%I
%S 1,3,6,4,6,9,8,5,9,13,20,9,10,8,6,10,53,9,48,28,18,20,35,18,76,10,9,8,
%T 7,68,20,15,20,53,30,9,58,48,78,28,19,18,63,20,68,35,28,18,46,108,76,
%U 10,158,9,52,8,87,133,18,68,51,20,46,35,78,20,17,138,35,30,230,20,72,58,76
%N Let T(n) = A000078(n+2), n >= 1; a(n) = smallest k such that n divides T(k).
%C Rank of apparition of n in the tetranacci numbers.  _T. D. Noe_, Dec 05 2005
%C This sequence is welldefined. Proof by _T. D. Noe_: for every prime p, Brenner proves we can find k(p) such that p divides the k(p)th term of nstep Fibonacci. Using Brenner's methods, we know that p will also divide every j*k(p)th term of the sequence for any j>0. We use this last fact to go to the general case: For integer m, we can find a term that m divides as follows: (1) factor m into primes: m = p1^e1 p2^e2...pr^er, (2) let K = m k(p1) k(p2)...k(pr) / (p1 p2 ... pr) (3) then m will divide the Kth term of the sequence. In general, K is much too large. However, it does show that every prime divides a term of every nstep Fibonacci sequence for n>1.  _T. D. Noe_, Dec 05 2005
%H J. L. Brenner, <a href="http://www.jstor.org/stable/2307216">Linear Recurrence Relations</a>, Amer. Math. Monthly, Vol. 61 (1954), 171173.
%H T. D. Noe and J. V. Post, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Noe/noe5.html">Primes in Fibonacci nstep and Lucas nStep Sequences</a>, J. Integer Seq. 8, Article 05.4.4, 2005.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TetranacciNumber.html">Tetranacci Number</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonaccinStepNumber.html">Fibonacci nStep Number</a>.
%F a(n) = Min{k: n  A000078(k)}.
%t n=4; Table[a=Join[{1}, Table[0, {n1}]]; k=0; While[k++; s=Mod[Plus@@a, i]; a=RotateLeft[a]; a[[n]]=s; s!=0]; k, {i, 100}] (* _T. D. Noe_, Dec 05 2005 *)
%Y Cf. A000078, A112269, A112305.
%K easy,nonn
%O 1,2
%A _Jonathan Vos Post_, Dec 02 2005
%E Corrected by _T. D. Noe_, Dec 05 2005
