A110493 Largest prime p such that p^2 divides binomial(2n,n), or 0 if binomial(2n,n) is squarefree.

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

Offset

0,4

Comments

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

Links

T. D. Noe, Table of n, a(n) for n = 0..10000

T. D. Noe, Plot of A110493

A. Granville and O. Ramaré, Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients, Mathematika 43 (1996), 73-107, [DOI].

Example

a(5) = 3 because binomial(10,5) = 252 = (2^2)(3^2)(7).

Maple

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:
map(f, [$0..100]); # Robert Israel, Sep 04 2019

Mathematica

Table[f=FactorInteger[Binomial[2n, n]]; s=Select[f, #[[2]]>1&]; If[s=={}, 0, s[[-1, 1]]], {n, 0, 100}]

Crossrefs

Cf. A110494 (least k such that prime(n)^2 divides binomial(2k, k)).

Cf. A059097, A110494.

Sequence in context: A152164 A263112 A368818 * A118234 A262771 A152039

Adjacent sequences: A110490 A110491 A110492 * A110494 A110495 A110496

Keyword

nonn,look

Author

T. D. Noe, Jul 22 2005

Extensions

a(0) prepended by T. D. Noe, Mar 27 2014

Status

approved