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%I #23 Apr 29 2018 02:07:31
%S 0,1,1,2,3,5,8,13,13,10,7,9,8,9,9,2,3,5,8,13,13,10,7,9,8,9,9,2,3,5,8,
%T 13,13,10,7,9,8,9,9,2,3,5,8,13,13,10,7,9,8,9,9,2,3,5,8,13,13,10,7,9,8,
%U 9,9,2,3,5,8,13,13,10,7,9,8,9,9
%N a(n) = sum of base-9 digits of a(n-1) + sum of base-9 digits of a(n-2).
%C The digital sum analog (in base 9) of the Fibonacci recurrence. - _Hieronymus Fischer_, Jun 27 2007
%C a(n) and Fib(n)=A000045(n) are congruent modulo 8 which implies that (a(n) mod 8) is equal to (Fib(n) mod 8) = A079344(n). Thus (a(n) mod 8) is periodic with the Pisano period A001175(8)=12. - _Hieronymus Fischer_, Jun 27 2007
%C For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=13=A131319(9) for the base p=9. - _Hieronymus Fischer_, Jun 27 2007
%H <a href="/index/Coi#Colombian">Index entries for Colombian or self numbers and related sequences</a>
%H <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).
%F Periodic from n=3 with period 12. - _Franklin T. Adams-Watters_, Mar 13 2006
%F From _Hieronymus Fischer_, Jun 27 2007: (Start)
%F a(n) = a(n-1)+a(n-2)-8*(floor(a(n-1)/9)+floor(a(n-2)/9)).
%F a(n) = floor(a(n-1)/9)+floor(a(n-2)/9)+(a(n-1)mod 9)+(a(n-2)mod 9).
%F a(n) = (a(n-1)+a(n-2)+8*(A010878(a(n-1))+A010878(a(n-2))))/9.
%F a(n) = Fib(n)-8*sum{1<k<n, Fib(n-k+1)*floor(a(k)/9)} where Fib(n)=A000045(n). (End)
%Y Cf. A000045, A010073, A010074, A010075, 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
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