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Least integers that satisfy sum(n>0,1/a(n)^z)=0, where a(1)=1, a(n+1)>a(n) and z=I*2.
3

%I #5 Mar 30 2012 18:36:38

%S 1,3,6,13,39,131,448,1547,5352,18524,64125,221990,768506,2660499,

%T 9210416,31885663,110385418,382144818,1322952481

%N Least integers that satisfy sum(n>0,1/a(n)^z)=0, where a(1)=1, a(n+1)>a(n) and z=I*2.

%C Sequence satisfies sum(n>0,1/a(n)^z)=0 by requiring that the modulus of the successive partial sums are monotonically decreasing in magnitude for the given z.

%F Limit a(n+1)/a(n) --> 3.4619139...

%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-A084815, A084817-A084818.

%K nonn

%O 1,2

%A _Paul D. Hanna_, Jun 04 2003