This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A137695 #51 Jul 21 2025 07:40:06
%S A137695 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,
%T A137695 13,13,255,33,23,21,19,17,15,15,511,41,27,25,23,21,19,17,17,1023,49,
%U A137695 31,29,27,25,23,21,19,19,2047,65,39,33,31,29,27,25,23,21,21,4095,81,47,37,35
%N 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.
%C A137695 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).
%C A137695 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.)
%C A137695 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
%D A137695 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]
%H A137695 M. F. Hasler, <a href="/A137695/b137695.txt">Table of n, a(n) for n = 1..1000</a>, Apr 08 2025
%H A137695 S. Klavzar et al., <a href="http://citeseer.ist.psu.edu/klavzar00framestewart.html">On the Frame-Stewart algorithm for the multi-peg Tower of Hanoi problem</a>, Discrete Applied Mathematics 120 (2002) 141-15 and references therein.
%F A137695 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) }.
%F A137695 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
%e A137695 The complete matrix would read (starting with row p = 3, column n = 1):
%e A137695 1_ 3 7 15 31 63 127 255 511 1023 ...
%e A137695 1 \_3_ 5 9 13 17 25 33 41 49 ...
%e A137695 1 3 \_5_ 7 11 15 19 23 27 31 ...
%e A137695 1 3 5 \_7_ 9 13 17 21 25 29 ...
%e A137695 1 3 5 7 \_9_ 11 15 19 23 27 ...
%e A137695 1 3 5 7 9 \_11_ 13 17 21 25 ...
%e A137695 1 3 5 7 9 11 \_13_ 15 19 23 ...
%e A137695 1 3 5 7 9 11 13 \_15_ 17 21 ...
%e A137695 1 3 5 7 9 11 13 15 \_17_ 19 ... _
%e A137695 (Since the columns become constant below the diagonal (symbolized by \_), we list only the part above it.)
%e A137695 First row: 3 pegs, equals A000225; 2nd row: 4 pegs, cf. A007664; 3rd row: 5 pegs, cf. A007665.
%t A137695 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 *)
%o A137695 (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
%Y A137695 Cf. A000225, A007664, A007665.
%K A137695 easy,nice,nonn,tabl
%O A137695 1,2
%A A137695 _M. F. Hasler_, Feb 09 2008, revised Feb 10 2008