Smallest product p_1*...*p_n of distinct primes such that p_i == 1 (mod p1*...*p_{i-1})

[Jianing Song]

Dear all,

I was wondering about the following sequence:
a(n) is the smallest product p_1*...*p_n of distinct primes such that p_i == 1 (mod p1*...*p_{i-1}) for i = 2..n.
a(n) is exactly the smallest number with the maximum number of groups among squarefree Numbers with n factors.
It is trivial that a(1) = 2, a(2) = 6 = 2*3, a(3) = 42 = 2*3*7, a(4) = 1806 = 2*3*7*43.

We have a(5) = 6525078 = 1806*(2*1806+1):
Write P(n,k) be the product of the k smallest prime factors of a(n), then clearly P(n,1) = 2, and P(n,i) >= P(n,i-1)*(P(n,i-1)+1) for i = 2..n.
Since we have a(5) <= 6525078, we have P(5,4) <= 2553, P(5,3) <= 50, and P(5,2) <= 6. Using P(n,i) == 1 (mod P(n,i-1)) for i = 2..n we clearly have P(5,2) = 6, P(5,3) = 42, P(5,4) = 1806, and a(5) = P(5,5) = 6525078.

Similarly, a(6) = 383048407282506 = 1806*10837*(1806*10837+1):
Since we have a(6) <= 425766435585918 = 6525078*(10*6525078+1), we have P(6,5) <= 20634108, P(6,4) <= 4541, P(6,3) <= 66, and P(6,2) <= 7. We again have P(6,2) = 6, P(6,3) = 42 (since 6*13 > 66), P(6,4) = 1806 (since 42*127 > 4541). Now P(6,5) can be 1806*3613, 1806*5419, 1806*10837. Checking the three cases we see that a(6) = P(6,6) corresponds to P(6,5) = 1806*10837.

I would appreciate any help of making a program to provide additional terms. Thanks in advance! :)

[Jianing Song]

Please kindly ignore "a(n) is exactly the smallest number with the maximum number of groups among squarefree numbers with n factors" since this is wrong. There is no such maximum number (from A000001, when p_2 == 1 (mod p_1) and p_3 == 1 (mod p_1*p_2) the number of groups of order p_1*p_2*p_3 is p_1+4 which depends on p_1).

[Robert Israel]

It seems to me that p_i =  A071580(i). 

Cheers,

Robert

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[Jianing Song]

Thanks. But this is not true, see my discussion on a(6). In particular, a(n) is not necessary divisible by its previous term. :)