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

%I A042996 #25 Aug 24 2024 05:58:55
%S A042996 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,
%T A042996 45,47,49,51,53,55,56,57,59,61,63,65,67,69,71,73,75,77,79,81,83,84,85,
%U A042996 87,89,90,91,93,95,97,99,101,103,105,107,109,111,113,115,117,119,121
%N A042996 Numbers k such that binomial(k, floor(k/2)) is divisible by k.
%C A042996 All the odd numbers are terms. - _Amiram Eldar_, Aug 24 2024
%H A042996 Ivan Neretin, <a href="/A042996/b042996.txt">Table of n, a(n) for n = 1..10000</a>
%e A042996 For n = 12, binomial(12,6) = 924 = 12*77 is divisible by 12, so 12 is in the sequence.
%e A042996 For n = 13, binomial(13,6) = 1716 = 13*132 is divisible by 13, so 13 is in the sequence.
%e A042996 From _David A. Corneth_, Apr 22 2018: (Start)
%e A042996 For n = 20, we wonder if 20 = 2^2 * 5 divides binomial(20, 10) = 20! / (10!)^2.
%e A042996 The exponent of 2 in the prime factorization of 20! is 10 + 5 + 2 + 1 = 18.
%e A042996 The exponent of 2 in the prime factorization of 10! is 5 + 2 + 1 = 8.
%e A042996 Therefore, the exponent of 2 in binomial(20, 10) is 18 - 2*8 = 2.
%e A042996 The exponent of 5 in the prime factorization of 20! is 4.
%e A042996 The exponent of 5 in the prime factorization of 10! is 2.
%e A042996 Therefore, exponent of 5 in binomial(20, 10) is 4 - 2*2 = 0.
%e A042996 So binomial(20, 10) is not divisible by 20, and 20 is not in the sequence. (End)
%t A042996 Select[Range[150],Divisible[Binomial[#,Floor[#/2]],#]&] (* _Harvey P. Dale_, Sep 15 2011 *)
%o A042996 (PARI) isok(n) = (binomial(n, n\2) % n) == 0; \\ _Michel Marcus_, Apr 22 2018
%Y A042996 Cf. A001405, A020475, A067315 (complement).
%K A042996 nonn
%O A042996 1,2
%A A042996 _Labos Elemer_