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The 35th prime number in the 'Fibonaccilike 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 Fibonaccilike 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 23 = 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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