A131294 - OEIS

%I #19 Apr 29 2018 02:09:43

%S 0,1,1,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,

%T 3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,

%U 3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3,3,2,3

%N a(n)=ds_3(a(n-1))+ds_3(a(n-2)), a(0)=0, a(1)=1; where ds_3=digital sum base 3.

%C The digital sum analog (in base 3) of the Fibonacci recurrence.

%C When starting from index n=3, periodic with Pisano period A001175(2)=3.

%C a(n) and Fib(n)=A000045(n) are congruent modulo 2 which implies that (a(n) mod 2) is equal to (Fib(n) mod 2)=A011655(n). Thus (a(n) mod 2) is periodic with the Pisano period A001175(2)=3 too.

%C For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=3=A131319(3) for the base p=3.

%H <a href="/index/Coi#Colombian">Index entries for Colombian or self numbers and related sequences</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 1).

%F a(n) = a(n-1)+a(n-2)-2*(floor(a(n-1)/3)+floor(a(n-2)/3)).

%F a(n) = floor(a(n-1)/3)+floor(a(n-2)/3)+(a(n-1)mod 3)+(a(n-2)mod 3).

%F a(n) = A002264(a(n-1))+A002264(a(n-2))+A010872(a(n-1))+A010872(a(n-2)).

%F a(n) = Fib(n)-2*sum{1<k<n, Fib(n-k+1)*floor(a(k)/3)}, where Fib(n)=A000045(n).

%e a(5)=3, since a(3)=2, ds_3(2)=2, a(4)=3=10(base 3),

%e ds_3(3)=1 and so a(5)=2+1.

%t nxt[{a_,b_}]:={b,Total[IntegerDigits[a,3]]+Total[IntegerDigits[b,3]]}; Transpose[NestList[nxt,{0,1},100]][[1]] (* _Harvey P. Dale_, Aug 02 2016 *)

%Y Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131295, A131296, A131297, A131318, A131319, A131320.

%K nonn,base

%O 0,4

%A _Hieronymus Fischer_, Jun 27 2007

%E Incorrect comment removed by _Michel Marcus_, Apr 29 2018