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Largest integer power of n which divides product of divisors of n.
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%I #23 Nov 12 2023 13:28:10

%S 1,2,3,4,5,36,7,64,9,100,11,1728,13,196,225,256,17,5832,19,8000,441,

%T 484,23,331776,25,676,729,21952,29,810000,31,32768,1089,1156,1225,

%U 1679616,37,1444,1521,2560000,41,3111696,43,85184,91125,2116,47

%N Largest integer power of n which divides product of divisors of n.

%C Product of the distinct parts in the divisor pairs (d,n/d) of n, where d < n/d. For example, the divisors of n = 4 are {1,2,4} with divisor pairs (1,4) and (2,2), but only the pair (1,4) has distinct parts, so a(4) = 1*4 = 4. - _Wesley Ivan Hurt_, Nov 10 2023

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>.

%F a(n) = n^A056924(n).

%F If n is a square a(n) = A007955(n)/sqrt(n), otherwise a(n) = A007955(n).

%e a(16)=256 since the factors of 16 are 1,2,4,8,16, their product is 1024 and the largest power of 16 which divides 1024 is 256.

%t lip[n_]:=Module[{pr=Times@@Divisors[n],pwr},pwr= Floor[ Log[n,pr]]; n^Last[Select[Range[pwr],Divisible[pr,n^#]&]]]; Join[{1},lip/@ Range[2,50]] (* _Harvey P. Dale_, Apr 02 2011 *)

%t a[n_] := n^Floor[DivisorSigma[0, n]/2]; Array[a, 50] (* _Amiram Eldar_, Jun 26 2022 *)

%o (Python)

%o from sympy import divisor_count

%o def A056925(n): return n**(divisor_count(n)//2) # _Chai Wah Wu_, Jun 25 2022

%o (PARI) a(n) = n^(numdiv(n)\2); \\ _Michel Marcus_, Nov 11 2023

%Y Cf. A007955, A056924.

%K nonn

%O 1,2

%A _Henry Bottomley_, Jul 12 2000