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

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

Offset: 1

Author

Michel Lagneau, Mar 27 2022

Keywords

• nonn
• hard
• more

Comments

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

Examples

			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).
		

Crossrefs

The sequence contains A281618 and A285282.

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

Programs

Maple

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:

PARI

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