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A049341
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a(n+1) = sum of digits of a(n) + a(n-1).
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4
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3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,1
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COMMENTS
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a(n+1) = a007953(a(n) + a(n-1)) for n > 0.
Terms of the simple continued fraction of 21447/[sqrt(1347705679)-29932]. [From Paolo P. Lava, Aug 06 2009]
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LINKS
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Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Michael Gilleland, Some Self-Similar Integer Sequences
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 1).
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FORMULA
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Period 8.
a(n)=1/224*{45*(n mod 8)+213*[(n+1) mod 8]-123*[(n+2) mod 8]+129*[(n+3) mod 8]+45*[(n+4) mod 8]+129*[(n+5) mod 8]-39*[(n+6) mod 8]-39*[(n+7) mod 8]} with n>=0 - Paolo P. Lava, Nov 27 2006
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EXAMPLE
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After 6,9 we get 6+9 = 15 -> 1+5 = 6.
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MATHEMATICA
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LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 1}, {3, 6, 9, 6, 6, 3, 9, 3}, 112] (* Ray Chandler, Aug 27 2015 *)
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PROG
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(Haskell)
a049341 n = a030132_list !! n
a049341_list =
3 : 6 : map a007953 (zipWith (+) a049341_list $ tail a049341_list)
-- Reinhard Zumkeller, Aug 20 2011
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CROSSREFS
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Cf. A030132, A030133, A049342.
Sequence in context: A019700 A151862 A067722 * A321943 A187082 A137991
Adjacent sequences: A049338 A049339 A049340 * A049342 A049343 A049344
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KEYWORD
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base,nonn
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AUTHOR
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Damir Olejar
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EXTENSIONS
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Definition improved by Reinhard Zumkeller, Aug 20 2011
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STATUS
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approved
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