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A307486
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a(0) = 3; a(n) = smallest k > 1 such that 1 + a(0)*a(1)*...*a(n-1)*k is composite.
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0
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3, 3, 3, 2, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
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OFFSET
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0,1
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COMMENTS
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a(n) = 4 for n = 4 and 39; a(n) = 3 for n = 0, 1, 2, 5, 6, 7, 14, 20, 25, 56, 90, 119, 316, 330, 1268, 1604, 1805, 1880, 1984, 2950, 3386, 3712, 4532, 4874, 8968, 18178, 19454, 23272, 45617, 51980, 52780, 60552, ...; a(n) = 2 for others n < 60552. Data from Amiram Eldar.
Similar tails have sequences with other initial terms that are natural numbers.
Conjecture: for any initial term a(0) > 0, a(n) > 3 only for finitely many n >= 0.
The question is how to prove that all these sequences are bounded, so bounded?
It seems that a(0) = 21 is the smallest initial term such that a(n) = 2 or 3 for every n > 0.
Note that if a(0) is a Sierpinski number A076336, then a(n) = 2 for every n > 0.
Carl Pomerance (in a letter to the author) wrote: Since the sequence a(0)a(1)...a(n-1) grows geometrically, the chance that any given k will give a prime is about 1/n, so the chance that both k = 2 and k = 3 will give primes is about 1/n^2, heuristically, which has a convergent sum. Thus, for this reason I would agree it's reasonable to conjecture that for any starting a(0), the sequence will eventually be made up of 2's and 3's, with 2's predominating. The number of 3's among the first n terms should be proportional to log n.
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LINKS
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MATHEMATICA
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a[0] = 3; a[n_] := a[n] = Module[{k = 2, p = Product[a[i], {i, 0, n - 1}]}, While[PrimeQ[1 + p*k], k++]; k]; Array[a, 100, 0] (* Amiram Eldar, Apr 10 2019 *)
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PROG
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(PARI) p=1; for (n=0, 100, for (k=2, oo, if (!isprime(1+p*k), print1 (k", "); p*=k; break))) \\ Rémy Sigrist, Apr 23 2019
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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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