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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)).
0

%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