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%I #19 Dec 20 2021 12:22:03
%S 1,1,2,1,2,1,2,0,2,2,3,0,3,2,4,-1,2,1,1,0,3,2,2,-2,2,-1,2,2,2,-1,3,-1,
%T 7,2,2,0,3,2,4,-2,3,-1,1,1,0,3,5,-1,4,1,1,-2,1,1,5,-2,4,-1,4,0,3,1,1,
%U -1,2,1,3,-1,6,-1,2,-5,7,0,1,-1,4,-2,8,-5,2,4,1,3,2,4,2,-3,1,0,2,-1,3,4,0,-6,2,-1,6,3,3,1,5,-6,9
%N a(n) = Sum_{k=1..n} (-1)^(k+1) * floor((n/k)^k).
%H Seiichi Manyama, <a href="/A350223/b350223.txt">Table of n, a(n) for n = 1..5000</a>
%e a(3) = [3/1] - [(3/2)^2] + [(3/3)^3] = 3 - 2 + 1 = 2.
%e a(4) = [4/1] - [(4/2)^2] + [(4/3)^3] - [(4/4)^4] = 4 - 4 + 2 - 1 = 1.
%t a[n_] := Sum[(-1)^(k + 1)*Floor[(n/k)^k], {k, 1, n}]; Array[a, 100] (* _Amiram Eldar_, Dec 20 2021 *)
%o (PARI) a(n) = sum(k=1, n, (-1)^(k+1)*(n^k\k^k));
%Y Cf. A347415, A350224.
%K sign
%O 1,3
%A _Seiichi Manyama_, Dec 20 2021