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A101290 Dissipation time for n under the following scheme: set f(0)=n, f(i)=0 for all i<>0, then iterate the evolutionary process by which f(i) is replaced by floor[(f(i-1)+f(i)+f(i+1))/3] for all i, with all calculations done in parallel. a(n) is the minimum number of iterations of this process required to reach f(i)=0 for all i, for n=0,1,2,3,... 1

%I #2 Mar 30 2012 17:37:53

%S 0,1,1,3,3,3,5,5,5,7,7,7,7,7,7,9,9,9,11,11,11,11,11,11,12,12,12,14,14,

%T 14,14,14,14,15,15,15,16,16,16,16,16,16,16,16,16,18,18,18,19,19,19,19,

%U 19,19,20,20,20,20,20,20,22,22,22,23,23,23,23,23,23,25,25,25,25,25,25

%N Dissipation time for n under the following scheme: set f(0)=n, f(i)=0 for all i<>0, then iterate the evolutionary process by which f(i) is replaced by floor[(f(i-1)+f(i)+f(i+1))/3] for all i, with all calculations done in parallel. a(n) is the minimum number of iterations of this process required to reach f(i)=0 for all i, for n=0,1,2,3,...

%e If n=10, the process evolves as follows on [ -3,3], showing that a(10)=7:

%e Start with: 0,0,0,10,0,0,0

%e Iteration 1: 0,0,3,3,3,0,0

%e Iteration 2: 0,1,2,3,2,1,0

%e Iteration 3: 0,1,2,2,2,1,0

%e Iteration 4: 0,1,1,2,1,1,0

%e Iteration 5: 0,0,1,1,1,0,0

%e Iteration 6: 0,0,0,1,0,0,0

%e Iteration 7: 0,0,0,0,0,0,0

%K nonn

%O 0,4

%A _John W. Layman_, Dec 21 2004

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Last modified September 18 23:40 EDT 2024. Contains 376002 sequences. (Running on oeis4.)