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A033120
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Base-2 digits of a(n) are, in order, the first n terms of the periodic sequence with initial period 1,0,1.
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1
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1, 2, 5, 11, 22, 45, 91, 182, 365, 731, 1462, 2925, 5851, 11702, 23405, 46811, 93622, 187245, 374491, 748982, 1497965, 2995931, 5991862, 11983725, 23967451, 47934902, 95869805, 191739611, 383479222, 766958445, 1533916891
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OFFSET
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1,2
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COMMENTS
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Minimal number of moves required, under the proviso of a classical tower-of-Hanoi game, to segregate an initial n-disc peg into even and odd numbered discs pegs. - Lekraj Beedassy, Sep 12 2006
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REFERENCES
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B. Averbach & O. Chein, "A Variant Of The Tower Of Brahma" in 'The Journal of Recreational Mathematics', pp. 48-55, vol. 33, no. 1, 2004-5, Baywood, NY.
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LINKS
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FORMULA
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a(3n) = (5*8^n - 5)/7, a(3n+1) = (10*8^n - 3)/7, a(3n+2) = (20*8^n - 6)/7.
G.f.: (1+x^2)/((1-x)(1-2x)(1+x+x^2)). (End)
The following recurrence produces this sequence: if(n==1) a(n)=1; else if(n%3==2) a(n)=a(n-1)*2; otherwise a(n)=a(n-1)*2+1. - Piotr Kakol, Jan 24 2011 (in an email message to N. J. A. Sloane).
a(n) = T(n-1) + 1 + T(n-3) + 1 + a(n-3), where T(n) = A000225(n) = 2^n-1 is the number of moves for a classic Tower of Hanoi with n discs.
a(n) = (5/8)*2^n + a(n-3).
a(n) = (5/7)*2^n - 2/3 - (1/21)*cos((2/3)*Pi*n) + (1/7)*sqrt(3)*sin((2/3)*Pi*n). (End)
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MATHEMATICA
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Table[FromDigits[PadRight[{}, n, {1, 0, 1}], 2], {n, 40}] (* Harvey P. Dale, Aug 26 2016 *)
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PROG
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(PARI) a(n)=if(n%3==0, 5*8^(n/3)-5, if(n%3==1, 10*8^((n-1)/3)-3, 20*8^((n-2)/3)-6))/7 \\ Ralf Stephan
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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