Let n = 2^m * Product_{i=1..k} p_i^e_i = 2^m * q with m >= 0, k >= 0, 2 < p_1 < ... < p_k primes and e_i >= 1, for all 1 <= i <= k. For each number n in this sequence k > 0, at least one e_i is odd, and for any two odd divisors f < g of n, 2^(m+1) * f < g. Let the odd divisors of n be 1 = d_1 < ... < d_2x = q where 2x = sigma_0(q). The z-th region of the symmetric spectrum of n has area a_z = 1/2 * (2^(m+1) - 1) *(d_z + d_(2x+1-z)), for 1 <= z <= 2x. Therefore, the sum of the area of the regions equals sigma(n). For a proof see Theorem 6 in the link of A071561. - Hartmut F. W. Hoft, Sep 09 2015, Sep 04 2018
