

A327565


Number of transfers of marbles between two sets until the first repetition.


2



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

1,1


COMMENTS

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^(m3) + 1.


LINKS

Table of n, a(n) for n=1..80.


FORMULA

For m > 1, first occurrence of a(n) = m is for n = A094373(m1) (conjectured).


EXAMPLE

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.


PROG

(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, rceil(r/2), r+ceil((2*nr)/2)); c++; f=!f); c} \\ Andrew Howroyd, Sep 17 2019


CROSSREFS

Cf. A094373; A327613 (three sets), A327614 (four sets).
Sequence in context: A305890 A305801 A075850 * A054437 A287821 A299757
Adjacent sequences: A327562 A327563 A327564 * A327566 A327568 A327569


KEYWORD

nonn


AUTHOR

Tristan Cam, Sep 17 2019


STATUS

approved



