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 A010077 a(n) = sum of digits of a(n-1) + sum of digits of a(n-2); a(0) = 0, a(1) = 1. 15
 0, 1, 1, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10, 2, 3, 5, 8, 13, 12, 7, 10, 8, 9, 17, 17, 16, 15, 13, 10, 5, 6, 11, 8, 10, 9, 10, 10 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS The digital sum analog (in base 10) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007 a(n) and Fib(n)=A000045(n) are congruent modulo 9 which implies that (a(n) mod 9) is equal to (Fib(n) mod 9) = A007887(n). Thus (a(n) mod 9) is periodic with the Pisano period A001175(9)=24. - Hieronymus Fischer, Jun 27 2007 a(n)==A004090(n) modulo 9 (A004090(n)=digital sum of Fib(n)). - 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)<=17=A131319(10) for the base p=10. - Hieronymus Fischer, Jun 27 2007 LINKS FORMULA Periodic from n=3 with period 24. - Franklin T. Adams-Watters, Mar 13 2006 a(n) = A030132(n-4) + A030132(n-3) for n>3. - Reinhard Zumkeller, Jul 04 2007 a(n)=a(n-1)+a(n-2)-9*(floor(a(n-1)/10)+floor(a(n-2)/10)). - Hieronymus Fischer, Jun 27 2007 a(n)=floor(a(n-1)/10)+floor(a(n-2)/10)+(a(n-1)mod 10)+(a(n-2)mod 10). - Hieronymus Fischer, Jun 27 2007 a(n)=A059995(a(n-1))+A059995(a(n-2))+A010879(a(n-1))+A010879(a(n-2)). - Hieronymus Fischer, Jun 27 2007 a(n)=Fib(n)-9*sum{1

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Last modified October 22 18:05 EDT 2019. Contains 328319 sequences. (Running on oeis4.)