OFFSET
1,1
COMMENTS
Every number a(n) has the form 2^(2*i + 1) * s^2, i>= 0 and s odd, the single middle divisor of a(n) is sqrt(a(n)/2), and sqrt(2*a(n)) - 1 = floor((sqrt(8*n + 1) - 1)/2) = A003056(a(n)).
The least number in the sequence with 3 odd prime divisors is a(126) = 1630818 = 2^1 * 3^2 * 7^2 * 43^2.
Conjecture: Let a(n) = 2^(2i+1) * s^2, i>=0 and s odd, be a number in the sequence.
(1) For any odd prime divisor p of s, number a(n) * p^2 is in the sequence.
(2) For any odd prime p not a divisor of s, number a(n) * p^2 is in the sequence if p satisfies sqrt(2*a(n)) < p < 2*a(n).
EXAMPLE
a(5) = 128 = 2^7 has 2^3 as its single middle divisor, and its symmetric representation of sigma consists of one part of width 1.
a(10) = 882 = 2 * 3^2 * 7^2 has 3 * 7 as its single middle divisor, its symmetric representation of sigma is the smallest in this sequence of maximum width 3, consists of one part, and has width 1 at the diagonal.
A table of ranges for the single odd prime factor p for numbers k in the sequence having the form 2^(2i+1) * p^(2j), i>=0 and j>0, indexed by exponent 2i+1 of 2 in number k. The lower bound is A014210(i+1) and the upper bound is A014234(2(i+1)) = A104089(i+1):
---------------------
2i+1 /---- p ----/
---------------------
1 3 .. 3
3 5 .. 13
5 11 .. 61
7 17 .. 251
9 37 .. 1021
...
MATHEMATICA
(* a2[ ] and its support functions are defined in A249223 *)
a365265Q[n_] := Module[{list=If[Divisible[n, Sqrt[n/2]], a2[n], {0}]}, Last[list]==1&&AllTrue[list, #>0&]]
a365265[{m_, n_}] := Select[Range[m, n], a365265Q]
a365265[{1, 75000}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Hartmut F. W. Hoft, Aug 29 2023
STATUS
approved