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Sequence related to discriminant of cyclotomic polynomials A004124.
5

%I #30 May 13 2024 05:24:14

%S 1,2,3,4,5,12,7,16,27,80,11,144,13,448,2025,256,17,1728,19,6400,35721,

%T 11264,23,20736,3125,53248,19683,200704,29,518400,31,65536,7144929,

%U 1114112,37515625,2985984,37,4980736,89813529,40960000,41,146313216,43,126877696

%N Sequence related to discriminant of cyclotomic polynomials A004124.

%C a(p) = p for primes p.

%D P. Ribenboim, Classical Theory of Algebraic Numbers, Springer, 2001, p. 297, eq.(1).

%H Alois P. Heinz, <a href="/A193679/b193679.txt">Table of n, a(n) for n = 1..1000</a>

%H Mohammad K. Azarian, <a href="http://www.ijpam.eu/contents/2007-36-2/9/9.pdf">On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials</a>, International Journal of Pure and Applied Mathematics, Vol. 36, No. 2, 2007, pp. 251-257. Mathematical Reviews, MR2312537. Zentralblatt MATH, Zbl 1133.11012.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CyclotomicPolynomial.html">Cyclotomic Polynomial</a>

%F a(n) = n^phi(n)/abs(discriminant(Phi(n,x))), n>=1, with the cyclotomic polynomials Phi(n,x) and the Euler totient function phi(n)=A000010(n).

%F a(n) = product(p^(phi(n)/(p-1)),p prime dividing n), n>=2, a(1)=1.

%F Conjecture: Dirichlet g.f. of log(a(n)): -zeta(s-1)*zeta'(s)/zeta(s)^2, where zeta'(s) is the derivative of zeta(s). This would give a(n) = exp(Sum_{d|n} Lambda(d)*phi(n/d)), with Lambda(n)=log(A014963) and phi(n)=A000010. - _Benedict W. J. Irwin_, Jul 14 2018

%e n=6: a(6) = 2^(2/(2-1))*3^(2/(3-1)) = 12.

%e Discriminant(Phi(6,x)) = -3 = - (6^phi(6))/a(6).

%p with(numtheory): A193679 := n -> n^phi(n)/abs(discrim(cyclotomic (n,x),x)); seq(A193679(i),i=1..49); # _Peter Luschny_, Aug 20 2011

%t a[n_] := n^EulerPhi[n]/Abs[Discriminant[Cyclotomic[n, x], x]]; Array[a, 44]

%t (* _Jean-François Alcover_, Mar 21 2017 *)

%t Table[Product[d^(-n*MoebiusMu[d]/d), {d, Divisors[n]}], {n, 1, 50}] (* _Vaclav Kotesovec_, May 12 2024 *)

%t Table[Product[p^(EulerPhi[n]/(p-1)), {p, Select[Divisors[n], PrimeQ[#]&]}], {n, 1, 50}] (* _Vaclav Kotesovec_, May 13 2024 *)

%o (PARI) a(n) = n^eulerphi(n)/abs(poldisc(polcyclo(n))); \\ _Michel Marcus_, Jul 14 2018

%Y Cf. A004124.

%K nonn

%O 1,2

%A _Wolfdieter Lang_, Aug 20 2011