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%I #10 Apr 11 2021 01:48:53
%S 1,7,18,51,147,423,1224,3543,10254
%N Least integers that satisfy Sum_{n>=1} 1/a(n)^z = 0, where a(1)=1, a(n+1) > a(n) and z = i*Pi/(4*log(2)).
%C Sequence satisfies Sum_{n>=1} 1/a(n)^z = 0 by requiring that the moduli of the successive partial sums are monotonically decreasing in magnitude for the given z.
%o (PARI) 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. A084812, A084813, A084814, A084815, A084816, A084817, A084818.
%K nonn,more
%O 1,2
%A _Paul D. Hanna_, Jun 04 2003