

A042996


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


4



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
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OFFSET

1,2


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.
Sequence in context: A092875 A155498 A069149 * A073040 A087268 A106765
Adjacent sequences: A042993 A042994 A042995 * A042997 A042998 A042999


KEYWORD

nonn


AUTHOR

Labos Elemer


STATUS

approved



