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%I #43 Jul 03 2024 16:39:41
%S 0,1,6,22,65,185,512,1402,3825,10412,28318,76995,209314,568995,
%T 1546713,4204428,11428848,31066858,84448506,229554871,623994868,
%U 1696193945,4610733216,12533272358,34068966559
%N a(n) is the least k such that Sum_{j=1..k} 1/(j^(1 + 1/j)) >= n.
%C As Sum_{j=1..k} 1/(j^(1 + 1/j)) diverges this sequence is well-defined.
%D R. P. Boas, Growth of partial sums of divergent series, Mathematics of Computation, 31 (1977), 257-264.
%H Michael Penn, <a href="https://www.youtube.com/watch?v=BXYc2DLXJo0">A notorious Calculus problem</a>, YouTube video, 2023.
%F a(n) >= A004080(n).
%F Conjecture: Limit_{n->oo} a(n)/a(n-1) = e. - _Hugo Pfoertner_, Jul 29 2023
%e Let f(m) = Sum_{j=1..m} 1/(j^(1 + 1/j)) and n = 2. Then 1.906406... f(5) < n = 2 <= f(6) = 2.030045839... . So 6 is the smallest m such that f(m) >= 2. Therefore a(2) = 6.
%t a={0}; sum=0; k=1; For[n=1, n<=8, n++, While[ sum<=n,If[(sum+=1/(k^(1+1/k)))>=n, AppendTo[a,k]]; k++]]; a (* _Stefano Spezia_, Jul 24 2023 *)
%o (PARI) a(n) = if(n == 0, return(0)); my(t = 0); for(i = 1, oo, t+= 1/(i^(1+1/i)); if(t >= n, return(i)))
%Y Cf. A002387, A004080.
%K nonn,more
%O 0,3
%A _David A. Corneth_, Jul 23 2023
%E a(13)-a(18) from _Hugo Pfoertner_, Jul 29 2023
%E a(19)-a(24) from _Sela Fried_, Jul 03 2024