A078414 a(n) = (a(n-1)+a(n-2))/7^k, where 7^k is the highest power of 7 dividing a(n-1)+a(n-2).

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

Offset: 1

Author

Yasutoshi Kohmoto, Dec 28 2002

Keywords

• nonn
• easy

Comments

From Vladimir Shevelev, Apr 01 2013; edited by Danny Rorabaugh, Feb 19 2016: (Start)

If we consider Fibonacci-like 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 non-periodic.

An open question: Is this sequence bounded?

Consider Fibonacci-like 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(n-1)+a(n-2) divided by the maximal possible powers of p and q.

Problem: For what sets of primes is the corresponding Fibonacci-like 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 Fibonacci-like sequence considered here modulo 2 has the form 0,1,1,0,1,1,... .

(End)

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..4000
• B. Avila, T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5.

Crossrefs

Cf. A000045, A078412, A214094, A214156, A216231, A216275, A216835.

Programs

Maple

a:= proc(n) option remember; local t, j;
      if n<3 then 1
    else t:= a(n-1)+a(n-2);
         while irem(t, 7, 'j')=0 do t:=j od; t
      fi
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jul 25 2012

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 *)

Formula

a(n) = A242603(a(n-1)+a(n-2)). - R. J. Mathar, Mar 13 2024

Extensions

Corrected by Harvey P. Dale, Jul 23 2012