A292441 Largest m such that m^2 divides A000984(n).

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, 4, 4, 4, 21, 14, 14, 6, 2, 2, 10, 140, 14, 126, 6, 60, 90, 12, 84, 84, 210, 30, 18, 12, 6, 36, 4, 4, 6, 4, 4, 12, 12, 132, 132, 440, 55, 330, 10, 10, 90, 30, 30, 180

Offset

0,4

Comments

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

Granville and Ramaré show that A006530(a(n)) > sqrt(n/5) if n >= 2082.

In particular a(n) -> infinity as n -> infinity. - Robert Israel, Sep 18 2017

Links

Robert Israel, Table of n, a(n) for n = 0..10000

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

Eric Weisstein's World of Mathematics, Erdős Squarefree Conjecture

Wikipedia, Kummer's theorem

Formula

a(n) > 1 for n > 4.

a(n) = A000188(A000984(n)). - Robert Israel, Sep 17 2017

Example

binomial(10,5)/7           =  252/7   = 36 = a(5)^2.
binomial(12,6)/(3*7*11)    =  924/231 =  4 = a(6)^2.
binomial(14,7)/(2*3*11*13) = 3432/858 =  4 = a(7)^2.

Maple

A000188:= n -> mul(t[1]^floor(t[2]/2), t = ifactors(n)[2]):
seq(A000188(binomial(2*n, n)), n=0..100); # Robert Israel, Sep 17 2017

Crossrefs

Cf. A000188, A000984, A006530, A292442.

Sequence in context: A222005 A351442 A325815 * A376242 A333726 A306549

Adjacent sequences: A292438 A292439 A292440 * A292442 A292443 A292444

Keyword

nonn

Author

Seiichi Manyama, Sep 16 2017

Status

approved