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%I #30 Jan 14 2023 10:51:15
%S 0,1,2,3,5,7,8,18,34,57,144,239,322,610,1134903170
%N Numbers m such that all prime factors of m^2+1 are Fibonacci numbers.
%C The Fibonacci numbers in the sequence include 1, 2, 3, 5, 8, 144, 610 and 1134903170.
%C 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
%e 57 is in the sequence because 57^2+1 = 2*5^3*13 and 2, 5 and 13 are Fibonacci numbers;
%e 1134903170 = Fibonacci(45) is in the sequence because 1134903170^2+1 = 433494437*2971215073 = Fibonacci(43)*Fibonacci(47).
%p with(numtheory):
%p A005478:={2, 3, 5, 13, 89, 233, 1597, 28657, 514229, 433494437, 2971215073, 99194853094755497,1066340417491710595814572169, 19134702400093278081449423917}:
%p for n from 0 to 11000 do:
%p y:=factorset(n^2+1):n0:=nops(y):
%p if A005478 intersect y = y
%p then
%p print(n):
%p else
%p fi:
%p od:
%o (PARI) isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
%o 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
%Y The sequence contains A281618 and A285282.
%Y Cf. A000032, A000045, A005478, A352290, A245236, A339461.
%K nonn,hard,more
%O 1,3
%A _Michel Lagneau_, Mar 27 2022