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a(n) = floor( Sum_{i=1..n} n^i/i ).
4

%I #8 Sep 08 2022 08:46:06

%S 1,4,16,97,840,9751,141364,2448224,49213158,1125229242,28823071636,

%T 817379051371,25417591832210,859893812841432,31439503159778411,

%U 1235301513693182001,51906185332750116476,2322562816174820224108,110253678955674860565875

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

%H Bruno Berselli, <a href="/A236772/b236772.txt">Table of n, a(n) for n = 1..100</a>

%e For n=5, floor(5 + 5^2/2 + 5^3/3 + 5^4/4 + 5^4) = floor(10085/12) = 840, therefore a(5) = 840.

%t Table[Floor[Sum[n^i/i, {i, n}]], {n, 20}]

%o (Magma) [Floor(&+[n^i/i: i in [1..n]]): n in [1..20]];

%K nonn

%O 1,2

%A _Bruno Berselli_, Feb 07 2014