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A120392
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a(1) is the least k such that p(1) = (k*3)^2 + k*3 - 1 is prime, then a(n+1) is the least k such that (k*p(n))^2 + k*p(n) - 1 = p(n+1) is prime.
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3
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1, 1, 1, 1, 13, 1, 101, 130, 109, 418, 388, 876, 5011, 11529
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OFFSET
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1,5
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COMMENTS
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The p(n) sequence starts 11, 131, 17291, 298995971, 15108361827832297751, ...
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LINKS
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EXAMPLE
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a(1) = 1 as 3^2 + 3 - 1 = 11 = p(1) is prime.
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MATHEMATICA
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f[0] = {0, 3}; f[n_] := f[n] = Module[{k = 1, p = f[n - 1][[2]]}, While[! PrimeQ[(k*p)^2 + k*p - 1], k++]; {k, (k*p)^2 + k*p - 1}]; Table[f[n][[1]], {n, 1, 10}] (* Amiram Eldar, Aug 28 2021 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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