OFFSET
1,2
COMMENTS
More precisely, start with 0,0,...,0,1 (with n-1 0's and a single 1); thereafter the next term is the digital root (A010888) of the sum of the previous n terms. This is a periodic sequence and a(n) is the length of the period.
Theorem: a(n) <= 9^n.
Conjecture: All entries >1 are divisible by 3.
Additional terms are a(242)=177144, a(243)=177879.
More: a(728)=1594320, a(729)=1596513, a(2186)=14348904, a(2187)=14355471, a(6560)=129140160, a(6561)=129159849, a(19682)=1162261464, a(19683)=1162320519. - Hans Havermann, Jan 30 2013, Feb 08 2013
The modulus-9 Pisano periods of Fibonacci numbers, k-th order sequences. - Hans Havermann, Feb 10 2013
Conjecture: a(3^n-1)=3^(2*n+1)-3, a(3^n)=3^(2*n+1)+3^(n+1)+3 - Fred W. Helenius (fredh(AT)ix.netcom.com), posting to MathFun, Feb 21 2013
LINKS
Hiroaki Yamanouchi, Table of n, a(n) for n = 1..100
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number
Eric Weisstein's World of Mathematics, Pisano Period
EXAMPLE
Digital roots of Fibonacci numbers (A030132) are 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,... Thus the period is 24 (1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9).
MAPLE
MATHEMATICA
f[n_] := f[n] = Block[{s = PadLeft[{1}, n], c = 1}, s = t = Nest[g, s, n]; While[t = g[t]; s != t, c++]; c]; g[lst_List] := Rest@Append[lst, 1 + Mod[-1 + Plus @@ lst, 9]]; Do[ Print[{n, f[n] // Timing}], {n, 100}]
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Paolo P. Lava, Robert G. Wilson v, Jan 21 2013
EXTENSIONS
a(23) from Hans Havermann, Jan 30 2013
a(24) from Hans Havermann, Feb 18 2013
a(28) from Robert G. Wilson v, Feb 21 2013
a(29)-a(30) from Hiroaki Yamanouchi, May 04 2015
STATUS
approved