1,1

Below is a general expression that can be used as a starting point for finding similar formulas, and three examples.

Be forewarned that not every possibility will work - additional conditions may apply -but it is easy to see that there are no doubt actually an infinite number of algorisms much like Rowland's that will have hundreds or thousands of primes between the 1's before a composite is encountered.

1) initialize the integers x,k,a,b and choose f(x), f(k).

2) repeat indefinitely a two-step process:

x := x + 1,

If GCD( f(x) , f(x - 1) - a* f(k) ) > 1, then k := k + b;

Examples:

A) f(x) := 5*x^2 + 5*x + 1, f(k) = k, x = 1, k = 2, a = 10, b = 1;

the first 20 values of the sequence that do not equal one: 11, 31, 61, 101, 151, 211, 281, 19, 41, 29, 661, 11, 911,1051,1201,1361,1531,59,1901

B) f(x) := x^2 - x + 41, f(k) := k, x = 1, k = 2, a = 3 , b = 1;

the first 20 values of the sequence that do not equal one: 47, 227, 71, 359, 113, 563, 173, 839, 251,1187,347, 1607, 461,2099,593, 2663, 743,3299, 911,4007

C) f(x) := 5*x^2 + 5*x + 1, f(k) = x^2 - x + 41, x = 1, k = 2, a = 2, b = 1;

the first 20 values of the sequence that do not equal one: 11, 1979, 2549, 11,4691, 11, 8929, 29, 11, 22051, 41, 19, 48619, 61751, 11, 229, 11, 144779, 175141, 11

Eric S. Rowland, A simple prime-generating recurrence, Abstracts Amer. Math. Soc., 29 (No. 1, 2008), p. 50 (Abstract 1035-11-986).

1) initialize the integers x,k,a,b and choose f(x), f(k). 2) repeat indefinitely a two-step process: x := x + 1, If GCD( f(x) , f(x - 1) - a* f(k) ) > 1, then k := k + b; In the above formual let f(x) := x^2 - x + 41, f(k) := k, x = 1, k = 2, a = 3 , b = 1.

Cf. A084662, A137613.

Sequence in context: A141874 A142203 A067986 this_sequence A102090 A033226 A142946

Adjacent sequences: A141534 A141535 A141536 this_sequence A141538 A141539 A141540

nonn,uned

Aldrich Stevens (aldrichstevens(AT)msn.com), Aug 15 2008