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a(n)=n^2-3*floor[n/sqrt(3)]^2
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%I #7 Nov 18 2024 13:53:53

%S 1,1,6,4,13,9,1,16,6,25,13,36,22,4,33,13,46,24,61,37,9,52,22,69,37,1,

%T 54,16,73,33,94,52,6,73,25,96,46,121,69,13,94,36,121,61,150,88,22,117,

%U 49,148,78,4,109,33,142,64,177,97,13,132,46,169,81,208,118,24,157,61,198

%N a(n)=n^2-3*floor[n/sqrt(3)]^2

%C a(n)=1 for n=A001075

%H Harvey P. Dale, <a href="/A180497/b180497.txt">Table of n, a(n) for n = 1..1000</a>

%e a(5)=13 since 5^2-3*floor[5/sqrt(3)]^2 =25-3*2^2=25-3*4=25-12=13

%t Table[n^2-3Floor[n/Sqrt[3]]^2,{n,70}] (* _Harvey P. Dale_, Nov 18 2024 *)

%Y Cf. A001075, A082532

%K nonn,changed

%O 1,3

%A _Carmine Suriano_, Sep 08 2010