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a(n) = floor(Sum_{m=1..n} m^(1/m)).
6

%I #15 Jun 17 2023 08:49:42

%S 1,2,3,5,6,7,9,10,11,13,14,15,16,18,19,20,21,22,23,25,26,27,28,29,30,

%T 31,33,34,35,36,37,38,39,40,42,43,44,45,46,47,48,49,50,51,53,54,55,56,

%U 57,58,59,60,61,62,63,64,65,67,68,69,70,71,72,73,74,75,76,77,78,79

%N a(n) = floor(Sum_{m=1..n} m^(1/m)).

%H Vaclav Kotesovec, <a href="/A098572/b098572.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..5000 from G. C. Greubel)

%F a(n) ~ n + log(n)^2/2 + c, where c = A363704 = sg1 + Sum_{k>=2} (-1)^k / k! * k-th derivative of zeta(k) = 0.9885496011422687506447541083399712644219986838..., where sg1 is the first Stieltjes constant (see A082633). - _Vaclav Kotesovec_, Jun 17 2023

%e floor(1^(1/1)+2^(1/2)+3^(1/3))=3 and floor(1^(1/1)+2^(1/2)+3^(1/3)+4^(1/4))=5.

%p A098572 := proc(p)

%p option remember;

%p add(root[i](i),i=1..p) ;

%p floor(%) ;

%p end proc:

%t Table[Floor[Sum[k^(1/k), {k, 1, n}]], {n, 1, 50}] (* _G. C. Greubel_, Feb 03 2018 *)

%o (PARI) for(n=1,30, print1(floor(sum(k=1,n, k^(1/k))), ", ")) \\ _G. C. Greubel_, Feb 03 2018

%o (Magma) [Floor((&+[k^(1/k): k in [1..n]])): n in [1..30]]; // _G. C. Greubel_, Feb 03 2018

%Y Cf. A351885, A363704.

%K easy,nonn

%O 1,2

%A Mark Hudson (mrmarkhudson(AT)hotmail.com), Sep 16 2004