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Search: id:A010074
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| A010074 |
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a(n) = sum of base 7 digits of a(n-1) + sum of base 7 digits of a(n-2). |
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+0
13
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| 0, 1, 1, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7
(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 7) of the Fibonacci recurrence. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n) and Fib(n)=A000045(n) are congruent modulo 6 which implies that (a(n) mod 6) is equal to (Fib(n) mod 6) = A082117(n-1) (for n>0). Thus (a(n) mod 6) is periodic with the Pisano period A001175(6)=24. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n)==A004090(n) modulo 6 (A004090(n)=digital sum of Fib(n)). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds (for n>2). Actually, a(n)<=11=A131319(7) for the base p=7. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
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FORMULA
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Periodic from n=3 with period 24. - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Mar 13 2006
a(n)=a(n-1)+a(n-2)-6*(floor(a(n-1)/7)+floor(a(n-2)/7)). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n)=floor(a(n-1)/7)+floor(a(n-2)/7)+(a(n-1)mod 7)+(a(n-2)mod 7). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n)=(a(n-1)+a(n-2)+6*(A010876(a(n-1))+A010876(a(n-2))))/7. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n)=Fib(n)-6*sum{1<k<n, Fib(n-k+1)*floor(a(k)/7)} where Fib(n)=A000045(n). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
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CROSSREFS
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Cf. A000045, A010073, A010075, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.
Sequence in context: A104647 A065115 A010075 this_sequence A116918 A116917 A121369
Adjacent sequences: A010071 A010072 A010073 this_sequence A010075 A010076 A010077
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KEYWORD
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nonn,base
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), Leonid Broukhis (leo(AT)mailcom.com)
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