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

1,2

Comments

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

Links

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

Example

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)

Mathematica

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

Prog

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

Crossrefs

Cf. A001405, A020475, A067315 (complement).

Sequence in context: A092875 A155498 A069149 * A073040 A087268 A106765

Adjacent sequences: A042993 A042994 A042995 * A042997 A042998 A042999

Keyword

nonn

Author

Labos Elemer

Status

approved