A356637 a(n) = A000265(A263931(n)).

1, 1, 1, 1, 1, 9, 3, 3, 45, 5, 1, 21, 7, 175, 675, 45, 45, 1485, 5775, 5775, 45045, 2145, 195, 8775, 2925, 5733, 22491, 833, 6545, 373065, 24871, 24871, 1566873, 3086265, 181545, 357903, 39767, 39767, 156975, 309925, 61985, 5020785, 239085, 20322225, 160730325

Offset: 0

Author

Peter Luschny, Sep 07 2022

Keywords

• nonn

Comments

Let n >= 5. If a(n) is squarefree, then 2 divides binomial(2*n, n) more than once and is the only prime that does so. There is only a finite number of such cases (see A059097).

An efficient algorithm for the calculation is available, which is based on prime factorization. See the SageMath implementation. The main application is the efficient calculation of the central binomial coefficient, which is the product of this sequence, the Glaisher/Gould sequence, and the upper primorial function (see the formula section).

Since the central binomial coefficient is a bisection of the swinging factorial A056040, and the swinging factorial, in turn, is the building block for an efficient algorithm for the computation of the factorial function, the terms of this sequence occur as factors in all these computations. See the links for details.

Examples

			Let n = 22 and consider the prime factorization of m = binomial(2*n, n):
2^3 * [3 * 5 * 13] * 23 * 29 * 31 * 37 * 41 * 43. Then a(22) = 3 * 5 * 13. This is what is left after the 'prime tail' A261130(n) and the 'prime head' A006519(m) = A001316(n) have been cut off.
		

Links

• Peter Luschny, Swing, divide and conquer the factorial, (excerpt).
• Peter Luschny, Fast Factorial Functions, a code repository.
• Eric Weisstein's World of Mathematics, Erdős Squarefree Conjecture.

Crossrefs

Cf. A000265, A263931, A261130, A006519, A000984, A001316, A059097, A056040.

Programs

Maple

A263931 := n -> binomial(2*n, n) / convert(select(isprime, {$n+1..2*n}), `*`):
A000265 := n -> n / 2^padic[ordp](n, 2):
seq(A000265(A263931(n)), n = 0..45);

SageMath

def A356637(n: int) -> int:
    m = 2 * n
    if m < 5: return 1
    sqrtm = isqrt(m) + 1
    R = prime_range(sqrtm, m // 3 + 1)
    factors = [x for x in R if is_odd(m // x)]
    for prime in prime_range(3, sqrtm):
        p: int = 1
        q: int = m
        while True:
            q //= prime
            if q == 0:
                break
            if q & 1 == 1:
                p *= prime
        if p > 1:
            factors.append(p)
    return product(factors)
print([A356637(n) for n in range(45)])

Formula

A000984(n) = a(n) * A001316(n) * A261130(n) for n >= 2.