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 A131295 a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4. 14
 0, 1, 1, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS The digital sum analog (in base 4) of the Fibonacci recurrence. When starting from index n=3, periodic with Pisano period A001175(3)=8. Also a(n)==A004090(n) modulo 3 (A004090(n)=digital sum of Fib(n)). For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(4) for the base p=4. a(n) and Fib(n)=A000045(n) are congruent modulo 3 which implies that (a(n) mod 3) is equal to (Fib(n) mod 3)=A082115(n-1) (for n>0). Thus (a(n) mod 3) is periodic with the Pisano period = A001175(3)=8 too. - Hieronymus Fischer LINKS Harvey P. Dale, Table of n, a(n) for n = 0..1000 FORMULA a(n)=a(n-1)+a(n-2)-3*(floor(a(n-1)/4)+floor(a(n-2)/4)). a(n)=floor(a(n-1)/4)+floor(a(n-2)/4)+(a(n-1)mod 4)+(a(n-2)mod 4). a(n)=A002265(a(n-1))+A002265(a(n-2))+A010873(a(n-1))+A010873(a(n-2)). a(n)=Fib(n)-3*sum{1

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Last modified September 18 05:32 EDT 2019. Contains 327165 sequences. (Running on oeis4.)