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A120396
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a(1) is the least k such that p(1) = (k*17)^2 + k*17 - 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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4, 4, 1, 46, 51, 197, 216, 225, 366, 1862, 3806, 116
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OFFSET
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1,1
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COMMENTS
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The p(n) sequence starts 4691, 352106459, 123978958821625139, ...
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LINKS
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EXAMPLE
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a(1) = 4 as (4*17)^2 + 4*17 - 1 = 4691 = p(1) is prime.
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MATHEMATICA
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f[0] = {0, 17}; 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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