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A229144 Partial sums of (Fibonacci numbers mod 3).

%I A229144 #15 Aug 18 2024 15:07:19
%S A229144 0,1,2,4,4,6,8,9,9,10,11,13,13,15,17,18,18,19,20,22,22,24,26,27,27,28,
%T A229144 29,31,31,33,35,36,36,37,38,40,40,42,44,45,45,46,47,49,49,51,53,54,54,
%U A229144 55,56,58,58,60,62,63,63,64,65,67,67,69,71,72,72,73,74,76,76,78,80,81,81,82,83,85,85,87,89
%N A229144 Partial sums of (Fibonacci numbers mod 3).
%H A229144 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 0, 0, 0, 0, 1, -1).
%F A229144 G.f.: (x+x^2+2*x^3+2*x^5+2*x^6+x^7)/((1-x^8)*(1-x)). [_Joerg Arndt_, Sep 15 2013]
%e A229144 The first F(n) are 0, 1, 1, 2, 3, 5, 8,... mod 3 this becomes 0, 1, 1, 2, 0, 2, 2,... so a(n) starts 0, 1, 2, 4 ,4, 6, 8, ...
%t A229144 Accumulate[Mod[Fibonacci[Range[0,80]],3]] (* or *) LinearRecurrence[{1,0,0,0,0,0,0,1,-1},{0,1,2,4,4,6,8,9,9},80] (* _Harvey P. Dale_, Aug 18 2024 *)
%o A229144 (JavaScript)
%o A229144 N=50;
%o A229144 f=new Array();
%o A229144 f[0]=0; f[1]=1;
%o A229144 for (i=2;i<N;i++)  f[i]=f[i-1]+f[i-2];
%o A229144 fs=0;
%o A229144 for (i=0;i<N;i++) { fs+=f[i]%3; document.write(fs+', '); }
%o A229144 (PARI) concat([0], Vec( (x+x^2+2*x^3+2*x^5+2*x^6+x^7)/((1-x^8)*(1-x)) + O(x^166) ) ) \\ _Joerg Arndt_, Sep 15 2013
%Y A229144 Cf. A000045, A082115.
%K A229144 nonn
%O A229144 0,3
%A A229144 _Jon Perry_, Sep 15 2013