A298747 If n = 2^(2*m+1)+j, 0 <= j < 3*2^(2*m+1), then a(n) = 2^(3*m)+j.

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

Offset

2,2

Comments

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

Links

Robert Israel, Table of n, a(n) for n = 2..10000

R. Israel, Answer to "What's the fastest growing function known to contain infinitely many primes?", Mathematics StackExchange.

Formula

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).

Maple

f:= proc(n) local m, j;
  m:= floor(log[4](n/2));
  2^(3*m)+n - 2^(2*m+1)
end proc:
map(f, [$2..100]); # Robert Israel, Jan 25 2018

Crossrefs

Sequence in context: A194866 A345254 A004726 * A327105 A356451 A129618

Adjacent sequences: A298744 A298745 A298746 * A298748 A298749 A298750

Keyword

nonn,easy

Author

Robert Israel, Jan 25 2018

Status

approved