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%I #19 Jan 11 2021 23:27:03
%S 2,2,3,1,3,2,3,2,5,1,5,2,4,2,5,1,5,2,4,2,6,1,6,2,4,2,6,1,6,2,4,2,6,1,
%T 7,2,5,2,6,1,6,2,4,2,7,1,7,2,5,2,7,1,6,2,5,2,7,1,6,2,4,2,8,1,9,2,5,2,
%U 6,1,6,2,4,2,7,1,9,2,6,2,7,1,7,2,5,2,7,1,6
%N Number of Fibonacci divisors of Lucas(n)^2 + 1.
%C Particular attention must be paid to the regularity properties of the number of divisors of Lucas(n)^2 + 1 observed for n < 156, when a(n) = 1 or 2. From this observation, we propose two conjectures verified for n < 156.
%C Conjecture 1: a(6*n+3) = 1.
%C Conjecture 2: a(6*n+1) = a(6*n+5) = 2.
%C The table in the links shows an array where terms are arranged in a table of 12 columns and 13 rows. We see the periods when a(n) = 1 and 2.
%H Michel Lagneau, <a href="/A339669/a339669.pdf">Table</a>
%e a(8) = 5 because the divisors of Lucas(8)^2 + 1 = 47^2 + 1 = 2210 are {1, 2, 5, 10, 13, 17, 26, 34, 65, 85, 130, 170, 221, 442, 1105, 2210} with 5 Fibonacci divisors: 1, 2, 5, 13 and 34.
%p with(combinat,fibonacci):nn:=100:F:={}:
%p Lucas:=n->2*fibonacci(n-1)+fibonacci(n):
%p for k from 0 to nn do:
%p F:=F union {fibonacci(k)}:
%p od:
%p for m from 0 to 90 do:
%p l:=Lucas(m)^2+1:d:=numtheory[divisors](l):n0:=nops(d):
%p lst:= F intersect d: n1:=nops(lst):printf(`%d, `,n1):
%p od:
%t Array[DivisorSum[LucasL[#]^2 + 1, 1 &, AnyTrue[Sqrt[5 #^2 + 4 {-1, 1}], IntegerQ] &] &, 89, 0] (* _Michael De Vlieger_, Dec 12 2020 *)
%o (PARI) a(n) = { my(l2 = 5*fibonacci(n)^2 + 4*(-1)^n + 1, k = 1, m = 2, res = 1, g); while(m <= l2, if(l2 % m == 0, res++); g = m; m += k; k = g; ); res } \\ _David A. Corneth_, Dec 12 2020
%Y Cf. A000032, A000045, A001254, A339461.
%K nonn
%O 0,1
%A _Michel Lagneau_, Dec 12 2020
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