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A101290
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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,...
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0
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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, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 18, 18, 18, 19, 19, 19, 19, 19, 19, 20, 20, 20, 20, 20, 20, 22, 22, 22, 23, 23, 23, 23, 23, 23, 25, 25, 25, 25, 25, 25
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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EXAMPLE
| If n=10, the process evolves as follows on [ -3,3], showing that a(10)=7:
Start with: 0,0,0,10,0,0,0
Iteration 1: 0,0,3,3,3,0,0
Iteration 2: 0,1,2,3,2,1,0
Iteration 3: 0,1,2,2,2,1,0
Iteration 4: 0,1,1,2,1,1,0
Iteration 5: 0,0,1,1,1,0,0
Iteration 6: 0,0,0,1,0,0,0
Iteration 7: 0,0,0,0,0,0,0
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CROSSREFS
| Sequence in context: A001650 A130175 A200266 * A080605 A130823 A101435
Adjacent sequences: A101287 A101288 A101289 * A101291 A101292 A101293
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KEYWORD
| nonn
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AUTHOR
| John W. Layman (layman(AT)math.vt.edu), Dec 21 2004
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