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a(n) = n-th positive divisor of ((2n)!/n!^2).
2

%I #15 Apr 03 2021 19:04:25

%S 1,2,4,7,6,7,11,11,20,34,17,34,25,19,19,34,27,25,25,26,33,41,36,45,47,

%T 49,98,62,40,86,88,82,57,65,67,148,142,67,92,134,61,66,56,68,71,68,63,

%U 71,83,97,106,134,77,95,166,186,222,183,113,119,103,103,107,178,134,164

%N a(n) = n-th positive divisor of ((2n)!/n!^2).

%H David A. Corneth, <a href="/A126786/b126786.txt">Table of n, a(n) for n = 1..10000</a>

%e (2*5)!/5!^2 = 252. The divisors of 252 are 1, 2, 3, 4, 6, 7, 9, 12, 14, 18, 21, 28, 36, 42, 63, 84, 126, 252. The 5th of these is 6. So a(5) = 6.

%t f[n_] := Divisors[(2n)!/n!^2][[n]];Array[f, 66] (* _Ray Chandler_, Feb 26 2007 *)

%Y Cf. A000984.

%K nonn

%O 1,2

%A _Leroy Quet_, Feb 18 2007

%E Extended by _Ray Chandler_, Feb 26 2007

%E More terms from _Emeric Deutsch_, Feb 25 2007