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

Author

Asher Auel, Apr 13 2000

Keywords

• nonn

Comments

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]

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

A054040(n) <= A011848(n+2), A054040(10000)=25007302 and A011848(n+2)=25007500. - Robert G. Wilson v, Apr 18 2003

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

Cf. A002387, A004080, A059750, A082915, A054044, A011848, A082915.

See A019529 for a different version.

Programs

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(zBenoit Cloitre, Sep 23 2012

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