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%I #23 Sep 04 2019 23:58:40
%S 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,
%T 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,
%U 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
%N Largest prime p such that p^2 divides binomial(2n,n), or 0 if binomial(2n,n) is squarefree.
%C 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.
%C Granville and Ramaré show that if n >= 2082 then a(n) >= sqrt(n/5). - _Robert Israel_, Sep 04 2019
%H T. D. Noe, <a href="/A110493/b110493.txt">Table of n, a(n) for n = 0..10000</a>
%H T. D. Noe, <a href="http://www.sspectra.com/math/A110493.gif">Plot of A110493</a>
%H A. Granville and O. Ramaré, <a href="http://www.dms.umontreal.ca/~andrew/PDF/ramare.pdf">Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients</a>, Mathematika 43 (1996), 73-107, <a href="http://dx.doi.org/10.1112/S0025579300011608">[DOI]</a>.
%e a(5) = 3 because binomial(10,5) = 252 = (2^2)(3^2)(7).
%p f:= proc(n) local F;
%p F:= select(t -> t[2]>=2, ifactors(binomial(2*n,n))[2]);
%p if F = [] then 0 else max(map(t -> t[1],F)) fi
%p end proc:
%p map(f, [$0..100]); # _Robert Israel_, Sep 04 2019
%t Table[f=FactorInteger[Binomial[2n, n]]; s=Select[f, #[[2]]>1&]; If[s=={}, 0, s[[-1,1]]], {n, 0, 100}]
%Y Cf. A110494 (least k such that prime(n)^2 divides binomial(2k, k)).
%Y Cf. A059097, A110494.
%K nonn,look
%O 0,4
%A _T. D. Noe_, Jul 22 2005
%E a(0) prepended by _T. D. Noe_, Mar 27 2014
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