A339315 a(n) is the smallest number k such that k^2+1 divided by its largest prime factor is equal to F(2*n-1) for n > 0, or 0 if no such k exists, where F(n) is the Fibonacci sequence.

1, 3, 8, 34, 55, 144, 610, 233, 12166, 2584, 4181, 68260, 46368, 75025, 3917414, 464656, 1346269, 16349962

Offset: 1

Author

Michel Lagneau, Nov 30 2020

Keywords

• nonn
• hard
• more

Comments

a(n) is the smallest number k such that A248516(k) = A001519(n) for n > 0, or 0 if no such k exists, where A001519(n) = F(2*n-1) (bisection of the Fibonacci sequence), with F(n) = A000045(n).

We observe that a(2 + 3m) = A001519(1 + 3m) = A000045(1 + 6m) for m = 2, 3, 4, 5. For n = 6, this property no longer works.

For k > 0, a(3k - 1) is odd, a(3k) and a(3k+1) are even.

We observe that a(n)^2 + 1 is the product of two prime Fibonacci numbers for n = 2, 3, 4, 6, 7.

The first 18 terms of the sequence are Fibonacci numbers, except a(9), a(12), a(15), a(16) and a(18).

The corresponding sequence b(n) = (a(n)^2+1)/ A001519(n) is 2, 5, 13, 89, 89, 233, 1597, 89, 92681, 1597, 1597, 162593, 28657, 28657, 29842993, 160373, 514229. We observe that a majority of terms of b(n) are prime Fibonacci numbers, except b(9), b(12), b(15) and b(16).

Examples

			a(4) = 34 because 34^2 + 1 = 13*89 = 1157, and 1157/89 = 13 = A248516(34) = A001519(4).
A curiosity: a(22) = 1134903170 = F(45) with F(45)^2 + 1 = F(43)*F(47) where F(43) and F(47) are prime Fibonacci numbers.
		

Crossrefs

Cf. A000045, A001519, A001906, A002522, A005478, A014442, A245236, A245306, A248516, A281618.

Programs

Maple

with(numtheory):with(combinat,fibonacci):
nn:=100:n0:=20:
for n from 1 to n0 do:
  ii:=0:
  for m from 1 to 10^10 while(ii=0) do:
   x:=m^2+1:y:=factorset(x):n1:=nops(y):
   z:=x/y[n1]:
    if z = fibonacci(2*n-1)
     then
     ii:=1:printf(`%d %d \n`,n,m):
     else
    fi:
  od:
od:

PARI

a(n) = {my(k=1, f=fibonacci(2*n-1)); while ((k^2+1)/vecmax(factor(k^2+1)[,1]) != f, k++); k;} \\ Michel Marcus, Nov 30 2020