login
Product of proper unitary divisors of n.
2

%I #22 May 17 2020 13:49:43

%S 1,1,1,1,1,6,1,1,1,10,1,12,1,14,15,1,1,18,1,20,21,22,1,24,1,26,1,28,1,

%T 27000,1,1,33,34,35,36,1,38,39,40,1,74088,1,44,45,46,1,48,1,50,51,52,

%U 1,54,55,56,57,58,1,216000,1,62,63,1,65,287496,1,68,69,343000,1,72,1,74,75,76,77,474552,1,80

%N Product of proper unitary divisors of n.

%H Antti Karttunen, <a href="/A290480/b290480.txt">Table of n, a(n) for n = 1..65537</a>

%F a(n) = A061537(n)/n.

%F a(n) = n^(2^(omega(n)-1)-1), where omega() is the number of distinct primes dividing n (A001221).

%F a(n) = 1 if n is a prime power.

%e a(12) = 12 because 12 has 6 divisors {1, 2, 3, 4, 6, 12} among which 3 are proper unitary {1, 3, 4} and 1*3*4 = 12.

%p with(numtheory):

%p a:= n-> mul(d, d=select(x-> igcd(x, n/x)=1, divisors(n) minus {n})):

%p seq(a(n), n=1..80); # _Alois P. Heinz_, Aug 03 2017

%t Table[Product[d, {d, Select[Divisors[n], GCD[#, n/#] == 1 &]}]/n, {n, 80}]

%t Table[n^(2^(PrimeNu[n] - 1) - 1), {n, 80}]

%o (Python)

%o from sympy import divisors, gcd, prod

%o def a(n): return prod(d for d in divisors(n) if gcd(d, n//d) == 1)//n

%o print([a(n) for n in range(1, 51)]) # _Indranil Ghosh_, Aug 04 2017

%o (PARI) A290480(n) = if(1==n,n,n^(2^(omega(n)-1)-1)); \\ _Antti Karttunen_, Aug 06 2018

%Y Cf. A001221, A007774 (fixed points), A007955, A007956, A034460, A061537, A062758, A063919, A078599, A087652, A126192, A136655, A183091.

%K nonn

%O 1,6

%A _Ilya Gutkovskiy_, Aug 03 2017