cp's OEIS Frontend

A054040 a(n) terms of series {1/sqrt(j)} are >= n.

%I A054040 #46 Jul 03 2024 14:23:06
%S A054040 1,3,5,7,10,14,18,22,27,33,39,45,52,60,68,76,85,95,105,115,126,138,
%T A054040 150,162,175,189,202,217,232,247,263,280,297,314,332,351,370,389,409,
%U A054040 430,451,472,494,517,540,563,587,612,637,662,688,715,741,769,797,825
%N A054040 a(n) terms of series {1/sqrt(j)} are >= n.
%C A054040 In many cases the first differences have the form {2k, 2k, 2k, 2k+1} (A004524). In such cases the second differences are {0, 0, 1, 1}. See A082915 for the exceptions. In as many as these, the first differences have the form {2k-1, 2k-1, 2k-1, 2k}. - _Robert G. Wilson v_, Apr 18 2003 [Corrected by _Carmine Suriano_, Nov 08 2013]
%C A054040 a(100)=2574, a(1000)=250731 & a(10000)=25007302 which differs from Sum{i=4..104}A004524(i)=2625, Sum{i=4..1004}A004524(i)=251250 & Sum{i=4..10004}A004524(i)=25012500. - _Robert G. Wilson v_, Apr 18 2003
%C A054040 A054040(n) <= A011848(n+2), A054040(10000)=25007302 and A011848(n+2)=25007500. - _Robert G. Wilson v_, Apr 18 2003
%H A054040 Sela Fried, <a href="/A054040/a054040_1.pdf">On the partial sums of the Zeta function Sum_{n>=1} 1/n^s for 0 < s < 1</a>, 2024.
%H A054040 Hugo Pfoertner, <a href="/A054040/a054040.png">Plot of a(n)-(1/4)*(n^2-2*zeta(1/2)*n)</a>, n=1..1000.
%F A054040 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
%e A054040 Let b(k) = 1 + 1/sqrt(2) + 1/sqrt(3) + ... + 1/sqrt(k):
%e A054040 .k.......1....2.....3.....4.....5.....6.....7
%e A054040 -------------------------------------------------
%e A054040 b(k)...1.00..1.71..2.28..2.78..3.23..3.64..4.01
%e A054040 For A019529 we have:
%e A054040 n=0: smallest k is a(0) = 1 since 1.00 > 0
%e A054040 n=1: smallest k is a(1) = 2 since 1.71 > 1
%e A054040 n=2: smallest k is a(2) = 3 since 2.28 > 2
%e A054040 n=3: smallest k is a(3) = 5 since 3.23 > 3
%e A054040 n=4: smallest k is a(4) = 7 since 4.01 > 4
%e A054040 For this sequence we have:
%e A054040 n=1: smallest k is a(1) = 1 since 1.00 >= 1
%e A054040 n=2: smallest k is a(2) = 3 since 2.28 >= 2
%e A054040 n=3: smallest k is a(3) = 5 since 3.23 >= 3
%e A054040 n=4: smallest k is a(4) = 7 since 4.01 >= 4
%t A054040 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}]
%o A054040 (PARI) a(n)=if(n<0,0,t=1;z=1;while(z<n,t++;z=z+1/sqrt(t));t) \\ _Benoit Cloitre_, Sep 23 2012
%Y A054040 Cf. A002387, A004080, A059750, A082915, A054044, A011848, A082915.
%Y A054040 See A019529 for a different version.
%K A054040 nonn
%O A054040 1,2
%A A054040 _Asher Auel_, Apr 13 2000
%E A054040 Definition and offset modified by _N. J. A. Sloane_, Sep 01 2009