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A102640
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Compute the greatest prime factors (GPFs, A006530()) of j + 2^n for j = 0, 1, ..., L. a(n) is the maximal length L of such a sequence in which the greatest prime factors are increasing with increasing j.
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4
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2, 2, 4, 2, 3, 2, 2, 2, 3, 2, 2, 4, 2, 3, 2, 2, 3, 4, 2, 3, 3, 6, 2, 3, 2, 4, 2, 2, 3, 4, 2, 3, 3, 2, 2, 4, 2, 4, 2, 2, 2, 4, 2, 3, 4, 4, 2, 4, 2, 3, 3, 2, 2, 3, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 3, 4, 2, 3, 4, 4, 2, 4, 2, 3, 2, 2, 4, 4, 2, 3, 3, 4, 2, 2, 2, 3, 2, 4, 2, 3, 2, 2, 3, 2, 2, 2, 3, 4, 2, 4, 2, 3, 2, 2, 3
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OFFSET
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1,1
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COMMENTS
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A006530(2^n)=2 is a local minimum. Going either upward or downward with the argument, the greatest prime factors are increasing for a while.
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LINKS
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EXAMPLE
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For n = 12: 2^10 = 4096. The greatest prime factors of 4096, 4097, 4098, 4099 are as follows: {2, 241, 683, 4099}. A006530(4100) = 41 is already smaller than A006530(4099). Thus the length of increasing GPF sequence is 4 = a(12).
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MATHEMATICA
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With[{nn = 12}, Table[Function[k, 1 + LengthWhile[#, # > 0 &] &@ Differences@ Array[FactorInteger[#][[-1, 1]] &, nn, k]][2^n], {n, 105}]] (* Michael De Vlieger, Jul 24 2017 *)
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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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