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+ The 35th prime number in the 'Fibonacci-like sequence' starting (3, 1). [Honaker, 2003]

Consider the 'Lucas sequence' starting with (1, 3). Which sequence will contains a greater number of primes as the number of terms approach infinity?

Let a(n) be the difference in the number of primes that occur between the Fibonacci-like sequences that start (3, 1) and (1, 3) respectively, up to the Nth prime. The sequence begins: 0, 0, 1, 0, -1, -1, -1, -1, 0, -1, -1, 0, 0, 0, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, ...

For example, a(5) = -1 because at the 5th prime, i.e., 11, there are 2 primes in the (3, 1) seq (i.e., 3, 5) and 3 primes in the (1, 3) seq (i.e., 3, 7, 11), hence 2-3 = -1.

(3, 1), 4, 5, 9, 14, 23, 37, 60, 97, 157, 254, 411, 665, 1076, ...

(1, 3), 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, ...

"One three, or not one three: that is the question." Act III, Scene I.




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