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Floor(sum_{i=1..n} (sum_{j=1..i} sqrt(1/j))).
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%I #14 May 02 2013 06:23:20

%S 1,2,4,7,11,14,18,23,27,32,38,43,49,55,62,68,75,82,90,97,105,113,121,

%T 130,138,147,156,166,175,185,194,204,214,225,235,246,257,267,279,290,

%U 301,313,325,336,349,361,373,385,398

%N Floor(sum_{i=1..n} (sum_{j=1..i} sqrt(1/j))).

%C The fact that a(n)/n diverges (it is greater than sqrt(n)) implies sum_{k>=1} 1/sqrt(k) is not Cesaro summable.

%H Balarka Sen, <a href="/A225154/b225154.txt">Table of n, a(n) for n = 1..500</a>

%F a(n) ~ 2*sum_{k=1..n} sqrt(k) ~ (4/3) n^(3/2).

%o (PARI) for(n=1,100,print1(floor(sum(i=1,n,sum(j=1,i,1/sqrt(j))))","))

%o (PARI) a(n)=sum(j=1,n,(n+1-j)/sqrt(j))\1 \\ _Charles R Greathouse IV_, May 02 2013

%Y Cf. A022819, A025224.

%K nonn,easy

%O 1,2

%A _Balarka Sen_, Apr 30 2013