OFFSET
1,1
COMMENTS
Let n = 2^m * q with m >= 0 and q odd, let row(n) = floor(sqrt(8*n+1) - 1)/2), and let 1 = d_1 < ... < d_h <= row(n) < d_(h+1) < ... < d_k = q be the k odd divisors of n.
The symmetric representation of sigma(n) consists of 3 parts precisely when there is a unique i, 1 <= i < h, such that 2^(m+1) * d_i < d_(i+1) and d_h <= row(n) < 2^(m+1) * d_h.
This property of the odd divisors of n is equivalent to the n-th row of the irregular triangle of A249223 consisting of a block of positive numbers, followed by a block of zeros, followed in turn by a block of positive numbers, i.e., determining the first part and the left half of the center part of the symmetric representation of sigma(n), resulting in 3 parts.
Let n be the product of two primes p and q satisfying 2 < p < q < 2*p. Then n satisfies the property above so that the odd numbers in A087718 form a subsequence.
EXAMPLE
a(4) = 35 = 5*7 is in the sequence since 1 < 2 < 5 < row(35) = 7 < 10;
a(8) = 70 = 2*5*7 is in the sequence since 1 < 4 < 5 < row(70) = 11 < 20;
140 = 4*5*7 is not in the sequence since 1 < 5 < 7 < 8 < row(140) = 16 < 20;
a(506) = 5950 = 2*25*7*17 is in the sequence since 1*4 < 5 is the only pair of odd divisors 1 < 5 < 7 < 17 < 25 < 35 < 85 < row(5950) = 108 satisfying the property (see A251820).
MATHEMATICA
segmentsSigma[n_] := Length[Select[SplitBy[a262045[n], #!=0&], First[#]!=0&]]
a279102[m_, n_] := Select[Range[m, n], segmentsSigma[#]==3&]
a279102[1, 700] (* sequence data *)
(* An equivalent, but slower computation is based on A237271 *)
a279102[m_, n_] := Select[Range[m, n], a237271[#]==3&]
a279102[1, 700] (* sequence data *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Hartmut F. W. Hoft, Dec 06 2016
STATUS
approved