OFFSET
1,1
COMMENTS
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.
From Chai Wah Wu, Jan 27 2020: (Start)
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)
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..57
FORMULA
Solutions to 2^(x-1) mod PrimePi(x-1) > 0 but 2^x mod PrimePi(x) = 0.
a(n) = A033844(n) for n > 1. - Chai Wah Wu, Jan 27 2020
MATHEMATICA
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 *)
PROG
(Python)
from sympy import prime
def A073799(n):
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Aug 12 2002
EXTENSIONS
Edited by Jon E. Schoenfield, Dec 10 2018
a(15)-a(18) from Amiram Eldar, Dec 11 2018
a(19)-a(33) from Chai Wah Wu, Jan 27 2020
STATUS
approved