[seqfan] How many primes should there be?

[Christian G.Bower]

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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