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A179206
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Smallest index k such that prime(k)*2^n * (prime(k)*2^n + 1) - 1 is prime.
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1
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1, 1, 1, 2, 1, 4, 3, 8, 1, 5, 5, 6, 6, 5, 1, 10, 16, 3, 13, 13, 37, 7, 1, 11, 1, 8, 18, 17, 6, 5, 15, 24, 12, 1, 16, 24, 7, 3, 9, 6, 17, 3, 15, 54, 24, 5, 16, 14, 8, 32, 1, 23, 17, 1, 62, 16, 23, 94, 65, 53, 70, 20, 9, 10, 9, 50, 12, 8, 12, 19, 155, 6, 46, 12, 2, 2, 12, 72, 20, 9
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OFFSET
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1,4
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COMMENTS
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Define partial sums S(N) = Sum_{n=1..N} n and T(N) = Sum_{n=1..N} a(n). Then lim_{N->infinity} T(N)/S(N) -> approx 0.435.
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LINKS
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EXAMPLE
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n=1: 2*2*(2*2 + 1) - 1 = 19 prime so k=1 as 2=prime(1).
n=2: 2*2^2*(2*2^2 + 1) - 1 = 71 prime so k=1 as 2=prime(1).
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MAPLE
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A179206 := proc(n) local k, pk; for k from 1 do pk := ithprime(k)*2^n ; if isprime(pk*(pk+1)-1) then return k; end if; end do: end proc:
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MATHEMATICA
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sik[n_]:=Module[{n2=2^n, k=1}, While[CompositeQ[(Prime[k]n2)(Prime[ k]n2+1)-1], k++]; k]; Array[sik, 80] (* Harvey P. Dale, Dec 07 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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