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A071829
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Smallest s > 0 such that Lpf(n) = Lpf(n+s) where Lpf(x) denotes the largest prime factor in x factorization.
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0
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2, 3, 4, 5, 3, 7, 8, 3, 5, 11, 6, 13, 7, 5, 16, 17, 6, 19, 5, 7, 11, 23, 3, 5, 13, 9, 7, 29, 10, 31, 32, 11, 17, 7, 12, 37, 19, 13, 5, 41, 7, 43, 11, 5, 23, 47, 6, 7, 10, 17, 13, 53, 18, 11, 7, 19, 29, 59, 15, 61, 31, 7, 64, 13, 11, 67, 17, 23, 14, 71, 9, 73, 37, 5, 19, 11, 13, 79
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OFFSET
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1,1
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LINKS
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MATHEMATICA
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ss[n_]:=Module[{s=1, f=FactorInteger[n][[-1, 1]]}, While[f!=FactorInteger[ s+n][[-1, 1]], s++]; s]; Array[ss, 80, 2] (* Harvey P. Dale, Aug 18 2015 *)
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PROG
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(PARI) for(n=2, 120, s=+1; while(abs(component(component(factor(n), 1), omega(n))-component(component(factor(n+s), 1), omega(n+s)))>0, s++); print1(s, ", "))
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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