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Search: id:A131294
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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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+0
15
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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
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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The digital sum analogue (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.
Also, a(n)==A004090(n) modulo 2 (A004090(n)=digital sum of Fib(n)).
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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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)=A002264(a(n-1))+A002264(a(n-2))+A010872(a(n-1))+A010872(a(n-2)).
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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CROSSREFS
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Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131295, A131296, A131297, A131318, A131319, A131320.
Adjacent sequences: A131291 A131292 A131293 this_sequence A131295 A131296 A131297
Sequence in context: A130633 A048198 A096006 this_sequence A102313 A007538 A025076
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KEYWORD
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nonn
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AUTHOR
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Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
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