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A010073
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a(n) = sum of base 6 digits of a(n-1) + sum of base 6 digits of a(n-2).
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11
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0, 1, 1, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6, 2, 3, 5, 8, 8, 6, 4, 5, 9, 9, 8, 7, 5, 7, 7, 4, 6, 5, 6, 6
(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 6) of the Fibonacci recurrence. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n)==A004090(n) modulo 5 (A004090(n)=digital sum of Fib(n)). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=9=A131319(6) for the base p=6. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n) and Fib(n)=A000045(n) are congruent modulo 5 which implies that (a(n) mod 5) is equal to (Fib(n) mod 5) = A082116(n) (for n>0). Thus (a(n) mod 6) is periodic with the Pisano period A001175(5)=20. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de),
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FORMULA
| Periodic from n=3 with period 20. - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Mar 13 2006
a(n)=a(n-1)+a(n-2)-5*(floor(a(n-1)/6)+floor(a(n-2)/6)). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n)=floor(a(n-1)/6)+floor(a(n-2)/6)+(a(n-1)mod 6)+(a(n-2)mod 6). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n)=(a(n-1)+a(n-2)+5*(A010875(a(n-1))+A010875(a(n-2))))/6. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
a(n)=Fib(n)-5*sum{1<k<n, Fib(n-k+1)*floor(a(k)/6)} where Fib(n)=A000045(n). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007
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CROSSREFS
| Cf. A000045, A010074, A010075, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.
Sequence in context: A112337 A141804 A121368 * A126167 A026260 A002153
Adjacent sequences: A010070 A010071 A010072 * A010074 A010075 A010076
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KEYWORD
| nonn,base
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Leonid Broukhis (leo(AT)mailcom.com)
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