Search: id:A030132
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%I A030132
%S A030132 0,1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9,1,1,2,3,5,8,4,3,7,
%T A030132 1,8,9,8,8,7,6,4,1,5,6,2,8,1,9,1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,
%U A030132 5,6,2,8,1,9,1,1,2,3,5,8,4,3,7,1,8,9,8,8,7,6,4,1,5,6,2,8,1,9,1,1,2
%N A030132 Digital root of Fibonacci(n).
%C A030132 Every other (a(0),a(1)) pair of nonzero digits enters a cycle of length 
               24, except for (3,3) which enters a cycle of length 8 and (9,9) which 
               is periodic of length 1. - Jonathan Vos Post (jvospost3(AT)gmail.com), 
               Dec 29 2005
%D A030132 S. Marivani and others, Problem 10974, Amer. Math. Monthly, 111 (No. 
               7, 2004), 628.
%H A030132 Colm Mulcahy, <a href="http://www.maa.org/columns/colm/cardcolm200706.html">
               Gibonacci Bracelets</a>.
%H A030132 Marc Renault, <a href="http://webspace.ship.edu/msrenault/fibonacci/fib.htm">
               The Fibonacci sequence modulo m</a>
%F A030132 a(n+1) = sum of digits of (a(n) + a(n-1)).
%F A030132 Periodic with period 24 given by {1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 
               8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9}
%F A030132 a(n+1) = sum of digits of (a(n) + a(n-1)). a(n+1) = A007953(a(n) + a(n-1)). 
               - Jonathan Vos Post (jvospost3(AT)gmail.com), Dec 29 2005
%F A030132 a(n) + a(n+1) = A010077(n+4); a(A017641(n)) = 9. - Reinhard Zumkeller 
               (reinhard.zumkeller(AT)gmail.com), Jul 04 2007
%Y A030132 Cf. A000045 (Fibonacci numbers), A010888 (digital roots), A004090, A007953.
%Y A030132 Cf. A030133.
%Y A030132 Sequence in context: A098906 A007887 A105472 this_sequence A130833 A004090 
               A104205
%Y A030132 Adjacent sequences: A030129 A030130 A030131 this_sequence A030133 A030134 
               A030135
%K A030132 nonn,base,nice
%O A030132 0,4
%A A030132 youngelder(AT)webtv.net (Ana)
%E A030132 Entry revised by N. J. A. Sloane (njas(AT)research.att.com), Aug 29 2004

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