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A343507 a(n) is the smallest nonnegative integer k such that (2*k)! / (k+n)!^2 is an integer. 1
0, 208, 3475, 8174, 252965, 3648835, 72286092, 159329607, 2935782889 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

So far, all the numbers a(n) + n are squarefree.

LINKS

Table of n, a(n) for n=1..9.

EXAMPLE

a(1) = 0: (2*0)! / (0+1)!^2 = 1/1 = 1.

From Jon E. Schoenfield, Apr 18 2021: (Start)

Let f(n,k) = (2*k)!/(k+n)!^2. Then a(n) is the smallest nonnegative k such that f(n,k) is an integer.

f(n,0) = (2*0)!/(0+n)!^2 = 1/n!^2, so the fraction begins at k=0 with a value of 1/n!^2, and each time k is incremented by 1, the fraction is multiplied by (2*k)*(2*k-1) and divided by (k+n)^2. Whenever k+n is a prime p, this division will cause the reduced fraction to have p in its denominator with multiplicity 1 (since the multiplicity of p in (k+n)!^2 is 2, but its multiplicity in (2*k)! is only 1). Further multiplications by (2*k)*(2*k-1)/(k+n)^2 using successive values of k will not remove the prime factor p from the reduced fraction's denominator until k reaches p. As a result, the interval [a(n)+1, a(n)+n] can never contain a prime.

For n=2, the factorizations of the numerator and denominator of the reduced fraction (2k)!/(k+n)!^2 are shown in the table below for the first several values of k and for the last few through k = 208. Large blocks of consecutive primes (denoted by ellipses), each with multiplicity one, accumulate in the numerator as k gets larger.

.

      |           Reduced fraction (2k)!/(k+2)!^2

      +-------------------------------------+---------------

    k | numerator                           | denominator

  ----+-------------------------------------+---------------

    0 | 1                                   | 2^2

    1 | 1                                   | 2 * 3^2

    2 | 1                                   | 2^3 * 3

    3 | 1                                   | 2^2 * 5

    4 | 7                                   | 2 * 3^2 * 5

    5 | 1                                   | 7

    6 | 3 * 11                              | 2^4 * 7

    7 | 11 * 13                             | 2^3 * 3^3

    8 | 11 * 13                             | 2 * 3^2 * 5

    9 | 13 * 17                             | 5 * 11

   10 | 13 * 17 * 19                        | 2^2 * 3^2 * 11

    . |                                     |

    . |                                     |

    . |                                     |

  205 | 2^3 * 5 * 7 * 11^2 * 13 * 17 * 19^2 | 23 * 103

      | * 31 * 37 * 43 * 53 * 71 * 73 * 79  |

      | * 107 * ... * 131 * 211 * ... * 409 |

      |                                     |

  206 | 3 * 5 * 7 * 11^2 * 17 * 19^2 * 31   | 2^3 * 13 * 23

      | * 37* 43 * 53 * 71 * 73 * 79        |

      | * 107 * ... * 137 * 211 * ... * 409 |

      |                                     |

  207 | 3^3 * 5 * 7^2 * 17 * 31 * 37        |

      | * 43 * 53 * 59 * 71 * 73 * 79       |

      | * 107 * ... * 137 * 211 * ... * 409 | 2^2 * 13

      |                                     |

  208 | 2 * 3 * 17 * 31 * 37 * 43           |

      | * 53 * 59 * 71 * ... * 83           |

      | * 107 * ... * 137 * 211 * ... * 409 | 1

.

The denominator first reaches 1 at k=208, so a(2)=208.

(End)

PROG

(PARI) f(n, k) = (2*k)! / (k+n)!^2;

isok(n, k) = denominator(f(n, k)) == 1;

a(n) = my(k=0); while (!isok(n, k), k++); k; \\ Michel Marcus, May 03 2021

(Python)

from fractions import Fraction

from sympy import factorial

def A343507(n):

    k, f = 0, Fraction(1, int(factorial(n))**2)

    while f.denominator != 1:

        k += 1

        f *= Fraction(2*k*(2*k-1), (k+n)**2)

    return k # Chai Wah Wu, May 03 2021

(Python)

from math import gcd

n = 0

while n >= 0:

    num, den, i, n = 1, 1, 1, n+1

    while i <= n:

        den, i = den*i*i, i+1

    k, kk = 0, 0

    while den > 1:

        k, kk = k+1, kk+2

        d = gcd(num, (n+k)*(n+k))*gcd(den, (kk-1)*kk)

        num, den = num*(kk-1)*kk//d, den*(n+k)*(n+k)//d

    print(n, k) # A.H.M. Smeets, May 03 2021

CROSSREFS

Cf. A001044, A010050.

Sequence in context: A235444 A255074 A172967 * A231111 A339762 A223252

Adjacent sequences:  A343504 A343505 A343506 * A343508 A343509 A343510

KEYWORD

nonn,hard,more

AUTHOR

Daniel Mizrahi, Apr 17 2021

STATUS

approved

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Last modified August 4 14:26 EDT 2021. Contains 346447 sequences. (Running on oeis4.)