%I
%S 0,2,0,4,5,0,0,9,12,7,15,0,10,22,8,0,29,0,34,35,0,39,42,0,0,25,52,18,
%T 0,0,64,65,0,23,0,25,0,82,84,0,89,45,95,0,0,11,21,112,114,57,0,119,60,
%U 125,0,44,0,135,0,14,142,0,22,155,0,0,55,0,58,87,0,179,184,0,189,192,0,0,50,102,209,0,215,0,219,222,112,0,23,232,234,239,244,245
%N Smallest index of a Lucas number (A000032) that is divisible by prime(n), if it exists, or 0 if it does not exist (for n > 1).
%C These a(n) values may be called Lucas "entry points".
%C a(1)=0, corresponding to prime(1)= 2, must be viewed as a special case here, because 0, in this case only, is the correct smallest index.
%C For all other zeros in a(n) (e.g., for a(3), a(6), a(7) corresponding to prime(3)=5, prime(6)=13, prime(7)=17), there are no Lucas numbers divisible by these primes. The full set of primes not divisible into any Lucas number are given by A053028.
%C For a(n)>0, a(n) is always equal to either (prime(n)1)/k or (prime(n)+1)/k, for some k>=1. This is similar to the Fibonacci entry points given by A001602.
%C For each value of a(n)>0 there is infinite, periodic subsequence within the Lucas numbers divisible by prime(n), given by Lucas(a(n)+2*i*a(n)), for all i>=0. Again this is analogous to the A001602 for Fibonacci. The periodicity of the subsequence is 2*a(n), twice the entry point vs. equal to the entry point for Fibonacci.
%C Conjecture: Infinite Lucas subsequences divisible by the powers of odd primes, p(n), for which a(n)>0, are given by:
%C Lucas(a(n) + a(n)*(p(n)1)*Sum[j=1..m1, p(n)^(j1)] + 2*i*a(n)*p(n)^(m1)) is divisible by p(n)^m, where p(n)>2, i>=0, m>1.
%C Note: the formula above also works for m=1 if the "Sum" is assumed to be zero when the upper summation index limit is less than initial summation index. See second Mathematica example below which works in this way for all m>=1, to demonstrate the rule for p = p(n) with corresponding a = a(n).
%C The divisibility of Lucas numbers by powers of 2 are limited to 2^1 and 2^2, as follows: Lucas(3*i) is divisible by 2 and Lucas(3+6i) is divisible by 4, for all i>=0.
%e For prime(10) = 29, we get a(10) = 7, because Lucas(7)= 29 is the first Lucas number divisible by 29. Also note 7 = (291)/4.
%e For prime(11) = 31, we get a(11) = 15, because Lucas(15) = 1364 is the first Lucas number divisible by 31. Also note 15 = (311)/2.
%t result={}; Do[iresult=0; Do[If[Divisible[LucasL[i], Prime[k]], iresult=i; Break[]], {i, 1, 2000}]; AppendTo[result, iresult], {k, 2, 200}]; result
%t p = 23; a = 12; m = 4; Table[Divisible[LucasL[a + a*(p  1)*Sum[p^(j  1), {j, 1, m  1}] + 2a*i*p^(m  1)], p^m], {i, 1, 100}]
%Y Cf. A000032, A001602, A053028.
%K nonn
%O 1,2
%A _Richard R. Forberg_, Jan 01 2016
