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A131296 a(n) = ds_5(a(n-1))+ds_5(a(n-2)), a(0)=0, a(1)=1; where ds_5=digital sum base 5.

%I A131296 #20 Sep 01 2024 12:31:14
%S A131296 0,1,1,2,3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,5,2,
%T A131296 3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,
%U A131296 5,2,3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,5,2,3,5,4,5,5,2,3
%N A131296 a(n) = ds_5(a(n-1))+ds_5(a(n-2)), a(0)=0, a(1)=1; where ds_5=digital sum base 5.
%C A131296 The digital sum analog (in base 5) of the Fibonacci recurrence.
%C A131296 When starting from index n=3, periodic with Pisano period A001175(4)=6.
%C A131296 a(n) and Fib(n)=A000045(n) are congruent modulo 4 which implies that (a(n) mod 4) is equal to (Fib(n) mod 4)=A079343(n). Thus (a(n) mod 4) is periodic with the Pisano period A001175(4)=6 too.
%C A131296 For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(5) for the base p=5.
%H A131296 Harvey P. Dale, <a href="/A131296/b131296.txt">Table of n, a(n) for n = 0..1000</a>
%H A131296 <a href="/index/Coi#Colombian">Index entries for Colombian or self numbers and related sequences</a>
%F A131296 a(n) = a(n-1)+a(n-2)-4*(floor(a(n-1)/5)+floor(a(n-2)/5)).
%F A131296 a(n) = floor(a(n-1)/5)+floor(a(n-2)/5)+(a(n-1)mod 5)+(a(n-2)mod 5).
%F A131296 a(n) = A002266(a(n-1))+A002266(a(n-2))+A010874(a(n-1))+A010874(a(n-2)).
%F A131296 a(n) = Fib(n)-4*sum{1<k<n, Fib(n-k+1)*floor(a(k)/5)}, where Fib(n)=A000045(n).
%e A131296 a(10)=3, since a(8)=5=10(base 5), ds_5(5)=1,
%e A131296 a(9)=2, ds_5(2)=2 and so a(10)=1+2.
%t A131296 nxt[{a_,b_}]:={b,Total[IntegerDigits[a,5]]+Total[IntegerDigits[b,5]]}; NestList[nxt,{0,1},100][[;;,1]] (* _Harvey P. Dale_, Sep 01 2024 *)
%Y A131296 Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131295, A131297, A131318, A131319, A131320.
%K A131296 nonn,base
%O A131296 0,4
%A A131296 _Hieronymus Fischer_, Jun 27 2007
%E A131296 Incorrect comment removed by _Michel Marcus_, Apr 29 2018