A042996 Numbers k such that binomial(k, floor(k/2)) is divisible by k.

1, 2, 3, 5, 7, 9, 11, 12, 13, 15, 17, 19, 21, 23, 25, 27, 29, 30, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 56, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 84, 85, 87, 89, 90, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121

Offset: 1

Author

Labos Elemer

Keywords

• nonn

Comments

All the odd numbers are terms. - Amiram Eldar, Aug 24 2024

Examples

			For n = 12, binomial(12,6) = 924 = 12*77 is divisible by 12, so 12 is in the sequence.
For n = 13, binomial(13,6) = 1716 = 13*132 is divisible by 13, so 13 is in the sequence.
From _David A. Corneth_, Apr 22 2018: (Start)
For n = 20, we wonder if 20 = 2^2 * 5 divides binomial(20, 10) = 20! / (10!)^2.
The exponent of 2 in the prime factorization of 20! is 10 + 5 + 2 + 1 = 18.
The exponent of 2 in the prime factorization of 10! is 5 + 2 + 1 = 8.
Therefore, the exponent of 2 in binomial(20, 10) is 18 - 2*8 = 2.
The exponent of 5 in the prime factorization of 20! is 4.
The exponent of 5 in the prime factorization of 10! is 2.
Therefore, exponent of 5 in binomial(20, 10) is 4 - 2*2 = 0.
So binomial(20, 10) is not divisible by 20, and 20 is not in the sequence. (End)
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001405, A020475, A067315 (complement).

Programs

Mathematica

Select[Range[150],Divisible[Binomial[#,Floor[#/2]],#]&] (* Harvey P. Dale, Sep 15 2011 *)

PARI

isok(n) = (binomial(n, n\2) % n) == 0; \\ Michel Marcus, Apr 22 2018