|
%I #26 Apr 19 2020 14:14:30
%S 0,1,1,2,3,5,8,6,7,13,13,12,11,9,6,8,7,8,8,2,3,5,8,6,7,13,13,12,11,9,
%T 6,8,7,8,8,2,3,5,8,6,7,13,13,12,11,9,6,8,7,8,8,2,3,5,8,6,7,13,13,12,
%U 11,9,6,8,7,8,8,2,3,5,8,6,7,13,13,12,11,9,6,8,7,8,8,2,3,5,8,6,7,13,13,12,11,9,6,8
%N a(n) = sum of base-8 digits of a(n-1) + sum of base-8 digits of a(n-2).
%C The digital sum analog (in base 8) of the Fibonacci recurrence. - _Hieronymus Fischer_, Jun 27 2007
%C a(n) and Fib(n)=A000045(n) are congruent modulo 7 which implies that (a(n) mod 7) is equal to (Fib(n) mod 7). Thus (a(n) mod 7) is periodic with the Pisano period A001175(7)=16. - _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(8) for the base p=8. - _Hieronymus Fischer_, Jun 27 2007
%H Harvey P. Dale, <a href="/A010075/b010075.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 16. - _Franklin T. Adams-Watters_, Mar 13 2006
%F From _Hieronymus Fischer_, Jun 27 2007: (Start)
%F a(n) = a(n-1)+a(n-2)-7*(floor(a(n-1)/8)+floor(a(n-2)/8)).
%F a(n) = floor(a(n-1)/8)+floor(a(n-2)/8)+(a(n-1)mod 8)+(a(n-2)mod 8).
%F a(n) = (a(n-1)+a(n-2)+7*(A010877(a(n-1))+A010877(a(n-2))))/8.
%F a(n) = Fib(n)-7*sum{1<k<n, Fib(n-k+1)*floor(a(k)/8)} where Fib(n)=A000045(n). (End)
%t nxt[{a_,b_}]:={b,Total[IntegerDigits[a,8]]+Total[IntegerDigits[b,8]]}; NestList[ nxt,{0,1},100][[All,1]] (* or *) PadRight[{0,1,1},100,{7,8,8,2,3,5,8,6,7,13,13,12,11,9,6,8}] (* _Harvey P. Dale_, Apr 19 2020 *)
%Y Cf. A000045, A010073, A010074, 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
|
