A339621 - OEIS

%I #31 Aug 10 2022 07:57:22

%S 1,3,6,8,1,16,1,8,19,3,1,3,6,42,1,3,1,8,19,3,1,50,6,8,1,3,1,8,6,3,1,

%T 16,6,8,103,3,1,8,6,3,1,3,6,8,14,3,1,55,6,3,1,3,6,8,1,126,1,21,6,3,14,

%U 3,6,8,1,3,1,8,6,3,391,3,6,21,1,3,1,8,6,3,1,37

%N Sum of Fibonacci divisors of n^2 + 1.

%C A Fibonacci divisor of a number k is a Fibonacci number that divides k. (The divisor 1 is only counted once.)

%C For n < 2*10^5, the subsequence of primes begins by 3, 19, 37, 97, 103, 131, 139, 239, 241, 283, 359, 487, 631, ...

%C The Fibonacci numbers of the sequence are 1, 3, 8, 21, 55, 144, 377, ...

%C Conjecture: If the sum of the Fibonacci divisors of m^2 + 1 is a Fibonacci number, then this number belongs to the sequence A001906(n) = F(2n) where F(n) is the Fibonacci sequence.

%C The sequence giving the least k such that the sum of Fibonacci divisors of k^2 + 1 is equal to F(2*n) for n > 0 begins with: 0, 1, 3, 57, 47, 15007, 1679553, ...

%H Michel Marcus, <a href="/A339621/b339621.txt">Table of n, a(n) for n = 0..10000</a>

%F a(A005574(n)) = 1 for n > 2.

%F a(n) = 3 when n^2 + 1 = 2*p, p prime and non-Fibonacci number.

%F a(n) = A005092(A002522(n)). - _Michel Marcus_, Aug 10 2022

%e a(3) = 8 because the divisors of 3^2 + 1 = 10 are {1, 2, 5, 10}, and the sum of the Fibonacci divisors is 1 + 2 + 5 = 8.

%p a:= n-> add(`if`(issqr(5*d^2+4) or issqr(5*d^2-4), d, 0)

%p , d=numtheory[divisors](n^2+1)):seq(a(n), n=0..100);

%t Array[DivisorSum[#^2 + 1, # &, AnyTrue[Sqrt[5 #^2 + 4 {-1, 1}], IntegerQ] &] &, 82, 0] (* _Michael De Vlieger_, Dec 10 2020 *)

%o (PARI) isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || issquare(k-8);

%o a(n) = sumdiv(n^2+1, d, if (isfib(d), d)); \\ _Michel Marcus_, Dec 10 2020

%Y Cf. A000045, A001906, A002522, A005092, A005574, A339461.

%K nonn

%O 0,2

%A _Michel Lagneau_, Dec 10 2020