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A327565
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Number of transfers of marbles between two sets until the first repetition.
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2
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2, 3, 4, 3, 5, 4, 4, 5, 6, 4, 5, 6, 5, 5, 6, 5, 7, 6, 5, 7, 6, 5, 7, 6, 6, 7, 6, 6, 7, 6, 6, 7, 8, 6, 7, 8, 6, 7, 8, 6, 7, 8, 6, 7, 8, 6, 7, 8, 7, 7, 8, 7, 7, 8, 7, 7, 8, 7, 7, 8, 7, 7, 8, 7, 9, 8, 7, 9, 8, 7, 9, 8, 7, 9, 8, 7, 9, 8, 7, 9
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OFFSET
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1,1
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COMMENTS
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There are initially n marbles in both sets. In the first turn, half of the marbles of set A are transferred to set B, rounding to the upper integer when halving. In the second turn, half of the marbles of set B are transferred back to set A, following the same rule. The game goes on back and forth until we reach a distribution already encountered.
a(n) is then the number of steps until the first repetition occurs.
First occurrence of a(n) = m > 1 in this sequence: 1, 2, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025, 2049...
Conjecture: for m > 2, the first occurrence of a(n) = m is for n = 2^(m-3) + 1.
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LINKS
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FORMULA
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For m > 1, first occurrence of a(n) = m is for n = A094373(m-1) (conjectured).
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EXAMPLE
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For n = 3, (SetA ; SetB):
(3 ; 3), ceiling(3/2)=2 marbles get transferred,
(1 ; 5), ceiling(5/2)=3 marbles get transferred,
(4 ; 2), ceiling(4/2)=2 marbles get transferred,
(2 ; 4), ceiling(4/2)=2 marbles get transferred,
(4 ; 2), this is a repetition, it took 4 steps to get there, so a(3) = 4.
For n = 9, (SetA ; SetB):
(9 ; 9), (4 ; 14), (11 ; 7), (5 ; 13), (12 ; 6), (6 ; 12), (12 ; 6) which is a repetition, so a(9) = 6.
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PROG
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(PARI) a(n)={my(v=vector(2*n+1), r=n, f=1, c=0); while(!v[1+r], v[1+r]=1; r=if(f, r-ceil(r/2), r+ceil((2*n-r)/2)); c++; f=!f); c} \\ Andrew Howroyd, Sep 17 2019
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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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