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a(1)=1; for n >= 2, choose a(n) so that |Sum_{k=1..n} 1/a(k)^z| < |Sum_{k=1..n-1} 1/a(k)^z|, where z = 1/2 + i*2*Pi.
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%I #9 Apr 11 2021 01:52:40

%S 1,4,5,11,14,19,29,43,59,78,99,123,150,179,212,247,286,327,372,419,

%T 470,523,580,639,702,767,836,907,982,1059,1140,1223,1310,1399,1492,

%U 1587,1686,1787,1892,1999,2110,2223,2340,2459,2582,2707,2836,2967,3102,3239

%N a(1)=1; for n >= 2, choose a(n) so that |Sum_{k=1..n} 1/a(k)^z| < |Sum_{k=1..n-1} 1/a(k)^z|, where z = 1/2 + i*2*Pi.

%o (PARI) z=1/2 + 2*Pi*I; S=0; w=1; a=0; for(n=1,100,b=a+1; while(abs(S+exp(-z*log(b)))>w,b++); S=S+exp(-z*log(b)); w=abs(S); a=b; print1(b,","))

%Y Cf. A084799-A084809, A084813-A084818.

%K nonn

%O 1,2

%A _Paul D. Hanna_, Jun 04 2003