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A263569
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Number of distinct prime divisors p of 2*n such that lpf(2*n - p) = p, where lpf = least prime factor (A020639).
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1
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0, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 3, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 4
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OFFSET
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1,3
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COMMENTS
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a(n) gives the number of times 2*n occurs in A061228.
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LINKS
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EXAMPLE
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a(10) = 1 since the distinct prime divisors of 2*10 = 20 are 2 and 5, A020639(20 - 2) = 2 and A020639(20 - 5) = 3, so only prime 2 is to be considered.
a(15) = 3 since the distinct prime divisors of 2*15 = 30 are 2, 3 and 5, A020639(30 - 2) = 2 and A020639(30 - 3) = 3 and A020639(30 - 5) = 5, so all three prime 2, 3 and 5 are to be considered.
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MATHEMATICA
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f[n_] := Select[First /@ FactorInteger[2 n], FactorInteger[2 n - #][[1, 1]] == # &]; Length /@ Table[f@ n, {n, 2, 105}] (* Michael De Vlieger, Oct 22 2015 *)
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PROG
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(PARI) a(n) = {my(f=factor(2*n)); sum(k=1, #f~, p=f[k, 1]; p == factor(2*n-p)[1, 1]); } \\ Michel Marcus, Oct 31 2015
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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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