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a(n) = floor( Sum_{k=1..n} k^(1/3) ).
1

%I #18 Mar 10 2022 10:56:48

%S 0,1,2,3,5,6,8,10,12,14,16,19,21,23,26,28,31,33,36,39,41,44,47,50,53,

%T 56,58,61,65,68,71,74,77,80,83,87,90,93,97,100,104,107,110,114,117,

%U 121,125,128,132,136,139,143,147,150,154,158,162,166,170,173,177,181,185,189,193,197

%N a(n) = floor( Sum_{k=1..n} k^(1/3) ).

%H Snehal Shekatkar, <a href="http://arxiv.org/abs/1204.0877">On the sum of the r'th roots of first n natural numbers</a>, arXiv:1204.0877 [math.NT], 2012-2013.

%e a(6) = 8 because 1^(1/3) + 2^(1/3) + 3^(1/3) + 4^(1/3) + 5^(1/3) + 6^(1/3) = 8.81667... .

%t a[n_] := Floor[ HarmonicNumber[n, -1/3]]; Array[ a, 66, 0]

%o (PARI) a(n) = floor(sum(k=0, n, k^(1/3))); \\ _Michel Marcus_, Mar 02 2022

%Y Cf. A031876, A025224.

%K easy,nonn

%O 0,3

%A _Robert G. Wilson v_, Mar 02 2022