A339461 -id:A339461 - cp's OEIS frontend

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

Michel Lagneau, Dec 12 2020

nonn

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.

			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.

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

Cf. A000032, A000045, A001254, A339461.

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:

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

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

A339621 Sum of Fibonacci divisors of n^2 + 1.

Original entry on oeis.org

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

Offset: 0

Michel Lagneau, Dec 10 2020

nonn

A Fibonacci divisor of a number k is a Fibonacci number that divides k. (The divisor 1 is only counted once.)
For n < 2*10^5, the subsequence of primes begins by 3, 19, 37, 97, 103, 131, 139, 239, 241, 283, 359, 487, 631, ...
The Fibonacci numbers of the sequence are 1, 3, 8, 21, 55, 144, 377, ...
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.
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, ...

			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.

Michel Marcus, Table of n, a(n) for n = 0..10000

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

a:= n-> add(`if`(issqr(5*d^2+4) or issqr(5*d^2-4), d, 0)
, d=numtheory[divisors](n^2+1)):seq(a(n), n=0..100);

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

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || issquare(k-8);
a(n) = sumdiv(n^2+1, d, if (isfib(d), d)); \\ Michel Marcus, Dec 10 2020

a(A005574(n)) = 1 for n > 2.
a(n) = 3 when n^2 + 1 = 2*p, p prime and non-Fibonacci number.
a(n) = A005092(A002522(n)). - Michel Marcus, Aug 10 2022

A340542 Number of Fibonacci divisors of Fibonacci(n)^2 + 1.

Original entry on oeis.org

1, 2, 2, 2, 3, 3, 3, 4, 4, 3, 4, 4, 3, 5, 5, 3, 5, 5, 3, 5, 5, 3, 5, 6, 4, 5, 6, 4, 5, 5, 3, 5, 5, 5, 7, 5, 5, 7, 5, 3, 5, 5, 3, 7, 7, 3, 7, 8, 4, 5, 6, 4, 5, 7, 5, 5, 7, 5, 5, 5, 3, 7, 7, 5, 9, 7, 5, 7, 5, 3, 5, 5, 3, 7, 7, 5, 9, 7, 5, 8, 6, 3, 6, 8, 5, 5, 7

Offset: 0

Michel Lagneau, Jan 12 2021

nonn

A Fibonacci divisor of a number k is a Fibonacci number that divides k.
It is interesting to compare this sequence with A339669.
We observe that a(2n) = A339669(2n) if n = 5*k + 2 or n = 5*k + 3, with k >= 0, because Lucas(2n)^2 = 5*Fibonacci(2n)^2 + 4 (see A005248: all nonnegative integer solutions of the Pell equation a(n)^2 - 5*b(n)^2 = +4 together with b(n)= A001906(n), n>=0. - Wolfdieter Lang, Aug 31 2004).
So, Lucas(2n)^2 + 1 = 5*(Fibonacci(2n)^2 + 1). Lucas(2n)^2 + 1 and Fibonacci(2n)^2 + 1 have the same Fibonacci divisors for n = 5*k + 2 or n = 5*k + 3. For the other values of n = 5*k, 5*k + 1 or 5*k + 4, 5 is a Fibonacci divisor of Lucas(2n)^2 + 1 but not of Fibonacci(2n)^2 + 1. So for these last three values of n, a(2n) = A339669(2n) - 1 (except for m = 1 and 2, 5*F(m) is never a Fibonacci number).

			a(13) = 5 because the 5 Fibonacci divisors of Fibonacci(13)^2 + 1 = 233^2 + 1 are 1, 2, 5, 89 and 610.
a(16) = 5 because the 5 Fibonacci divisors of Fibonacci(16)^2 + 1 = 987^2 + 1 are 1, 2, 5, 610, and 1597.
Remark: the 5 Fibonacci divisors of Lucas(16)^2 + 1 = 2207^2 + 1 are 1, 2, 5, 610, and 1597, the index 16 = 2*8 with 8 of the form 5*k + 3.

Michel Marcus, Table of n, a(n) for n = 0..200

Cf. A000032, A000045, A005248, A010056, A339461, A339669.

Cf. A005086, A245306.

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

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8)); \\ A010056
a(n) = sumdiv(fibonacci(n)^2+1, d, isfib(d)); \\ Michel Marcus, Jan 12 2021

a(n) = A005086(A245306(n)). - Michel Marcus, Aug 10 2022

A352290 Numbers m such that the greatest prime factor of m^2 + 1 is a Fibonacci number.

Original entry on oeis.org

1, 2, 3, 5, 7, 8, 18, 34, 55, 57, 89, 123, 144, 233, 239, 322, 377, 411, 500, 568, 610, 746, 788, 843, 987, 1487, 1542, 1568, 1636, 2207, 2584, 2707, 3173, 3639, 3793, 3804, 3817, 4050, 4181, 4217, 4594, 4662, 5270, 5778, 6107, 6613, 8595, 8972, 10341, 10569

Offset: 1

Michel Lagneau, Mar 11 2022

nonn

A281618 is a subsequence.
The corresponding greatest prime Fibonacci factors of the sequence are 2, 5, 5, 13, 5, 13, 13, 89, 89, 13, 233, 89, 233, ...
The Fibonacci numbers of the sequence are 1, 2, 3, 5, 8, 34, 55, 89, 144, 233, 377, 610, 987, 2584, 4181, 10946, 17711, ... (subsequence of A000045).
The Lucas numbers of the sequence are 1, 2, 3, 7, 18, 123, 322, 843, 2207, 5778, 39603, 103682, ... (subsequence of A000032).
The prime numbers of the sequence are 2, 3, 5, 7, 89, 233, 239, 1487, 2207, 2707, 3793, 4217, 11789, 11981, 13763, ... including the prime Fibonacci numbers 2, 3, 5, 89, 233, 1066340417491710595814572169, ... (subsequence of A005478).

			18 is in the sequence because 18^2 + 1 = 5^2*13 and 13 is a Fibonacci number.

Cf. A000032, A000045, A002522, A005478, A014442, A248516, A338762, A339461.

q:= n-> (t-> ormap(issqr, [t+4, t-4]))(5*max(numtheory[factorset](n^2+1))^2):
select(q, [$1..12000])[];  # Alois P. Heinz, Mar 11 2022

With[{f = Fibonacci[Range[21]], m = f[[-1]]}, Select[Range[m], MemberQ[f, FactorInteger[#^2 + 1][[-1, 1]]] &]] (* Amiram Eldar, Mar 11 2022 *)

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
isok(m) = isfib(vecmax(factor(m^2+1)[,1])); \\ Michel Marcus, Mar 11 2022

A350707 Numbers m such that all prime factors of m^2+1 are Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 8, 18, 34, 57, 144, 239, 322, 610, 1134903170

Offset: 1

Michel Lagneau, Mar 27 2022

The Fibonacci numbers in the sequence include 1, 2, 3, 5, 8, 144, 610 and 1134903170.

The sequence includes terms of the form sqrt(f(n) - 1) and sqrt(5 * f(n) - 1), where f(n) = Fibonacci(A281087(n)) * Fibonacci(A281087(n)+2) = A140362(n). - Daniel Suteu, Mar 29 2022

			57 is in the sequence because 57^2+1 = 2*5^3*13 and 2, 5 and 13 are Fibonacci numbers;
1134903170 = Fibonacci(45) is in the sequence because 1134903170^2+1 = 433494437*2971215073 = Fibonacci(43)*Fibonacci(47).

The sequence contains A281618 and A285282.

Cf. A000032, A000045, A005478, A352290, A245236, A339461.

with(numtheory):
A005478:={2, 3, 5, 13, 89, 233, 1597, 28657, 514229, 433494437, 2971215073, 99194853094755497,1066340417491710595814572169, 19134702400093278081449423917}:
for n from 0 to 11000 do:
   y:=factorset(n^2+1):n0:=nops(y):
   if A005478 intersect y = y
       then
       print(n):
       else
     fi:
od:

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
isok(m) = my(f=factor(m^2+1)); for (i=1, #f~, if (!isfib(f[i,1]), return(0))); return(1); \\ Michel Marcus, Mar 29 2022

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

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A339621 Sum of Fibonacci divisors of n^2 + 1.

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A340542 Number of Fibonacci divisors of Fibonacci(n)^2 + 1.

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A352290 Numbers m such that the greatest prime factor of m^2 + 1 is a Fibonacci number.

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A350707 Numbers m such that all prime factors of m^2+1 are Fibonacci numbers.

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