

A078414


a(n) = (a(n1)+a(n2))/7^k, where 7^k is the highest power of 7 dividing a(n1)+a(n2).


8



1, 1, 2, 3, 5, 8, 13, 3, 16, 19, 5, 24, 29, 53, 82, 135, 31, 166, 197, 363, 80, 443, 523, 138, 661, 799, 1460, 2259, 3719, 122, 3841, 3963, 7804, 1681, 1355, 3036, 4391, 1061, 5452, 6513, 11965, 18478, 4349, 3261, 7610, 1553, 187, 1740, 1927, 3667, 5594, 27
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OFFSET

1,3


COMMENTS

If we consider Fibonaccilike numbers {F_p(n)} without positive multiples of p, where p is a fixed prime, then {F_2(n)} has period of length 1, {F_3(n)} has period of length 3, {F_5(n)} has period of length 6. This sequence is the first which does not have a trivial period and, probably, even is nonperiodic.
An open question: Is this sequence bounded?
Consider Fibonaccilike sequences without multiples of several primes, defined analogously: e.g., for {F_(p,q)(n)}, a(0)=0, a(1)=1, for n>=2, a(n)=a(n1)+a(n2) divided by the maximal possible powers of p and q.
Problem: For what sets of primes is the corresponding Fibonaccilike sequence without multiples of these primes periodic?
Examples: sequence {F_(7,11,13)(n)} has period of length 12: 0, 1, 1, 2, 3, 5, 8, 1, 9, 10, 19, 29, 48, 1, 1, 2, 3, 5,...; {F_(11,13,19)(n)} has period of length 9; {F_(13,19,23)(n)} has period of length 12; {F_(17,19,23,29)(n)} has period of length 15; {F_(19,23,31,53,59,89)(n)} has period of length 24; {F_(23,29,73,233)(n)} has period of length 18.
Don Reble noted that lengths of all such periods could only be multiples of 3 because every Fibonaccilike sequence considered here modulo 2 has the form 0,1,1,0,1,1,... .
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LINKS



FORMULA



MAPLE

a:= proc(n) option remember; local t, j;
if n<3 then 1
else t:= a(n1)+a(n2);
while irem(t, 7, 'j')=0 do t:=j od; t
fi
end:


MATHEMATICA

nxt[{a_, b_}]:=Module[{n=IntegerExponent[a+b, 7]}, {b, (a+b)/7^n}]; Transpose[ NestList[nxt, {1, 1}, 60]][[1]] (* Harvey P. Dale, Jul 23 2012 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



