, Towers of Hanoi

Wikipedia, Four pegs and beyond [Has further references]

FORMULA

a(n) = min{ 2 a(k) + 2^(n-k) - 1 ; k < n}, which is always odd. - M. F. Hasler, Feb 06 2008

a(n)=sum(2^A003056(i), i=0..n-1) - Daniele Parisse (daniele.parisse(AT)m.eads.net), May 09 2003

A007664(n) = 1 + (n + A003056(n) - 1 - A003056(n)*(A003056(n) + 1)/2)*2^A003056(n) - Daniele Parisse (daniele.parisse(AT)m.dasa.de), Feb 06 2001

a(n) = 1 + (n - 1 - A003056(n)*(A003056(n) - 1)/2)*2^A003056(n) - Daniele Parisse (daniele.parisse(AT)t-online.de), Jul 07 2007

MAPLE

A007664:=proc(n) option remember; min(seq(2*A007664(k)+2^(n-k)-1, k=0..n-1)) end; A007664(0):=0; # - M. F. Hasler, Feb 06 2008

A007664 := n -> 1 + (n - 1 - A003056(n)*(A003056(n) - 1)/2)*2^A003056(n); A003056 := n -> round(sqrt(2*n+2))-1; # - M. F. Hasler, Feb 06 2008

PROGRAM

(PARI) A007664(n) = (n - 1 - (n=A003056(n))*(n-1)/2)*2^n +1

A003056(n) = (sqrt(2*n+2)-.5)\1 \- M. F. Hasler, Feb 06 2008

(PARI) print_7664(n, s=0, t=1, c=1, d=1)=while(n-->=0, print1(s+=t, ", "); c--&next; c=d++; t<<=1)

(PARI) A007664(n, c=1, d=1, t=1)=sum(i=c, n, i>c&(t<<=1)&c+=d++; t) \- M. F. Hasler, Feb 06 2008

CROSSREFS

Cf. A007665, A003056, A000225 (analogue for 3 pegs), A137688 (first differences).

Sequence in context: A061571 A049690 A080075 this_sequence A114395 A075314 A152737

Adjacent sequences: A007661 A007662 A007663 this_sequence A007665 A007666 A007667

KEYWORD

nonn,nice

AUTHOR

N. J. A. Sloane (njas(AT)research.att.com), Mira Bernstein and Robert G. Wilson v (rgwv(AT)rgwv.com)

EXTENSIONS

Edited, corrected and extended by M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Feb 06 2008

Further edits by N. J. A. Sloane (njas(AT)research.att.com), Feb 08 2008

Upper bound updated with a reference by Max Alekseyev (maxale(AT)gmail.com), Nov 23 2008

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Last modified December 5 08:23 EST 2009. Contains 170348 sequences.


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