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Partial sums of A062968.
0

%I #15 Sep 18 2017 13:12:07

%S 1,2,4,6,10,13,19,24,31,38,48,55,67,78,90,102,118,131,149,164,182,201,

%T 223,240,263,286,310,333,361,384,414,441,471,502,534,562,598,633,669,

%U 702,742,777,819,858,898,941,987,1026,1073,1118,1166,1213,1265,1312

%N Partial sums of A062968.

%C A062968(n) counts fractions of the format i/j with 1<=j<n and (i,j) relatively prime.

%C The partial sum gives the number of "essentially" distinct values on the unit circle for all roots up to the n-th. This relates to the problem of decomposing the generating function of the restricted partitions of n, A026820, into partial fractions.

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

%F a(n) = Sum_{i=1..n} A062968(i).

%F a(n) = Sum_{i=1..n} i - floor(n/(i+1)). - _Wesley Ivan Hurt_, Sep 13 2017

%F G.f.: x*(2 - x)/(1 - x)^3 - (1/(1 - x))*Sum_{k>=1} x^k/(1 - x^k). - _Ilya Gutkovskiy_, Sep 18 2017

%e A062968(1)=1 counts the fraction 0/1.

%e A062968(2)=1 counts 1/2.

%e A062968(3)=2 counts {1/3,2/3}.

%e A062968(4)=2 counts {1/4,3/4} skipping 2/4 which could be reduced to 1/2.

%e A062968(5)=4 counts {1/5,2/5,3/5,4/5}. The value a(5)=1+1+2+2+4=10 counts all these distinct fractions {0/1,1/2,1/3,2/3,..,4/5}, which represent the phases of the roots of the polynomials 1-x^j, j=1..5.

%t Table[n + 1 - DivisorSigma[0, n], {n, 1, 54}] // Accumulate (* _Jean-François Alcover_, Jun 24 2013 *)

%o (PARI) A062968(n)={ return(n+1-numdiv(n)) ; }

%o A(n)={ return(sum(i=1,n,A062968(i))) ; }

%o { for(n=1,100,print1(A(n),", ")) ; }

%Y Cf. A062968, A007305, A002088.

%K easy,nonn

%O 1,2

%A _R. J. Mathar_, Jul 09 2008