A126283
Largest number k for which the n-th prime is the median of the largest prime dividing the first k integers.
Original entry on oeis.org
4, 18, 40, 76, 116, 182, 246, 330, 426, 532, 652, 770, 904, 1058, 1210, 1386, 1560, 1752, 1956, 2162, 2394, 2640, 2894, 3150, 3422, 3680, 3984, 4302, 4628, 4974, 5294, 5650, 5914, 6006, 6372, 6746, 7146, 7536, 7938, 8386, 8794, 9222, 9702, 10156
Offset: 1
a(1)=4 because the median of {2,3,2} = {2, *2*,3} is 2 (the * surrounds the median) and for any number greater than 4 the median is greater than 2.
a(1)=18 because the median of {2,3,2,5,3,7,2,3,5,11,3,13,7,5,2,17,3} = {2,2,2,2,3,3,3,3, *3*,5,5,5,7,7,11,13,17}.
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t = Table[0, {100}]; lst = {}; Do[lpf = FactorInteger[n][[ -1, 1]]; AppendTo[lst, lpf]; mdn = Median@lst; If[PrimeQ@ mdn, t[[PrimePi@mdn]] = n], {n, 2, 10^4}]; t
A308905
Number of numbers k such that exactly half the numbers in [1..k] are prime(n)-smooth.
Original entry on oeis.org
2, 1, 1, 4, 5, 1, 4, 1, 3, 1, 1, 2, 1, 2, 7, 1, 4, 4, 3, 2, 5, 3, 6, 6, 1, 4, 1, 3, 2, 5, 3, 3, 2, 2, 2, 5, 4, 7, 8, 7, 2, 6, 5, 3, 13, 10, 1, 9, 2, 6, 3, 2, 8, 4, 4, 1, 11, 3, 3, 1, 7, 2, 4, 1, 1, 5, 4, 2, 10, 5, 4, 6, 9, 7, 1, 3, 8, 8, 6, 6, 1, 3, 4, 2, 2, 2
Offset: 1
For n=1: prime(1)=2, and the 2-smooth numbers are 1, 2, 4, 8, 16, 32, ... (A000079, the powers of 2), so for k = 1..10, the number of 2-smooth numbers in the interval [1..k] increases as follows:
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Number m
2-smooth of 2-smooth
numbers numbers
k in [1..k] in [1..k] m/k
== ============ =========== ===============
1 {1} 1 1/1 = 1.000000
2 {1, 2} 2 2/2 = 1.000000
3 {1, 2} 2 2/3 = 0.666667
4 {1, 2, 4} 3 3/4 = 0.750000
5 {1, 2, 4} 3 3/5 = 0.600000
6 {1, 2, 4} 3 3/6 = 0.500000 = 1/2
7 {1, 2, 4} 3 3/7 = 0.428571
8 {1, 2, 4, 8} 4 4/8 = 0.500000 = 1/2
9 {1, 2, 4, 8} 4 4/9 = 0.444444
10 {1, 2, 4, 8} 4 4/10 = 0.400000
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It is easy to show that, for all k > 8, fewer than half of the numbers in [1..k] are 2-smooth, so there are only 2 values of k, namely, k=6 and k=8, at which exactly half of the numbers in the interval [1..k] are 2-smooth numbers, so a(1)=2.
For n=2: prime(2)=3, and the 3-smooth numbers are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, ... (A003586). It can be shown that k=20 is the only number k such that exactly half of the numbers in the interval [1..k] are 3-smooth. Since there is only 1 such number k, a(2)=1.
Showing 1-2 of 2 results.
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