%I #27 Apr 29 2018 02:07:03

%S 0,1,1,2,3,5,8,7,3,4,7,5,6,11,11,10,9,7,4,5,9,8,5,7,6,7,7,2,3,5,8,7,3,

%T 4,7,5,6,11,11,10,9,7,4,5,9,8,5,7,6,7,7,2,3,5,8,7,3,4,7,5,6,11,11,10,

%U 9,7,4,5,9,8,5,7,6,7,7

%N a(n) = sum of base-7 digits of a(n-1) + sum of base-7 digits of a(n-2).

%C The digital sum analog (in base 7) of the Fibonacci recurrence. - _Hieronymus Fischer_, Jun 27 2007

%C a(n) and Fib(n)=A000045(n) are congruent modulo 6 which implies that (a(n) mod 6) is equal to (Fib(n) mod 6) = A082117(n-1) (for n>0). Thus (a(n) mod 6) is periodic with the Pisano period A001175(6)=24. - _Hieronymus Fischer_, Jun 27 2007

%C For general bases p>2, the inequality 2<=a(n)<=2p-3 holds (for n>2). Actually, a(n)<=11=A131319(7) for the base p=7. - _Hieronymus Fischer_, Jun 27 2007

%H Vincenzo Librandi, <a href="/A010074/b010074.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 Periodic from n=3 with period 24. - _Franklin T. Adams-Watters_, Mar 13 2006

%F From _Hieronymus Fischer_, Jun 27 2007: (Start)

%F a(n) = a(n-1)+a(n-2)-6*(floor(a(n-1)/7)+floor(a(n-2)/7)).

%F a(n) = floor(a(n-1)/7)+floor(a(n-2)/7)+(a(n-1)mod 7)+(a(n-2)mod 7).

%F a(n) = (a(n-1)+a(n-2)+6*(A010876(a(n-1))+A010876(a(n-2))))/7.

%F a(n) = Fib(n)-6*sum{1<k<n, Fib(n-k+1)*floor(a(k)/7)} where Fib(n)=A000045(n). (End)

%t nxt[{a_,b_}]:={b,Total[IntegerDigits[a,7]]+Total[IntegerDigits[b,7]]}; Transpose[NestList[nxt,{0,1},80]][[1]] (* _Harvey P. Dale_, Oct 12 2013 *)

%Y Cf. A000045, A010073, A010075, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

%K nonn,base

%O 0,4

%A _N. J. A. Sloane_, _Leonid Broukhis_

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