A056059 GCD of largest square and squarefree part of central binomial coefficients.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 6, 2, 1, 1, 1, 3, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 6, 3, 1, 1, 1, 2, 3, 6, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 3, 6, 6, 3, 1, 2, 2, 1, 2, 1, 3, 6, 1, 1, 1, 2, 1, 2

Offset

1,14

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000

Formula

a(n) = A055229(A001405(n)), where A055229(n) = gcd(A008833(n), A007913(n)).

Example

n=14, binomial(14,7) = 3432 = 8*3*11*13. The largest square divisor is 4, squarefree part is 858. So a(14) = gcd(4,858) = 2.

Mathematica

Table[GCD[First@ Select[Reverse@ Divisors@ #, IntegerQ@ Sqrt@ # &], Times @@ Power @@@ Map[{#1, Mod[#2, 2]} & @@ # &, FactorInteger@ #]] &@ Binomial[n, Floor[n/2]], {n, 80}] (* Michael De Vlieger, Feb 18 2017, after Zak Seidov at A007913 *)

Prog

(PARI)
A001405(n) = binomial(n, n\2);
A055229(n) = { my(c=core(n)); gcd(c, n/c); } \\ Charles R Greathouse IV, Nov 20 2012
A056059(n) = A055229(A001405(n)); \\ Antti Karttunen, Jul 20 2017
(Python)
from sympy import binomial, gcd
from sympy.ntheory.factor_ import core
def a001405(n): return binomial(n, n//2)
def a055229(n):
    c = core(n)
    return gcd(c, n//c)
def a(n): return a055229(a001405(n))
print([a(n) for n in range(1, 151)]) # Indranil Ghosh, Jul 20 2017

Crossrefs

Cf. A000188, A001405, A007913, A008833, A034974, A046098, A055229, A055231.

Cf. A056056, A056057, A056058, A056060, A056061.

A056201 is the cube of this sequence.

Sequence in context: A369008 A238015 A257679 * A355915 A357900 A357732

Adjacent sequences: A056056 A056057 A056058 * A056060 A056061 A056062

Keyword

nonn

Author

Labos Elemer, Jul 26 2000

Extensions

Formula clarified by Antti Karttunen, Jul 20 2017

Status

approved