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A131294
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a(n)=ds_3(a(n-1))+ds_3(a(n-2)), a(0)=0, a(1)=1; where ds_3=digital sum base 3.
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14
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0, 1, 1, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3
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OFFSET
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0,4
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COMMENTS
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The digital sum analog (in base 3) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(2)=3.
a(n) and Fib(n)=A000045(n) are congruent modulo 2 which implies that (a(n) mod 2) is equal to (Fib(n) mod 2)=A011655(n). Thus (a(n) mod 2) is periodic with the Pisano period A001175(2)=3 too.
For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=3=A131319(3) for the base p=3.
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LINKS
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FORMULA
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a(n) = a(n-1)+a(n-2)-2*(floor(a(n-1)/3)+floor(a(n-2)/3)).
a(n) = floor(a(n-1)/3)+floor(a(n-2)/3)+(a(n-1)mod 3)+(a(n-2)mod 3).
a(n) = Fib(n)-2*sum{1<k<n, Fib(n-k+1)*floor(a(k)/3)}, where Fib(n)=A000045(n).
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EXAMPLE
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a(5)=3, since a(3)=2, ds_3(2)=2, a(4)=3=10(base 3),
ds_3(3)=1 and so a(5)=2+1.
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MATHEMATICA
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nxt[{a_, b_}]:={b, Total[IntegerDigits[a, 3]]+Total[IntegerDigits[b, 3]]}; Transpose[NestList[nxt, {0, 1}, 100]][[1]] (* Harvey P. Dale, Aug 02 2016 *)
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CROSSREFS
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Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131295, A131296, A131297, A131318, A131319, A131320.
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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