A054040 a(n) terms of series {1/sqrt(j)} are >= n.
1, 3, 5, 7, 10, 14, 18, 22, 27, 33, 39, 45, 52, 60, 68, 76, 85, 95, 105, 115, 126, 138, 150, 162, 175, 189, 202, 217, 232, 247, 263, 280, 297, 314, 332, 351, 370, 389, 409, 430, 451, 472, 494, 517, 540, 563, 587, 612, 637, 662, 688, 715, 741, 769, 797, 825
Offset: 1
Keywords
Examples
Let b(k) = 1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k): .k.......1....2.....3.....4.....5.....6.....7 ------------------------------------------------- b(k)...1.00..1.71..2.28..2.78..3.23..3.64..4.01 For A019529 we have: n=0: smallest k is a(0) = 1 since 1.00 > 0 n=1: smallest k is a(1) = 2 since 1.71 > 1 n=2: smallest k is a(2) = 3 since 2.28 > 2 n=3: smallest k is a(3) = 5 since 3.23 > 3 n=4: smallest k is a(4) = 7 since 4.01 > 4 For this sequence we have: n=1: smallest k is a(1) = 1 since 1.00 >= 1 n=2: smallest k is a(2) = 3 since 2.28 >= 2 n=3: smallest k is a(3) = 5 since 3.23 >= 3 n=4: smallest k is a(4) = 7 since 4.01 >= 4
Links
- Sela Fried, On the partial sums of the Zeta function Sum_{n>=1} 1/n^s for 0 < s < 1, 2024.
- Hugo Pfoertner, Plot of a(n)-(1/4)*(n^2-2*zeta(1/2)*n), n=1..1000.
Crossrefs
Programs
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Mathematica
f[n_] := Block[{k = 0, s = 0}, While[s < n, k++; s = N[s + 1/Sqrt[k], 50]]; k]; Table[f[n], {n, 1, 60}] -
PARI
a(n)=if(n<0,0,t=1;z=1;while(z
Formula
Let f(n) = (1/4)*(n^2-2*zeta(1/2)*n) then we have a(n) = f(n) + O(1). More precisely we claim that for n >= 2 we have a(n) = floor(f(n)+c) where c > Max{a(n)-f(n) : n>=1} = a(153) - f(153) = 1.032880076066608813953... and we believe we can take c = 1.033. - Benoit Cloitre, Sep 23 2012
Extensions
Definition and offset modified by N. J. A. Sloane, Sep 01 2009
Comments