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%I #12 Oct 07 2017 07:35:42
%S 4,8,15,6,9,7,29,19,30,78,15,20,36,83,30,34,65,69,101,133,32,19,271,
%T 110,20,187,14,185,106,173,587,80,12,35,11,224,72,38,42,315,101,26,73,
%U 172,383,27,84,362,35,250,37,29,507,305,55,38,178,332,62,537,778,459,31,124
%N Least index k such that the n-th prime divides the k-th tribonacci number.
%C The tribonacci numbers are indexed so that trib(0) = trib(1) = 0, trib(2) = 1, for n>2: trib(n) = trib(n-1) + trib(n-2) + trib(n-3). See A112618 for another version.
%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 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">MathWorld: Tribonacci Number</a>
%F a(n) = minimum k such that prime(n) | A000073(k) and A000073(k) >= prime(n). a(n) = minimum k such that A000040(n) | A000073(k) and A000073(k) >= A000040(n).
%e a(1) = 4 because prime(1) = 2 and tribonacci( 4) = 2.
%e a(2) = 8 because prime(2) = 3 and tribonacci( 8) = 24 = 3 * 2^3.
%e a(3) = 15 because prime(3) = 5 and tribonacci(15) = 1705 = 5 *(11 * 31).
%e a(4) = 6 because prime(4) = 7 and tribonacci( 6) = 7.
%e a(5) = 9 because prime(5) = 11 and tribonacci( 9) = 44 = 11 * 4.
%e a(6) = 7 because prime(6) = 13 and tribonacci( 7) = 13.
%e a(7) = 29 because prime(7) = 17 and tribonacci(29) = 8646064 = 17 *(2^4 * 7 * 19 * 239).
%t a[0] = a[1] = 0; 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 Cf. A000040, A000045, A000073, A000204, A001644, A053028, A106299, A112312.
%Y Cf. also A112618 = this sequence minus 1.
%K easy,nonn
%O 1,1
%A _Jonathan Vos Post_, Nov 29 2005
%E Corrected and extended by _Robert G. Wilson v_, Dec 01 2005
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