cp's OEIS Frontend

Dihedral perfect numbers: degree n such that dihedral group D_{2n} has order equal to its total number of subgroups, i.e. A007503(n)=2n. - Lekraj Beedassy, Oct 14 2004

If 2^(m+1)+2m+1 is prime then 2^m*(2^(m+1)+2m+1) is in the sequence (see A105330 & A105331). This fact is a result of the following interesting theorem that I found (take k=0). Theorem: If m & k are integers and 2^(m+1)+2m+1-k is prime then 2^m*(2^(m+1)+2m+1-k) is a solution of the equation sigma(x)+ tau(x)=2x+k; the proof is easy. - Farideh Firoozbakht, Apr 27 2005

Allowing n = 2^k pq, for primes p < q, it is fairly easy to discover that p must be in the range 2^(k+1) to 2^(k+2) and there is only one possible q for each p. Exhaustive search can be used to find the primes p. The sequence up to a(23) has all possible solutions for k=1, 3 and 7. There is one solution for k=10: 91707741184; one solution for k=12: 14451706793984; and three solutions for k=15: 2258918614925312, 1153167823398797312, 4611826823562493952. There are no other n = 2^k pq for k up to 26. - T. D. Noe, Jun 19 2008

Assuming n has the form 2^k pqr, for primes p < q < r, there are only a finite number of triples (p,q,r) possible for each k. For k=3, the sequence already has 89397016 and 173706136. For k=4, 6801628304 is the only solution. For k=7, the search finds 102898828936832, 141573123151232, 220346295412352, 619057492909952, and 3585817801980032. For k=9, 989473186649763328 is the only solution. There are no other solutions for k <= 13. - T. D. Noe, Feb 12 2010

This sequence is a subsequence of A083874 (see A083874 & A105330).

This is because these numbers satisfy tau(n) + sigma(n) = 2n when n = 2^k * p with p is prime; for instance tau(14) + sigma(14) = 4 + 24 = 28 = 2 x 14. [See References.] - Bernard Schott, Apr 07 2017

Primes in A061761. The values of k-1 are in A105330. These primes, multiplied by 2^(k-1), produce A105331, which are a subset of the dihedral perfect numbers, A083874.