A137695 Tower of Hanoi with p pegs: A(p,n) = number of moves needed for n disks, using Frame's or Stewart's algorithm, read by columns of the upper right triangle of rows p >= 3, columns n >= p-2.

1, 3, 3, 7, 5, 5, 15, 9, 7, 7, 31, 13, 11, 9, 9, 63, 17, 15, 13, 11, 11, 127, 25, 19, 17, 15, 13, 13, 255, 33, 23, 21, 19, 17, 15, 15, 511, 41, 27, 25, 23, 21, 19, 17, 17, 1023, 49, 31, 29, 27, 25, 23, 21, 19, 19, 2047, 65, 39, 33, 31, 29, 27, 25, 23, 21, 21, 4095, 81, 47, 37, 35

Offset: 1

Author

M. F. Hasler, Feb 09 2008, revised Feb 10 2008

Keywords

• easy
• nice
• nonn
• tabl

Comments

In the cited paper by Klavzar et al. it is proved that Frame's algorithm and Stewart's algorithm, as well as several variations, all yield the same number of moves needed for the n disk, p peg problem, given by the formula for A(p,n).

This sequence lists the elements of the upper right triangle of the matrix having as rows the number of moves required, depending on the number of disks, for a given number of pegs. (The first row refers to 3 pegs, etc.) The lower left triangle of the matrix is uninteresting, since all elements below a given diagonal element are equal to that element, namely 2n-1. (For p>n, each disk can be moved to a separate peg.)

However, the article by Klavžar and Milutinović, "Simple explicit formulas for the Frame-Stewart numbers", points out that there is (at least as of 2002) no proof that this algorithm is optimal. - N. J. A. Sloane, Sep 10 2018

Examples

			The complete matrix would read (starting with row p = 3, column n = 1):
  1_  3   7  15  31   63  127  255  511 1023 ...
  1 \_3_  5   9  13   17   25   33   41   49 ...
  1   3 \_5_  7  11   15   19   23   27   31 ...
  1   3   5 \_7_  9   13   17   21   25   29 ...
  1   3   5   7 \_9_  11   15   19   23   27 ...
  1   3   5   7   9 \_11_  13   17   21   25 ...
  1   3   5   7   9   11 \_13_  15   19   23 ...
  1   3   5   7   9   11   13 \_15_  17   21 ...
  1   3   5   7   9   11   13   15 \_17_  19 ...                     _
(Since the columns become constant below the diagonal (symbolized by  \_), we list only the part above it.)
First row: 3 pegs, equals A000225; 2nd row: 4 pegs, cf. A007664; 3rd row: 5 pegs, cf. A007665.
		

References

• Sandi Klavžar and Uroš Milutinović. "Simple explicit formulas for the Frame-Stewart numbers." Annals of Combinatorics 6.2 (2002): 157-167; 0218-0006/02/020157-11. [This is an 11-page article. There is a free article on the web with the same authors and title, but which is only two pages long. - N. J. A. Sloane, Sep 10 2018]

Links

• M. F. Hasler, Table of n, a(n) for n = 1..1000, Apr 08 2025
• S. Klavzar et al., On the Frame-Stewart algorithm for the multi-peg Tower of Hanoi problem, Discrete Applied Mathematics 120 (2002) 141-15 and references therein.

Crossrefs

Cf. A000225, A007664, A007665.

Programs

Mathematica

s[n_, p_] := (k = 0; While[ n > Binomial[p - 3 + k++, p - 2] ] ; k - 2); x[n_, p_] := (snp = s[n, p]; Sum[2^t*Binomial[p - 3 + t, p - 3], {t, 0, snp - 1}] + 2^snp*(n - Binomial[p - 3 + snp, p - 2])); Flatten[Table[x[n, p], {n, 1, 12}, {p, 3, n + 2}] ] (* Jean-François Alcover, Jun 01 2011 *)

PARI

A137695(p, n=0/* if n=0, give p-th term of the "flat" sequence */)={n || p = 2+p+(1-n=1+sqrtint(2*p-2-sqrtint(2*p-2)))*n\2; if(p>n+1, 2*n-1, p==3, 2^n-1, my(s=1, t=0); while( n>binomial(p-2+t++,p-2), s+=2^t*binomial(p-3+t, p-3)); s+2^t*(n-binomial(p-3+t,p-2)))} \\ Edited by M. F. Hasler, Apr 08 2025

Formula

A(p,n) = Sum_{t=0..s-1} 2^t*binomial(p-3+t, p-3) + 2^s*(n-binomial(p-3+s, p-2)) where s = max { k in Z | n > binomial(p-3+k,p-2) }.

A(p,n) = 2n-1 if n > p+1, else 2^n-1 if p = 3, else see above. - M. F. Hasler, Apr 08 2025