A364415 a(n) is the least k such that Sum_{j=1..k} 1/(j^(1 + 1/j)) >= n.
0, 1, 6, 22, 65, 185, 512, 1402, 3825, 10412, 28318, 76995, 209314, 568995, 1546713, 4204428, 11428848, 31066858, 84448506, 229554871, 623994868, 1696193945, 4610733216, 12533272358, 34068966559
Offset: 0
Examples
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.
References
- R. P. Boas, Growth of partial sums of divergent series, Mathematics of Computation, 31 (1977), 257-264.
Links
- Michael Penn, A notorious Calculus problem, YouTube video, 2023.
Programs
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Mathematica
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 *) -
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)))
Formula
a(n) >= A004080(n).
Conjecture: Limit_{n->oo} a(n)/a(n-1) = e. - Hugo Pfoertner, Jul 29 2023
Extensions
a(13)-a(18) from Hugo Pfoertner, Jul 29 2023
a(19)-a(24) from Sela Fried, Jul 03 2024
Comments