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Numbers n for which gcd{k=1..n-1} binomial(2*n, 2*k) = 2n-1.
3

%I #15 Dec 13 2015 07:55:03

%S 10,12,21,22,24,30,34,36,40,51,52,55,57,69,70,76,82,84,87,90,96,99,

%T 100,106,112,114,115,117,120,129,132,136,141,142,147,154,156,159,166,

%U 174,177,180,184,187,192,195,201,205,210,216,217,220,222,225,231,232,234,240,244,246,250,252,255,261,262,274,279,282,285,286,294,297,300

%N Numbers n for which gcd{k=1..n-1} binomial(2*n, 2*k) = 2n-1.

%H Antti Karttunen, <a href="/A265403/b265403.txt">Table of n, a(n) for n = 1..2014</a>

%t Select[Range@ 300, GCD @@ Array[Function[k, Binomial[2 #, 2 k]], {# - 1}] == 2 # - 1 &] (* _Michael De Vlieger_, Dec 11 2015 *)

%o (PARI) isok(n) = (n>1) && gcd(vector(n-1, k, binomial(2*n, 2*k))) == 2*n-1; \\ _Michel Marcus_, Dec 08 2015, edited by _Antti Karttunen_, Dec 11 2015 (see A265388 for why).

%Y Cf. A265388.

%Y Cf. also A265402.

%K nonn

%O 1,1

%A _Antti Karttunen_, Dec 08 2015