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 A010076 a(n) = sum of base-9 digits of a(n-1) + sum of base-9 digits of a(n-2). 12
 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, 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, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS The digital sum analog (in base 9) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007 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 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 LINKS Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1). FORMULA Periodic from n=3 with period 12. - Franklin T. Adams-Watters, Mar 13 2006 From Hieronymus Fischer, Jun 27 2007: (Start) a(n) = a(n-1)+a(n-2)-8*(floor(a(n-1)/9)+floor(a(n-2)/9)). a(n) = floor(a(n-1)/9)+floor(a(n-2)/9)+(a(n-1)mod 9)+(a(n-2)mod 9). a(n) = (a(n-1)+a(n-2)+8*(A010878(a(n-1))+A010878(a(n-2))))/9. a(n) = Fib(n)-8*sum{1

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Last modified September 15 20:11 EDT 2019. Contains 327086 sequences. (Running on oeis4.)