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A073799
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Numbers that begin a run of consecutive integers k such that PrimePi(k) divides 2^k.
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3
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2, 7, 19, 53, 131, 311, 719, 1619, 3671, 8161, 17863, 38873, 84017, 180503, 386093, 821641, 1742537, 3681131, 7754077, 16290047, 34136029, 71378569, 148948139, 310248241, 645155197, 1339484197, 2777105129, 5750079047, 11891268401, 24563311309, 50685770167, 104484802057, 215187847711
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OFFSET
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1,1
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COMMENTS
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It seems that each term is a bit larger than twice the previous one.
Runs have lengths 3, 4, 4, 6, 6, 2, 8, 2, 2, 6, 18, 18, 30, 8, 24, 6, 2, 18, ..., respectively.
Theorem: a(1) = 2 and a(n) = A033844(n) for n > 1. For n > 1, the length of the n-th run is prime(2^n+1)-prime(2^n) = A051439(n)-A033844(n) = A074325(n).
Proof: Let r > 1. If p = prime(2^r), then primepi(p) = 2^r.
primepi(p-1) = 2^r - 1. Since r > 1, 2^r - 1 > 2 and odd and thus does not divide any power of 2.
In addition 2^r < p and thus divides 2^p. This means that p is a term. Let q be such that p < q < prime(2^r+1). Then primepi(q) = 2^r and divides 2^q. Since primepi(q-1) = 2^r and divides 2^(q-1), this means that q does not start a run and thus is not a term.
Let w be such that prime(2^r+1) <= w < prime(2^(r+1)). Then 2^r + 1 <= primepi(w) < 2^(r+1) and does not divide any power of 2. This means that w is not a term.
(End)
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LINKS
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FORMULA
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Solutions to 2^(x-1) mod PrimePi(x-1) > 0 but 2^x mod PrimePi(x) = 0.
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MATHEMATICA
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aQ[k_] := Divisible[2^k, PrimePi[k]]; s = {}; len = {}; n = 2; While[Length[s] < 10, While[! aQ[n], n++]; n1 = n; While[aQ[n], n++]; If[n > n1, AppendTo[s, n1]; AppendTo[len, n - n1]]; n++]; s (* Amiram Eldar, Dec 11 2018 *)
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PROG
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(Python)
from sympy import prime
return 2 if n == 1 else prime(2**n) # Chai Wah Wu, Jan 27 2020
(PARI) a(n) = if(n==1, 2, prime(2^n)); \\ Jinyuan Wang, Mar 01 2020
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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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