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%I #19 Apr 29 2018 02:10:09
%S 0,1,1,2,3,5,8,13,11,4,5,9,14,13,7,10,17,17,14,11,5,6,11,7,8,15,13,8,
%T 11,9,10,19,19,18,17,15,12,7,9,16,15,11,6,7,13,10,13,13,6,9,15,14,9,
%U 13,12,5,7,12,9,11,10,11,11,2,3,5,8,13,11,4,5,9,14,13,7,10,17,17,14,11
%N a(n) = ds_11(a(n-1))+ds_11(a(n-2)), a(0)=0, a(1)=1; where ds_11=digital sum base 11.
%C The digital sum analog (in base 11) of the Fibonacci recurrence.
%C When starting from index n=3, periodic with Pisano period A001175(10)=60.
%C a(n) and Fib(n)=A000045(n) are congruent modulo 10 which implies that (a(n) mod 10) is equal to (Fib(n) mod 10)=A003893(n). Thus (a(n) mod 10) is periodic with the Pisano period A001175(10)=60 too.
%C a(n)==A074867(n) modulo 10 (A074867(n)=digital product analog base 10 of the Fibonacci recurrence).
%C For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=19=A131319(11) for the base p=11.
%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)-10*(floor(a(n-1)/11)+floor(a(n-2)/11)).
%F a(n) = floor(a(n-1)/11)+floor(a(n-2)/11)+(a(n-1)mod 11)+(a(n-2)mod 11).
%F a(n) = Fib(n)-10*sum{1<k<n, Fib(n-k+1)*floor(a(k)/11)}, where Fib(n)=A000045(n).
%e a(10)=5, since a(8)=11=10(base 11), ds_11(11)=1,
%e a(9)=4, ds_11(4)=4 and so a(10)=1+4.
%t nxt[{a_,b_}]:={b,Total[IntegerDigits[a,11]]+Total[IntegerDigits[b,11]]}; NestList[nxt,{0,1},80][[All,1]] (* or *) PadRight[{0,1,1},80,{10,11,11,2,3,5,8,13,11,4,5,9,14,13,7,10,17,17,14,11,5,6,11,7,8,15,13,8,11,9,10,19,19,18,17,15,12,7,9,16,15,11,6,7,13,10,13,13,6,9,15,14,9,13,12,5,7,12,9,11}] (* _Harvey P. Dale_, Jul 24 2017 *)
%Y Cf. A131297 A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131295, A131296, 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
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