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%I #5 Apr 01 2020 08:24:23
%S 2,3,5,8,12,17,22,30,38,49,60,74,90,108,129,152,177,206,238,272,310,
%T 352,397,447,500,558,620,686,758,834,916,1003,1096,1194,1298,1409,
%U 1526,1649,1779,1916,2060,2212,2371,2538,2713,2896,3087,3287,3496,3714,3941,4178
%N Least k such that f(k) < 0, where f(1) = n and f(k) = f(k-1)*2^(1/k) - 1 for k > 1.
%C If f(k) = f(k-1)*c^(1/k) - 1 for k > 1 and c > e = 2.718281828..., then as long as f(1) is large enough, f(k) can always be positive.
%e For n=0, f(1) = 0 and f(2) = -1 < 0. Thus, a(0) = 2.
%o (PARI) a(n) = {my(k=1, f=n); while(f>0, f=f*sqrtn(2, k++)-1); k+!n; }
%Y Cf. A333342.
%K nonn
%O 0,1
%A _Jinyuan Wang_, Mar 31 2020
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