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a(n) = the 2n-th positive divisor of n!.
1

%I #20 Apr 25 2022 12:40:00

%S 24,15,16,16,20,24,27,27,30,30,33,36,40,40,44,44,48,50,52,52,55,57,63,

%T 65,68,68,70,70,75,77,80,84,87,87,90,92,95,95,98,98,100,104,108,108,

%U 111,114,116,119,121,121,124,126,129,132,135,135,138,138,141,144,147,150

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

%H David A. Corneth, <a href="/A132420/b132420.txt">Table of n, a(n) for n = 4..10003</a>

%F a(n) = A079210(n, 2*n). - _Michel Marcus_, Apr 25 2022

%e The positive divisors of 5! = 120 are 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120. a(5) is the 10th of these, which is 15.

%t a = {}; For[n = 4, n < 70, n++, i = 1; c = 0; While[c < 2n, If[Mod[n!, i] == 0, c++ ]; i++ ]; AppendTo[a, i - 1]]; a (* _Stefan Steinerberger_, Nov 24 2007 *)

%t Table[Divisors[n!][[2n]],{n,4,70}] (* _Harvey P. Dale_, Apr 24 2022 *)

%o (PARI) a(n) = { my(t = 0, nf = n!); for(i = n+1, oo, if(nf % i == 0, t++; if(t == n, return(i)) ) ) } \\ _David A. Corneth_, Apr 25 2022

%o (Python)

%o from math import factorial

%o def a(n):

%o nf, t = factorial(n), 0

%o for i in range(n+1, nf+1):

%o if nf%i == 0: t += 1

%o if t == n: return i

%o print([a(n) for n in range(4, 66)]) # _Michael S. Branicky_, Apr 25 2022 after _David A. Corneth_

%Y Cf. A000142, A079210.

%K nonn

%O 4,1

%A _Leroy Quet_, Nov 20 2007

%E Corrected and extended by _Stefan Steinerberger_, Nov 24 2007