cp's OEIS Frontend

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.

A183125 Magnetic Tower of Hanoi, total number of moves, generated by a certain algorithm, yielding a "forward moving" non-optimal solution of the [NEUTRAL ; NEUTRAL ; NEUTRAL] pre-colored puzzle.

Original entry on oeis.org

0, 1, 4, 11, 30, 83, 236, 687, 2026, 6027, 18008, 53927, 161654, 484803, 1454212, 4362399, 13086914, 39260411, 117780848, 353342103, 1060025806, 3180076851, 9540229916, 28620689039, 85862066330, 257586198123, 772758593416, 2318275779207, 6954827336486, 20864482008227, 62593446023348, 187780338068607, 563341014204274, 1690023042611163
Offset: 0

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Author

Uri Levy, Jan 08 2011

Keywords

Comments

The Magnetic Tower of Hanoi puzzle is described in link 1 listed below. The Magnetic Tower is pre-colored. Pre-coloring is [NEUTRAL ; NEUTRAL ; NEUTRAL], given in [Source ; Intermediate ; Destination] order. The solution algorithm producing the presented sequence is NOT optimal. The particular "61" algorithm solving the puzzle at hand is not explicitly presented in any of the referenced papers. For the optimal solution of the Magnetic Tower of Hanoi puzzle with the given pre-coloring configuration (the "natural" or "free" Magnetic Tower) see A183117 and A183118. Optimal solutions are discussed and their optimality is proved in link 2 listed below.
Large N limit of the sequence is 0.5*(197/324)*3^N ~ 0.5*0.61*3^N. Series designation: S61(n).

References

  • Uri Levy, The Magnetic Tower of Hanoi, Journal of Recreational Mathematics, Volume 35 Number 3 (2006), 2010, pp 173.

Crossrefs

A183123 is an integer sequence generated by another non-optimal algorithm solving the "free" [NEUTRAL ; NEUTRAL ; NEUTRAL] Magnetic Tower of Hanoi puzzle.
A003462 "Partial sums of A000244" is the sequence (also) describing the total number of moves solving [RED ; BLUE ; BLUE] or [RED ; RED ; BLUE] pre-colored Magnetic Tower of Hanoi puzzle.

Programs

  • GAP
    a:=[30, 83, 236, 687, 2026];; for n in [6..30] do a[n]:=5*a[n-1]-6*a[n-2] -2*a[n-3]+7*a[n-4]-3*a[n-5]; od; Concatenation([0, 1, 4, 11], a); # G. C. Greubel, Dec 04 2018
  • Magma
    I:=[0,1,4,11,30,83,236,687,2026]; [n le 9 select I[n] else 5*Self(n-1)-6*Self(n-2)-2*Self(n-3)+7*Self(n-4)-3*Self(n-5): n in [1..35]]; // Vincenzo Librandi, Dec 04 2018
    
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!( (-4*x^8 -2*x^6 +x^4 -3*x^3 -x^2 +x)/(3*x^5 -7*x^4 +2*x^3 +6*x^2 -5*x +1))); // G. C. Greubel, Dec 04 2018
    
  • Maple
    seq(coeff(series((-4*x^8-2*x^6+x^4-3*x^3-x^2+x)/(3*x^5-7*x^4+2*x^3+6*x^2-5*x+1),x,n+1), x, n), n = 0 .. 35); # Muniru A Asiru, Dec 04 2018
  • Mathematica
    Join[{0, 1, 4, 11}, LinearRecurrence[{5, -6, -2, 7, -3}, {30, 83, 236, 687, 2026}, 30]] (* Jean-François Alcover, Dec 04 2018 *)
    CoefficientList[Series[(- 4 x^8 - 2 x^6 + x^4 - 3 x^3 - x^2 + x) / (3 x^5 - 7 x^4 + 2 x^3 + 6 x^2 - 5 x + 1), {x, 0, 33}], x] (* Vincenzo Librandi, Dec 04 2018 *)
  • PARI
    my(x='x+O('x^30)); concat([0], Vec((-4*x^8 -2*x^6 +x^4 -3*x^3 -x^2 +x)/(3*x^5 -7*x^4 +2*x^3 +6*x^2 -5*x +1))) \\ G. C. Greubel, Dec 04 2018
    
  • Sage
    s=((-4*x^8 -2*x^6 +x^4 -3*x^3 -x^2 +x)/(3*x^5 -7*x^4 +2*x^3 +6*x^2 -5*x +1)).series(x, 30); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 04 2018
    

Formula

G.f.: (-4*x^8 -2*x^6 +x^4 -3*x^3 -x^2 +x)/(3*x^5 -7*x^4 +2*x^3 +6*x^2 -5*x +1).
a(n) = +5*a(n-1)-6*a(n-2)-2*a(n-3)+7*a(n-4)-3*a(n-5).
(a(n) = S61(n) as in referenced paper):
a(n) = 3*a(n-1) - 2*n^2 + 17*n - 43 ; n even ; n >= 6.
a(n) = 3*a(n-1) - 2*n^2 + 17*n - 42 ; n odd ; n >= 5.
a(n) = S64(n-1) + S64(n-2) + S75(n-3) + 4*3^(n-3) + 2 ; n >= 3.
S64(n) and S75(n) refer to the integer sequences described by A183121 and A183119 respectively.
a(n) = 0.5*(197/324)*3^n + n^2 - 5.5*n + 91/8; n even; n >= 4.
a(n) = 0.5*(197/324)*3^n + n^2 - 5.5*n + 93/8; n odd; n >= 5.

Extensions

More terms from Jean-François Alcover, Dec 04 2018