Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
a(n) = Sum_{k=0..n} (-1)^k*floor(k^(1/4)).
1

%I #19 Nov 06 2018 21:18:46

%S 0,-1,0,-1,0,-1,0,-1,0,-1,0,-1,0,-1,0,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,

%T -1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,

%U 1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-2,1,-2,1,-2,1,-2,1,-2,1,-2,1,-2,1,-2,1

%N a(n) = Sum_{k=0..n} (-1)^k*floor(k^(1/4)).

%H Antti Karttunen, <a href="/A262352/b262352.txt">Table of n, a(n) for n = 0..16383</a>

%H Antti Karttunen, <a href="/A262352/a262352.txt">Data supplement: n, a(n) computed for n = 0..65537</a>

%F a(n) = floor(n^(1/4))*(-1)^n/2-((-1)^(floor(n^(1/4))+1)+1)/4.

%e Letting [] denote the floor function, a(7) = [0^(1/4)] - [1^(1/4)] + [2^(1/4)] - [3^(1/4)] + [4^(1/4)] - [5^(1/4)] + [6^(1/4)] - [7^(1/4)] = 0 - 1 + 1 - 1 + 1 - 1 + 1 - 1 = -1.

%t Print[Table[Sum[(-1)^k*Floor[k^(1/4)],{k,0,n}],{n,0,100}]] ;

%o (PARI) a(n)=floor(n^(1/4))*(-1)^n/2-((-1)^(floor(n^(1/4))+1)+1)/4

%o (PARI) a(n)=sum(k=0,n,(-1)^k*floor(k^(1/4)))

%o (PARI) A262352(n) = sum(k=0,n,((-1)^k)*sqrtnint(k, 4)); \\ _Antti Karttunen_, Nov 06 2018

%Y Cf. A270370, A268173, A022554, A031876, A032512, A032513, A032514, A032515, A032516, A032517, A032518, A032519, A032520, A032521.

%K sign,easy

%O 0,82

%A _John M. Campbell_, Mar 24 2016

%E More terms from _Antti Karttunen_, Nov 06 2018