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A014587
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Nim function for Take-a-Factorial-Game (a subtraction game).
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4
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0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2
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OFFSET
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0,3
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COMMENTS
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Concerning the January 1997 dissertation of Achim Flammenkamp, his home page (currently http://wwwhomes.uni-bielefeld.de/cgi-bin/cgiwrap/achim/index.cgi) has the link shown below, and a comment that a book was published in July 1997 by Hans-Jacobs-Verlag, Lage, Germany with the title Lange Perioden in Subtraktions-Spielen (ISBN 3-932136-10-1). This is an enlarged study (more than 200 pages) of his dissertation. - N. J. A. Sloane, Jul 25 2019
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, E26.
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LINKS
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FORMULA
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Conjecture: Appears to be periodic with period of length 25 = [0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3] starting with the initial term (there is no preamble). - Michel Dekking, Jul 26 2019
This conjecture is false, because moving from 10! = 3628800 to 0 is a legal move, and so a(3628800) cannot be zero. A similar argument shows that for no value of P is this sequence periodic with period P starting at term 0 (for a(P!) cannot be zero). - Nathan Fox, Jul 28 2019.
The first counterexample to the conjecture above is a(5050) = 4. - Pontus von Brömssen, Jul 09 2022
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PROG
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(Sage)
res = []
fact = [1]
while fact[-1] <= max : fact.append(factorial(len(fact)))
for i in range(max+1) :
moves = list({res[i-f] for f in fact if f <= i})
moves.sort()
k = len(moves)
mex = next((j for j in range(k) if moves[j] != j), k)
res.append(mex)
return res
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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