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A134191
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Impure numbers in the Collatz (3x+1) iteration.
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2
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2, 4, 5, 8, 10, 11, 13, 14, 16, 17, 20, 22, 23, 26, 28, 29, 31, 32, 34, 35, 38, 40, 41, 44, 46, 47, 49, 50, 52, 53, 56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 74, 76, 77, 80, 82, 83, 85, 86, 88, 89, 91, 92, 94, 95, 98, 100, 101, 103, 104, 106, 107, 110, 112, 113, 116, 118
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OFFSET
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1,1
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COMMENTS
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Let f(k) be the trajectory of the Collatz iteration of the number k. Then Shaw calls a number n impure if n is in f(k) for some k < n. Shaw has an algorithm for finding congruences that the impure numbers satisfy.
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LINKS
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FORMULA
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EXAMPLE
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The Collatz trajectory of 3 is (3,10,5,16,8,4,2,1), showing that the numbers 4,5,8,10,16 are impure.
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MATHEMATICA
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c[n_] := If[EvenQ[n], n/2, 3n + 1]; nn=1000; t=Table[0, {nn}]; Do[If[t[[n]]==0, m=n; While[m=c[m]; If[nn>=m>n && t[[m]]==0, t[[m]]=n]; m>nn || t[[m]]>0]], {n, nn}]; Flatten[Position[t, _?(#>0&)]]
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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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