cp's OEIS Frontend

A266587 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).

%I A266587 #20 Feb 27 2023 05:10:24
%S A266587 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 A266587 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 A266587 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 A266587 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 A266587 These a(n) values may be called Lucas "entry points".
%C A266587 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 A266587 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 A266587 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 A266587 For each value of a(n) > 0 there is an 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 A266587 Conjecture: Infinite Lucas subsequences divisible by the powers of odd primes, p(n), for which a(n) > 0, are given by:
%C A266587 Lucas(a(n) + a(n)*(p(n)-1)*(Sum_{j=1..m-1} p(n)^(j-1)) + 2*i*a(n)*p(n)^(m-1)) is divisible by p(n)^m, where p(n) > 2, i >= 0, m > 1.
%C A266587 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 A266587 The divisibility of Lucas numbers by powers of 2 is 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 A266587 For prime(10) = 29, we get a(10) = 7, because Lucas(7)= 29 is the first Lucas number divisible by 29. Also note 7 = (29-1)/4.
%e A266587 For prime(11) = 31, we get a(11) = 15, because Lucas(15) = 1364 is the first Lucas number divisible by 31. Also note 15 = (31-1)/2.
%t A266587 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 A266587 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 A266587 Cf. A000032, A001602, A053028.
%Y A266587 Essentially the same as A194363.
%K A266587 nonn
%O A266587 1,2
%A A266587 _Richard R. Forberg_, Jan 01 2016