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%I #14 Jul 30 2018 18:07:29
%S 0,1,1,2,3,5,5,4,3,4,4,2,3,5,5,4,3,4,4,2,3,5,5,4,3,4,4,2,3,5,5,4,3,4,
%T 4,2,3,5,5,4,3,4,4,2,3,5,5,4,3,4,4,2,3,5,5,4,3,4,4,2,3,5,5,4,3,4,4,2,
%U 3,5,5,4,3,4,4,2,3,5,5,4,3,4,4,2,3,5,5,4,3,4,4,2,3,5,5,4,3,4,4,2,3
%N a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4.
%C The digital sum analog (in base 4) of the Fibonacci recurrence.
%C When starting from index n=3, periodic with Pisano period A001175(3)=8.
%C Also a(n)==A004090(n) modulo 3 (A004090(n)=digital sum of Fib(n)).
%C For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(4) for the base p=4.
%C a(n) and Fib(n)=A000045(n) are congruent modulo 3 which implies that (a(n) mod 3) is equal to (Fib(n) mod 3)=A082115(n-1) (for n>0). Thus (a(n) mod 3) is periodic with the Pisano period = A001175(3)=8 too. - _Hieronymus Fischer_
%H Harvey P. Dale, <a href="/A131295/b131295.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Coi#Colombian">Index entries for Colombian or self numbers and related sequences</a>
%F a(n)=a(n-1)+a(n-2)-3*(floor(a(n-1)/4)+floor(a(n-2)/4)).
%F a(n)=floor(a(n-1)/4)+floor(a(n-2)/4)+(a(n-1)mod 4)+(a(n-2)mod 4).
%F a(n)=A002265(a(n-1))+A002265(a(n-2))+A010873(a(n-1))+A010873(a(n-2)).
%F a(n)=Fib(n)-3*sum{1<k<n, Fib(n-k+1)*floor(a(k)/4)}, where Fib(n)=A000045(n).
%e a(8)=3, since a(6)=5=11(base 4), ds_4(5)=2,
%e a(7)=4=10(base 4), ds_4(4)=1 and so a(8)=2+1.
%t nxt[{a_,b_}]:={b,Total[IntegerDigits[a,4]]+Total[IntegerDigits[b,4]]}; NestList[ nxt,{0,1},110][[All,1]] (* _Harvey P. Dale_, Jul 30 2018 *)
%Y Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131296, A131297, A131318, A131319, A131320.
%K nonn,base
%O 0,4
%A _Hieronymus Fischer_, Jun 27 2007
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