OFFSET
1,1
COMMENTS
Let k >= 1. In the multiplicative basis Q^(k) = {p^(k+1)^j, p runs through A000040, j=0,1,...} every positive integer m has a unique factorization of the form m = Product_{q is in Q^(k)} q^(m_q), where m_q is in {0,1,...,k}. In particular, in the case of k=1, we have the unique factorization over distinct terms of A050376. Notice that the standard prime basis is the limiting value for k tending to infinity, and, by the definition, Q^(infinity)=A000040. The number d is called a k-divisor of m if the exponents d_q in its factorization in the basis Q^(k) do not exceed m_q. A number m is called k-perfect if it equals to the sum of its proper positive k-divisors. Conjecture: a(11)=0. Note that we also know of n-perfect numbers for n = 12, 14, 15, 16, and 18.
LINKS
S. Litsyn and V. S. Shevelev, On factorization of integers with restrictions on the exponent, INTEGERS: Electronic Journal of Combinatorial Number Theory, 7 (2007), #A33, 1-36.
FORMULA
m = Product_{q is in Q^(k)} q^(m_q) is a k-perfect number iff Product_{q is in Q^(k)} (q^((m_q)+1)-1)/(q-1) = 2*m.
EXAMPLE
In case of n=2, we have the basis ("2-primes"): 2,3,5,7,8,11,13,... By the formula, we construct from the left m and from the right 2*m. By the condition, m begins from "2-prime" 8. From the right we have 8+1=3^2, therefore from the left we have 8*3^2 and from the right 3^2*(3^3-1)/(3-1)=3^2*13. Thus from the left it should be 8*3^2*13 and from the right 3^2*13*14. Finally, from the left we obtain m=8*3^2*13*7=6552 and from the right we have 2*m=3^2*13*14*8. By the construction, it is the smallest 2-perfect number of the required form. Thus a(2)=6552.
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Vladimir Shevelev, Jun 14 2010, Jun 18 2010
STATUS
approved