cp's OEIS Frontend

Intersection of A249763 and A023194.

The next term, if it exists, must be greater than 5*10^12.

Each term > 2 is a square.

From Chai Wah Wu, Jul 13 2016: (Start)

Every term > 2 is of the form p^(2m) with p prime and m > 1. Proof: from the discussion in A023194, a term is of the form p^(2m). An odd term cannot be of the form n = p^2. If p = 6k+1, then sigma(n) = 36k^2 + 18k + 3 is composite. If p = 6k-1, then sigma(n) + 2 = 36k^2 - 6k + 3 is composite. Finally, 4 is not a term.

This could be the reason this sequence is so much sparser than A274963.

(End)

Terms cannot be of the form 2^(2m) since sigma(2^(2m)) + 2 = 2^(2m+1) + 1 is divisible by 3. - Altug Alkan, Jul 14 2016

Terms cannot be of the form 3^(2m) since sigma(3^(3m)) + 2 = 3(3^(2m) + 1)/2 is divisible by 3, i.e., all terms are of the form (6*k+1)^(2m) or (6*k-1)^(2m) - Chai Wah Wu, Aug 06 2016

Terms cannot be of the form p^6 since if p = 6*k+1, then sigma((6*k+1)^6) + 2 = 9*(5184*k^6 + 6048*k^5 + 3024*k^4 + 840*k^3 + 140*k^2 + 14*k + 1) and if p = 6*k-1 then sigma((6*k-1)^6) + 2 = 3*(15552*k^6 - 12960*k^5 + 4752*k^4 - 936*k^3 + 108*k^2 - 6*k + 1). Also note that terms cannot be of the form p^8 since if p = 6*k-1 then sigma((6*k-1)^8) = (1 - 6*k + 36*k^2)*(1 - 18*k + 432*k^2 - 4104*k^3 + 19440*k^4 - 46656*k^5 + 46656*k^6) and if p = 6*k+1 then sigma((6*k+1)^8) = 9*(186624*k^8 + 279936*k^7 + 186624*k^6 + 72576*k^5 + 18144*k^4 + 3024*k^3 + 336*k^2 + 24*k + 1). The least term that is of the form p^10 is 2089^10. So this partially explains why numbers of the form p^4 appear in this sequence most of the time in limited range. - Altug Alkan, Jul 15 2016

From Chai Wah Wu, Jul 20 2016: (Start)

If p^m > 2 is a term, then m == 4 mod 6 and p == 1 mod 6. Proof: Let q(k) be sigma(p^m) expressed as a polynomial in k. If p = 6k-1, then q(k) = 1 + (6k-1) + (6k-1)^2 + ... + (6k-1)^m.

The constant term of q(k) is 1-1+1-1+...-1+1 = 1 whereas the other coefficients are multiples of 6, i.e., q(k) = 1 + 6k*(...), thus sigma(p^m) + 2 is a multiple of 3.

Suppose p = 6k+1, then q(k) = 1 + (6k+1) + (6k+1)^2 + ... + (6k+1)^m. The constant term is m+1 and the other coefficients are multiples of 6, i.e., q(k) = (m+1) + 6k*(...).

This means that if m = 6r+2, then sigma(p^m) is a multiple of 3 and if m = 6r, then sigma(p^m) + 2 is a multiple of 3. End of Proof.

The following table lists the minimal k for r <= 4.

r | smallest k such that (6k+1)^(6r+4) is a term (A275237)

------------------------------------------------------------

0 | 1

1 | 348

2 | 436

3 | 6018

4 | 5880

For every prime p = 6k+1, does there exist r >= 0 such that(6k+1)^(6r+4) is a term?

(End)

Altug Alkan found that sigma((6k+1)^34) (i.e., the r = 5 case) is always composite (see comment in A275237). - Chai Wah Wu, Jul 21 2016

If x is a number such that sigma(x) and sigma(x)+2 are both prime (A274962), then x = 2 or x is of the form (6k+1)^(6r+4) where 6k+1 is prime.

For p = 6*k+1, sigma(p^34) = (46656*k^6 + 54432*k^5 + 27216*k^4 + 7560*k^3 + 1260*k^2 + 126*k + 7) * (1296*k^4 + 1080*k^3 + 360*k^2 + 60*k + 5) * c(k), thus a(5) = -1. - Altug Alkan , Jul 21 2016

Similarly a(12) = a(19) = a(23) = a(26) = a(33) = a(34) = -1. Furthermore, for all r > 0, a(5*r) = -1 since sigma((6k+1)^(30r+4)) = ((6*k+1)^(6*r) + ((6*k+1)^(6*r) -1)/(6*k))*(1296*k^4*(6*k + 1)^(24*r) + 864*k^3*(6*k + 1)^(24*r) + 216*k^3*(6*k + 1)^(18*r) + 216*k^2*(6*k + 1)^(24*r) + 108*k^2*(6*k + 1)^(18*r) + 36*k^2*(6*k + 1)^(12*r) + 24*k*(6*k + 1)^(24*r) + 18*k*(6*k + 1)^(18*r) + 12*k*(6*k + 1)^(12*r) + 6*k*(6*k + 1)^(6*r) + (6*k + 1)^(24*r) + (6*k + 1)^(18*r) + (6*k + 1)^(12*r) + (6*k + 1)^(6*r) + 1). - Chai Wah Wu, Jul 21 2016