A056651 Numbers k such that binomial(k,floor(k/2)) has no non-unitary square divisors: all of their square divisors are unitary ones.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 31, 32, 35, 36, 37, 39, 40, 41, 43, 47, 48, 49, 55, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 75, 79, 80, 95, 96, 97, 111, 129, 130, 131, 132, 133, 143, 144, 151, 161, 163, 167, 191, 192, 193

Offset: 1

Author

Labos Elemer, Aug 09 2000

Keywords

• nonn

Comments

This property is weaker than "squarefreedom", but shows how central binomial coefficients are "poor of squares".

Numbers k such that binomial(k,floor(k/2)) is cubefree (A004709). - Amiram Eldar, Jul 22 2024

Examples

			223 is a term because x = binomial(223,111) has 35 prime divisors. 33 arises at power 1. Only 2 and 13 has powers 2 > 1. So square divisors of x are {1, 4, 169, 676} ={s}. All of them are also unitary divisors since GCD(s,x/s) = 1 holds for them.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..129 (no others < 10^8)

Crossrefs

Cf. A001405, A004709, A046098, A056175, A110495.

Programs

Mathematica

Select[Range[0, 11000], AllTrue[FactorInteger[Binomial[#, Floor[#/2]]][[;;, 2]], #1 <= 2 &] &] (* Amiram Eldar, Jul 22 2024 *)

PARI

is(n) = if(n <= 1, 1, vecmax(factor(binomial(n, floor(n/2)))[, 2]) < 3); \\ Amiram Eldar, Jul 22 2024