[seqfan] Proof that all terms>=41 of A188579 are primes

[Vladimir Shevelev]

Let N>=41 be composite number. Let p>=2 be the smallest prime divisor of N such that N has the form N=p^t*m, where t>=1 and all prime divisors of m are more than p. Note that m is odd.  Distinguish several cases:
1) m>=7;
2) p=2, m=3;
3) p=2, m=5;
4) p=3, m=5;
5) N=2^t, t>=6;
6) N=3^t, t>=4;
7) N=5^t, t>=3;
8) N=p^t, t>=2, p>=7.
In case 1) put k=p^t. Note that, since t>=1, then 2 <= k <= (N-2)/2. Indeed,
(N-2)/2 = (p^t*m-2)/2 >= (p^t*(p+1)-2)/2 >= p^(t-1)*(p+1) - 1 >= p^t = k.
Now the number  N - k = p^t*(m-1) has at least the following 4 divisors: d_1 = p^t, d_2 = 2*p^t,
 d_3 = p^t* (m-1)/2,  d_4 = N - k = p^t*(m-1). Since m>=7, then d_1 < d_2 < d_3 < d_4 and k divides N - d_i,  i = 1,2,3,4.
Thus N is not in the sequence. 
Below we exhaust all cases when m is either 5 or 3 or 1.
In case 2), the nearest N >=41 of the form  3*2^t  is 48. Thus t>=4.  Put k=3*2^(t-4), then  N  - k = 45*2^(t-4) has 
at least 4 divisors d_1 = 3*2^(t-4), d_2 =  9*2^(t-4), d_3 =  15*2^(t-4), d_4 =  45*2^(t-4),  such that k | N - d_i, i = 1,2,3,4.
Thus N is not in the sequence.
In case 3), the nearest N >=41 of the form  5*2^t  is 80. Thus t>=4.  Put k=5*2^(t-4), then  N  - k = 75*2^(t-4) has 
at least 4 divisors d_1 = 5*2^(t-4), d_2 =  15*2^(t-4), d_3 =  25*2^(t-4), d_4 =  75*2^(t-4),  such that k | N - d_i, i = 1,2,3,4.
Thus N is not in the sequence.
In case 4), the nearest N>=45 of the form 5*3^t is 45.  Thus t>=2.  Put k=5*3^(t-2), then  N  - k = 40*3^(t-2) has 
at least 4 divisors d_1 = 5*3^(t-2), d_2 =  10*3^(t-2), d_3 =  20*3^(t-2), d_4 =  40*3^(t-2),  such that k | N - d_i, i = 1,2,3,4.
Thus N is not in the sequence.
In cases 5)-7) put respectively k = 2^(t-4), {d_1,d_2,d_3,d_4} = {i*2^(t-4), i = 1,3,5,15}, 
 k = 3^(t-2), {d_1,d_2,d_3,d_4} = {i*3^(t-2), i = 1,2,4,8} and  k = 5^(t-2), {d_1,d_2,d_3,d_4} = {i*5^(t-2), i = 1,2,3,4}.
Thus N is not in the sequence.
Finally, in case 8) put k = p^(t-1), then  N  - k = p^(t-1)*(p-1) has at least 4 divisors d_1 = p^(t-1),  d_2 =2* p^(t-1),
d_3 =  p^(t-1)*(p-1)/2,  d_4 = p^(t-1)*(p-1), such that k | N - d_i, i = 1,2,3,4.
Thus N is not in the sequence.
This completes proof.

Regards, 
Vladimir

 Shevelev Vladimir‎