A327565 Number of transfers of marbles between two sets until the first repetition.

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

Offset: 1

Author

Tristan Cam, Sep 17 2019

Keywords

• nonn

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^(m-3) + 1.

Examples

			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.
		

Crossrefs

Cf. A094373; A327613 (three sets), A327614 (four sets).

Programs

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

Formula

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