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%I #14 Oct 03 2017 03:12:41
%S 3,7,14,5,8,6,28,18,29,77,14,19,35,82,29,33,64,68,100,132,31,18,270,
%T 109,19,186,13,184,105,172,586,79,11,34,10,223,71,37,41,314,100,25,72,
%U 171,382,26,83,361,34,249,36,28,506,304,54,37,177,331,61,536,777,458,30,123
%N Let T(n) = A000073(n+1), n >= 1; a(n) = smallest k such that prime(n) divides T(k).
%C 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.
%H T. D. Noe, <a href="/A112618/b112618.txt">Table of n, a(n) for n=1..1000</a>
%H J. L. Brenner, <a href="http://www.jstor.org/stable/2307216">Linear Recurrence Relations</a>, Amer. Math. Monthly, Vol. 61 (1954), 171-173.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TribonacciNumber.html">Tribonacci Number</a>
%F a(n) = A112305(prime(n)).
%e 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 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_ *)
%Y Equals A112312(n)-1.
%K nonn
%O 1,1
%A _T. D. Noe_, Dec 05 2005
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