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%I #8 Sep 02 2014 09:38:27
%S 1,2,1,2,1,2,1,2,1,2,1,4,1,2,1,2,1,2,1,2,1,2,1,6,1,2,1,2,1,2,1,2,1,2,
%T 1,4,1,2,1,2,1,2,1,2,1,2,1,8,1,2,1,2,1,2,1,2,1,2,1,7,1,2,1,2,1,2,1,2,
%U 1,2,1,6,1,2,1,2,1,2,1,2,1,2,1,4,1,2,1,2,1,2,1,2,1,2,1,10,1,2,1,2,1,2,1,2,7
%N Number of q-partial fraction summands of the reciprocal of n-th cyclotomic polynomial.
%C Let Phi(n,q) be the n-th cyclotomic polynomial in q. The q-partial fraction decomposition of 1/Phi(n,q) is a representation of 1/Phi(n,q) as a finite sum of functions v(q)/(1-q^m)^t, such that m<=n and degree(v)<phi(m) (Euler's totient function A000010).
%H Augustine O. Munagi, <a href="http://www.emis.de/journals/INTEGERS/papers/h25/h25.Abstract.html">Computation of q-partial fractions</a>, INTEGERS: Electronic Journal Of Combinatorial Number Theory, 7 (2007), #A25.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CyclotomicPolynomial.html">Cyclotomic Polynomial</a>
%e (i) a(3)=1 because 1/Phi(3,q)=(1-q)/(1-q^3);
%e (ii) a(6)=2 because 1/Phi(6,q)=(-1-q)/(1-q^3) + (2+2q)/(1-q^6).
%Y Cf. A051664 (Number of terms in n-th cyclotomic polynomial).
%K nonn
%O 1,2
%A _Augustine O. Munagi_, Jul 12 2007
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