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If n = 2^(2*m+1)+j, 0 <= j < 3*2^(2*m+1), then a(n) = 2^(3*m)+j.
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%I #9 Feb 20 2018 22:51:39

%S 1,2,3,4,5,6,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,

%T 27,28,29,30,31,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,

%U 82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108

%N If n = 2^(2*m+1)+j, 0 <= j < 3*2^(2*m+1), then a(n) = 2^(3*m)+j.

%C An easily computed sequence, growing faster than linearly, that provably contains infinitely many primes.

%H Robert Israel, <a href="/A298747/b298747.txt">Table of n, a(n) for n = 2..10000</a>

%H R. Israel, <a href="https://math.stackexchange.com/questions/2619773/whats-the-fastest-growing-function-known-to-contain-infinitely-many-primes/2619820#2619820">Answer to "What's the fastest growing function known to contain infinitely many primes?"</a>, Mathematics StackExchange.

%F G.f.: x^2/(1-x)^2 + Sum_{m>=1} (7*2^(3*m-3)-3*2^(2*m-1))*x^(2*4^m)/(1-x).

%p f:= proc(n) local m,j;

%p m:= floor(log[4](n/2));

%p 2^(3*m)+n - 2^(2*m+1)

%p end proc:

%p map(f, [$2..100]); # _Robert Israel_, Jan 25 2018

%K nonn,easy

%O 2,2

%A _Robert Israel_, Jan 25 2018