OFFSET
1,1
COMMENTS
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
REFERENCES
Eric S. Rowland, A simple prime-generating recurrence, Abstracts Amer. Math. Soc., 29 (No. 1, 2008), p. 50 (Abstract 1035-11-986).
FORMULA
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.
CROSSREFS
KEYWORD
nonn,uned
AUTHOR
Aldrich Stevens (aldrichstevens(AT)msn.com), Aug 15 2008
STATUS
approved