A112305 -id:A112305 - cp's OEIS frontend

A112618 Let T(n) = A000073(n+1), n >= 1; a(n) = smallest k such that prime(n) divides T(k).

3, 7, 14, 5, 8, 6, 28, 18, 29, 77, 14, 19, 35, 82, 29, 33, 64, 68, 100, 132, 31, 18, 270, 109, 19, 186, 13, 184, 105, 172, 586, 79, 11, 34, 10, 223, 71, 37, 41, 314, 100, 25, 72, 171, 382, 26, 83, 361, 34, 249, 36, 28, 506, 304, 54, 37, 177, 331, 61, 536, 777, 458, 30, 123

Offset: 1

T. D. Noe, Dec 05 2005

nonn

Brenner proves that every prime divides some tribonacci number T(n). For the similar 3-step Lucas sequence A001644, there are primes (A106299) that do not divide any term.

			Sequence T(n) starts 1,1,2,4,7,13,24,44. For the primes 2,3,7,11,13, it is easy to see that a(1)=3, a(2)=7, a(4)=5, a(5)=8, a(6)=6.

T. D. Noe, Table of n, a(n) for n=1..1000
J. L. Brenner, Linear Recurrence Relations, Amer. Math. Monthly, Vol. 61 (1954), 171-173.
Eric Weisstein's World of Mathematics, Tribonacci Number

Equals A112312(n)-1.

a[0]=0; a[1]=a[2]=1; a[n_]:=a[n]=a[n-1]+a[n-2]+a[n-3]; f[n_]:= Module[{k=2, p=Prime[n]}, While[Mod[a[k], p] != 0, k++ ]; k]; Array[f, 64] (* Robert G. Wilson v *)

a(n) = A112305(prime(n)).

A112374 Let T(n) = A000078(n+2), n >= 1; a(n) = smallest k such that n divides T(k).

Original entry on oeis.org

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, 7, 68, 20, 15, 20, 53, 30, 9, 58, 48, 78, 28, 19, 18, 63, 20, 68, 35, 28, 18, 46, 108, 76, 10, 158, 9, 52, 8, 87, 133, 18, 68, 51, 20, 46, 35, 78, 20, 17, 138, 35, 30, 230, 20, 72, 58, 76

Offset: 1

Jonathan Vos Post, Dec 02 2005

Rank of apparition of n in the tetranacci numbers. - T. D. Noe, Dec 05 2005

This sequence is well-defined. 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 n-step 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 K-th term of the sequence. In general, K is much too large. However, it does show that every prime divides a term of every n-step Fibonacci sequence for n>1. - T. D. Noe, Dec 05 2005

J. L. Brenner, Linear Recurrence Relations, Amer. Math. Monthly, Vol. 61 (1954), 171-173.
T. D. Noe and J. V. Post, Primes in Fibonacci n-step and Lucas n-Step Sequences, J. Integer Seq. 8, Article 05.4.4, 2005.
Eric Weisstein's World of Mathematics, Tetranacci Number.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.

Cf. A000078, A112269, A112305.

n=4; Table[a=Join[{1}, Table[0, {n-1}]]; 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 *)

a(n) = Min{k: n | A000078(k)}.

Corrected by T. D. Noe, Dec 05 2005

cp's OEIS Frontend

A112618 Let T(n) = A000073(n+1), n >= 1; a(n) = smallest k such that prime(n) divides T(k).

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