OFFSET
0,2
COMMENTS
a(n) = number of products of half-odd-primes <= 2^n. E.g., a(2) = 7 since 1, 3/2, (3/2)^2, (3/2)^3, (3/2)*(5/2), 5/2, 7/2 are all <= 2^2. - David W. Wilson
m is sufficiently large precisely when 2^m > 3^(m-n), i.e., when m >= floor(n*log(3)/log(1.5)) = A117630(n+1) = A126281(n) for n > 1. (Robert G. Wilson v asks if this conjecture holds in a comment to A126281.) - David A. Corneth, Apr 09 2015
From Robert G. Wilson v, Apr 13 2020: (Start)
This sequence shows a sufficiently large row of A126279 read backwards or a sufficiently large column of A126279 read vertically.
log(y) ~ a + b*x + c*x^2, where a=1.1422, b=0.7419, and c=-0.00035, with an r^2 of 1.0. (End)
[But what is y? - Editors, Jun 15 2021]
LINKS
Martin Raab, Table of n, a(n) for n = 0..41 (Terms 0..35 from Robert G. Wilson v)
EXAMPLE
Between 1 and 2^m there is just one number with m prime factors, namely 2^m, so a(0) = 1.
For m >= 3, up to 2^m there are 2 numbers with m-1 prime factors, 2^(m-1) and 3*2^(m-2), so a(1) = 2.
MATHEMATICA
AlmostPrimePi[k_Integer, n_] := Module[{a, i}, a[0] = 1; If[k == 1, PrimePi[n], Sum[PrimePi[n/Times @@ Prime[ Array[a, k - 1]]] - a[k - 1] + 1, Evaluate[ Sequence @@ Table[{a[i], a[i - 1], PrimePi[(n/Times @@ Prime[Array[a, i - 1]])^(1/(k - i + 1))]}, {i, k - 1}]] ]]]; (* Eric W. Weisstein, Feb 07 2006 *)
Table[ AlmostPrimePi[Floor[n(1 + 1/Sqrt@2)] + 2, 2^(n + Floor[n(1 + 1/Sqrt@2)]) + 2]], {n, 2, 30}] (* Robert G. Wilson v, Feb 21 2006 *)
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Bernd-Rainer Lauber (br.lauber(AT)surf1.de), Jan 21 2000
EXTENSIONS
More terms from David W. Wilson, Feb 01 2000
a(24)-a(29) from Robert G. Wilson v, Feb 21 2006
STATUS
approved