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A110493
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Largest prime p such that p^2 divides binomial(2n,n), or 0 if binomial(2n,n) is squarefree.
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6
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0, 0, 0, 2, 0, 3, 2, 2, 3, 2, 2, 2, 2, 5, 5, 3, 3, 3, 5, 5, 3, 2, 2, 5, 5, 7, 7, 7, 2, 2, 2, 2, 7, 7, 7, 3, 2, 2, 5, 7, 7, 7, 3, 5, 5, 3, 7, 7, 7, 5, 3, 3, 3, 3, 2, 2, 3, 2, 2, 3, 3, 11, 11, 11, 11, 11, 5, 5, 5, 5, 5, 5, 11, 11, 11, 11, 11, 3, 5, 5, 3, 7, 7, 11, 11, 13, 13, 13, 13, 13, 13, 5, 5, 5, 11, 11
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OFFSET
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0,4
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COMMENTS
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Binomial(2n,n) is squarefree for only n = 0, 1, 2, 4. Sequence A059097 lists n such that a(n) = 0 or 2. The plot shows the quadratic nature of this sequence. Sequence A110494 makes the quadratic behavior clearer.
Granville and Ramaré show that if n >= 2082 then a(n) >= sqrt(n/5). - Robert Israel, Sep 04 2019
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LINKS
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EXAMPLE
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a(5) = 3 because binomial(10,5) = 252 = (2^2)(3^2)(7).
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MAPLE
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f:= proc(n) local F;
F:= select(t -> t[2]>=2, ifactors(binomial(2*n, n))[2]);
if F = [] then 0 else max(map(t -> t[1], F)) fi
end proc:
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MATHEMATICA
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Table[f=FactorInteger[Binomial[2n, n]]; s=Select[f, #[[2]]>1&]; If[s=={}, 0, s[[-1, 1]]], {n, 0, 100}]
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CROSSREFS
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Cf. A110494 (least k such that prime(n)^2 divides binomial(2k, k)).
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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