A339669 Number of Fibonacci divisors of Lucas(n)^2 + 1.

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, 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, 6, 1, 6, 2, 4, 2, 7, 1, 9, 2, 6, 2, 7, 1, 7, 2, 5, 2, 7, 1, 6

Offset: 0

Author

Michel Lagneau, Dec 12 2020

Keywords

• nonn

Comments

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.

Conjecture 1: a(6*n+3) = 1.

Conjecture 2: a(6*n+1) = a(6*n+5) = 2.

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.

Examples

			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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..17800
• Michel Lagneau, Table

Crossrefs

Cf. A000032, A000045, A001254, A339461.

Programs

Maple

with(combinat,fibonacci):nn:=100:F:={}:
Lucas:=n->2*fibonacci(n-1)+fibonacci(n):
for k from 0 to nn do:
  F:=F union {fibonacci(k)}:
od:
   for m from 0 to 90 do:
    l:=Lucas(m)^2+1:d:=numtheory[divisors](l):n0:=nops(d):
    lst:= F intersect d: n1:=nops(lst):printf(`%d, `,n1):
   od:

Mathematica

Array[DivisorSum[LucasL[#]^2 + 1, 1 &, AnyTrue[Sqrt[5 #^2 + 4 {-1, 1}], IntegerQ] &] &, 89, 0] (* Michael De Vlieger, Dec 12 2020 *)

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