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

Author

T. D. Noe, Dec 05 2005

Keywords

• nonn

Comments

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.

Examples

			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.
		

Links

• 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

Crossrefs

Equals A112312(n)-1.

Programs

Mathematica

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 *)

Formula

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