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A131297
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a(n)=ds_11(a(n-1))+ds_11(a(n-2)), a(0)=0, a(1)=1; where ds_11=digital sum base 11.
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13
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0, 1, 1, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 17, 17, 14, 11, 5, 6, 11, 7, 8, 15, 13, 8, 11, 9, 10, 19, 19, 18, 17, 15, 12, 7, 9, 16, 15, 11, 6, 7, 13, 10, 13, 13, 6, 9, 15, 14, 9, 13, 12, 5, 7, 12, 9, 11, 10, 11, 11, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 17, 17, 14, 11
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| The digital sum analogue (in base 11) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(10)=60.
a(n) and Fib(n)=A000045(n) are congruent modulo 10 which implies that (a(n) mod 10) is equal to (Fib(n) mod 10)=A003893(n). Thus (a(n) mod 10) is periodic with the Pisano period A001175(10)=60 too.
Also, a(n)==A004090(n) modulo 10 (A004090(n)=digital sum of Fib(n)).
a(n)==A074867(n) modulo 10 (A074867(n)=digital product analogue base 10 of the Fibonacci recurrence).
For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=19=A131319(11) for the base p=11.
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FORMULA
| a(n)=a(n-1)+a(n-2)-10*(floor(a(n-1)/11)+floor(a(n-2)/11)).
a(n)=floor(a(n-1)/11)+floor(a(n-2)/11)+(a(n-1)mod 11)+(a(n-2)mod 11).
a(n)=Fib(n)-10*sum{1<k<n, Fib(n-k+1)*floor(a(k)/11)}, where Fib(n)=A000045(n).
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EXAMPLE
| a(10)=5, since a(8)=11=10(base 11), ds_11(11)=1,
a(9)=4, ds_11(4)=4 and so a(10)=1+4.
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CROSSREFS
| Cf. A131297 A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131295, A131296, A131318, A131319, A131320.
Sequence in context: A105995 A104701 A074867 * A010077 A065076 A069638
Adjacent sequences: A131294 A131295 A131296 * A131298 A131299 A131300
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KEYWORD
| nonn,base
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AUTHOR
| Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
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