[seqfan] How many primes should there be?
Christian G.Bower
bowerc at usa.net
Fri May 15 15:54:17 CEST 1998
Since sequences A007908 (1 12 123 1234...) and A000422 (1 21 321...)
have so few primes, let's figure out how many primes they should
have.
Only every 6th number from A007908 has a chance at being prime.
1234567
12345678910111213
12345678910111213141516171819
...
(all others are divisible by 2 or 3)
The probability that a number z chosen at random is about 1/log(z)
If I know that z not divisible by 2 or 3, then probability z is
prime should be 3/log(z).
Using this I've calculated the expected number of primes in the
sequence from 1 to n.
(In this, n is the nth prime candidate, k is 6*n+1, the index in the
sequence n corresponds to:)
n k E(n)
10 61 0.489141
100 601 0.700195
1000 6001 0.850933
2000 12001 0.889626
3000 18001 0.9104
4000 24001 0.924393
5000 30001 0.93496
10000 60001 0.966763
20000 120001 0.997553
30000 180001 1.01436
100000 600001 1.06073
500000 3000001 1.11743
1000000 6000001 1.13979
2000000 12000001 1.16164
5000000 30000001 1.18856
Here are values for A000422 where every third number (4321 7654321...)
is a candidate. Note that Eric (or Ralf) found a prime at term 82 and
no others in the first 2000 terms. Here k=3*n+1.
n k E(n)
10 31 1.12075
24 82 1.3244
100 301 1.60092
1000 3001 1.9316
2000 6001 2.01221
3000 9001 2.05796
4000 12001 2.08961
5000 15001 2.11285
10000 30001 2.18029
20000 60001 2.2439
100000 300001 2.37932
500000 1500001 2.49867
1000000 3000001 2.54524
2000000 6000001 2.58995
5000000 15000001 2.64717
Based on this, the first prime in A007908... should occur around
180000, but possibly much higher.
The second prime in A000422 would be expected around 6000.
- --------------
My last posting about the number of k such that Smarandache(k)=n
was wrong.
The number of values k S(k)=n is just d(n!)-d((n-1)!).
where d(n) is the number of divisors of n.
The sequence should be 1 2 4 8 14 30 36 64 110...
Christian
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