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A056651
Numbers k such that binomial(k,floor(k/2)) has no non-unitary square divisors: all of their square divisors are unitary ones.
2
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
1,3
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
LINKS
T. D. Noe, Table of n, a(n) for n = 1..129 (no others < 10^8)
EXAMPLE
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.
MATHEMATICA
Select[Range[0, 11000], AllTrue[FactorInteger[Binomial[#, Floor[#/2]]][[;; , 2]], #1 <= 2 &] &] (* Amiram Eldar, Jul 22 2024 *)
PROG
(PARI) is(n) = if(n <= 1, 1, vecmax(factor(binomial(n, floor(n/2)))[, 2]) < 3); \\ Amiram Eldar, Jul 22 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Aug 09 2000
STATUS
approved