From Daniel Forgues, May 24 2013: (Start)
With a(1) = 2, we have 1/2 + 1/2 = (1 + 1)/2 = 1;
with a(2) = 6 = 2 * 3, we have
1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = (1*3 + 3)/(2*3) = (1 + 1)/2 = 1;
with a(3) = 42 = 6 * 7, we have
1/2 + 1/3 + 1/7 + 1/42 = (21 + 14 + 6 + 1)/42 =
(3*7 + 2*7 + 7)/(6*7) = (3 + 2 + 1)/6 = 1;
with a(4) = 1806 = 42 * 43, we have
1/2 + 1/3 + 1/7 + 1/43 + 1/1806 = (903 + 602 + 258 + 42 + 1)/1806 =
(21*43 + 14*43 + 6*43 + 43)/(42*43) = (21 + 14 + 6 + 1)/42 = 1;
with a(5) = 47058 (not oblong number), we have
1/2 + 1/3 + 1/11 + 1/23 + 1/31 + 1/47058 =
(23529 + 15686 + 4278 + 2046 + 1518 + 1)/47058 = 1.
For n = 1 to 8, a(n) has n prime factors:
a(1) = 2
a(2) = 2 * 3
a(3) = 2 * 3 * 7
a(4) = 2 * 3 * 7 * 43
a(5) = 2 * 3 * 11 * 23 * 31
a(6) = 2 * 3 * 11 * 23 * 31 * 47059
a(7) = 2 * 3 * 11 * 17 * 101 * 149 * 3109
a(8) = 2 * 3 * 11 * 23 * 31 * 47059 * 2217342227 * 1729101023519
If a(n)+1 is prime, then a(n)*[a(n)+1] is also primary pseudoperfect. We have the chains: a(1) > a(2) > a(3) > a(4); a(5) > a(6). (end)
A primary pseudoperfect number (greater than 2) is oblong if and only if it is not the initial member of a chain.  Daniel Forgues, May 29 2013
If a(n)1 is prime, then a(n)*(a(n)1) is a Giuga number (A007850). This occurs for a(2), a(3), and a(5). See A235139 and the link "The padic order . . .", Theorem 8 and Example 1.  Jonathan Sondow, Jan 06 2014
