A178942 -id:A178942 - cp's OEIS frontend

A179921 a(n) = prime(n) if n<=3; for n>3, a(n) is the smallest prime >a(n-1), such that the denominator of fraction (a(n-1)-a(n-2))/(a(n)-a(n-1)) did not appear earlier.

2, 3, 5, 7, 13, 23, 31, 53, 67, 79, 113, 131, 151, 193, 233, 271, 307, 353, 379, 409, 457, 557, 613, 691, 761, 809, 883, 907, 1013, 1069, 1123, 1181, 1213, 1279, 1361, 1423, 1483, 1571, 1657, 1709, 1811, 1933, 1997, 2087, 2179, 2273, 2341, 2459

Offset: 1

Vladimir Shevelev, Jan 12 2011

nonn

Using Dirichlet's theorem on arithmetic progressions, it is easy to prove that the sequence is infinite. The sequence of the corresponding denominators begins with 2,1,3,5,4,11,7,6,17, ...

			The first four terms 2,3,5,13 give three denominators: 2,1,3. Then a(5) is not in {17, 19}, since (13-5)/(17-13) = 2/1, (13-5)/(19-13) = 4/3 and denominators 1 and 3 already appeared earlier. Since (13-5)/(23-13) = 4/5 and 5 is not yet in the denominator sequence, a(5) = 23.

Cf. A168253, A178942, A179210, A179234, A179256, A179328.

Edited by Alois P. Heinz, Jan 12 2011

cp's OEIS Frontend

A179921 a(n) = prime(n) if n<=3; for n>3, a(n) is the smallest prime >a(n-1), such that the denominator of fraction (a(n-1)-a(n-2))/(a(n)-a(n-1)) did not appear earlier.

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