A014587 Nim function for Take-a-Factorial-Game (a subtraction game).

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

Offset: 0

Author

Achim Flammenkamp

Keywords

• nonn

Comments

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

References

• R. K. Guy, Unsolved Problems in Number Theory, E26.

Links

• Eric M. Schmidt, Table of n, a(n) for n = 0..10000
• Achim Flammenkamp, Lange Perioden in Subtraktions-Spielen, Dissertation, Dept. Math., University of Bielefeld, Germany.

Crossrefs

Cf. A014586-A014589, A355556.

Programs

Sage

def A014587(max) :
    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
# Eric M. Schmidt, Jul 20 2013, corrected Eric M. Schmidt, Apr 24 2019

Formula

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