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%I #28 May 10 2018 03:27:51
%S 1,1,1,2,1,6,2,2,3,2,2,2,2,10,30,12,3,6,10,10,6,2,2,60,30,42,42,28,2,
%T 4,4,4,21,14,14,6,2,2,10,140,14,126,6,60,90,12,84,84,210,30,18,12,6,
%U 36,4,4,6,4,4,12,12,132,132,440,55,330,10,10,90,30,30,180
%N Largest m such that m^2 divides A000984(n).
%C a(n) is the product of p^floor(m(n,p)/2) over primes p<n, where m(n,p) is the number of carries when adding n to itself in base p. - _Robert Israel_, Sep 17 2017
%C Granville and Ramaré show that A006530(a(n)) > sqrt(n/5) if n >= 2082.
%C In particular a(n) -> infinity as n -> infinity. - _Robert Israel_, Sep 18 2017
%H Robert Israel, <a href="/A292441/b292441.txt">Table of n, a(n) for n = 0..10000</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>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ErdosSquarefreeConjecture.html">Erdős Squarefree Conjecture</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Kummer%27s_theorem">Kummer's theorem</a>
%F a(n) > 1 for n > 4.
%F a(n) = A000188(A000984(n)). - _Robert Israel_, Sep 17 2017
%e binomial(10,5)/7 = 252/7 = 36 = a(5)^2.
%e binomial(12,6)/(3*7*11) = 924/231 = 4 = a(6)^2.
%e binomial(14,7)/(2*3*11*13) = 3432/858 = 4 = a(7)^2.
%p A000188:= n -> mul(t[1]^floor(t[2]/2), t = ifactors(n)[2]):
%p seq(A000188(binomial(2*n,n)),n=0..100); # _Robert Israel_, Sep 17 2017
%Y Cf. A000188, A000984, A006530, A292442.
%K nonn
%O 0,4
%A _Seiichi Manyama_, Sep 16 2017
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