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a(n) = least prime > (5/3)*a(n - 1)*a(n - 3)/a(n - 2), with a(1) = 2, a(2) = 3, a(3) = 5.
1

%I #4 Nov 17 2024 07:32:10

%S 2,3,5,7,11,17,19,23,37,53,59,71,107,149,167,211,317,419,467,593,887,

%T 1171,1307,1657,2477,3257,3637,4621,6899,9059,10133,12889,19207,25169,

%U 28151,35809,53377,69941,78203,99487,148301,194309,217253,276359,411967

%N a(n) = least prime > (5/3)*a(n - 1)*a(n - 3)/a(n - 2), with a(1) = 2, a(2) = 3, a(3) = 5.

%C If the recurrence is generalized to a(n) > r*a(n - 1)*a(n - 3)/b(n - 2), then then 5/3 is the least value of r for which (a(n)) is strictly increasing.

%t {a[1], a[2], a[3]} = {2, 3, 5}; r = 5/3;

%t a[n_] := a[n] = NextPrime[r*a[n - 1] a[n - 3]/a[n - 2]];

%t Table[a[n], {n, 1, 100}]

%Y Cf. A000040, A377543.

%K nonn

%O 1,1

%A _Clark Kimberling_, Nov 13 2024