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A089645
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Penny Flipping, or Flipping Coins: Given a stack of n coins, flip the top coin, then the stack of the top two coins, then the stack of the top three etc... starting again with the top coin after flipping all n coins. A flip of m coins reverses their order and inverts their state. This is the number of flips required to restore the stack to its original configuration.
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0
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2, 3, 9, 11, 24, 35, 28, 31, 80, 60, 121, 119, 116, 195, 75, 79, 204, 323, 228, 199, 146, 264, 529, 504, 200, 675, 540, 251, 840, 899, 186, 191, 1088, 748, 1225, 324, 740, 1140, 1521, 1079, 1680, 336, 1204, 484, 540, 460, 1692, 1151, 734, 2499
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OFFSET
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1,1
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COMMENTS
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Here "original configuration" seems to mean each coin in original orientation and either original or reverse order; for the original order and either original or reverse orientation n*A002326(n) flips required, while for both original order and original orientation n*A003558(n) required. - Henry Bottomley, Jan 19 2007
Birtwistle (1973) attributes the problem to Iain Bride and John Gilder of UMIST (University of Manchester Institute of Science and Technology).
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REFERENCES
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G. M. Birtwistle, Simula Begin, Auerbach Publishers, Philadelphia, 1973 [Uses this problem to illustrate the power of the Simula language]
Popular Computing (Calabasas, CA), Penny Flipping, Vol. 3 (No. 23, Feb 1973), pages PC23-10 to PC23-13) and Vol. 3 (No. 29, Aug 1975), pages PC29-6 to PC29-8. Gives first 32 terms.
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LINKS
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FORMULA
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EXAMPLE
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For 3 coins (starting with HHH) the flips move the stack through the sequence: HHH -1-> THH -2-> THH -3-> TTH -1-> HTH -2-> HTH -3-> THT -1-> HHT -2-> TTT -3-> HHH. (-n-> indicates n coins are flipped)
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MATHEMATICA
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b[n_] := MultiplicativeOrder[2, 2n+1];
c[n_] := MultiplicativeOrder[2, 2n+1, {-1, 1}];
a[1] = 2; a[n_] := If[b[n] == c[n], n*b[n], n*b[n]/2 - 1];
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CROSSREFS
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KEYWORD
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easy,nonn,nice
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AUTHOR
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Richard Forster (gbrl01(AT)yahoo.co.uk), Jan 02 2004
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STATUS
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approved
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