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Number of vertices on the axis of symmetry of the symmetric representation of sigma(n) with subparts.
3

%I #21 Dec 14 2021 10:06:31

%S 2,2,1,2,1,3,1,2,2,1,1,3,1,1,3,2,1,2,1,3,1,1,1,3,2,1,1,3,1,3,1,2,1,1,

%T 3,2,1,1,1,3,1,3,1,1,3,1,1,3,2,2,1,1,1,3,1,3,1,1,1,3,1,1,3,2,1,3,1,1,

%U 1,3,1,4,1,1,1,1,3,1,1,3,2,1,1,3,1,1,1,3,1,3,3,1,1,1,1,3,1,2,3,2,1,1,1,3,1

%N Number of vertices on the axis of symmetry of the symmetric representation of sigma(n) with subparts.

%C The number of middle divisors of n is equal to a(n) - 1.

%C For the definition of "subparts" see A279387.

%H Antti Karttunen, <a href="/A348406/b348406.txt">Table of n, a(n) for n = 1..65537</a>

%F a(n) = 1 + A067742(n).

%e For n = 2, 6 and 10 the symmetric representation of sigma(n) with subparts respectively looks like this:

%e .

%e . _ _ _

%e . _| | | | | |

%e . 2 |_ _| | | | |

%e . _ _| | | |

%e . | _ _| | |

%e . _ _ _| |_| _ _| |

%e . 6 |_ _ _ _| | _ _|

%e . _ _|_|

%e . | _|

%e . _ _ _ _ _| |

%e . 10 |_ _ _ _ _ _|

%e .

%e For n = 2 there are two vertices on the axis of symmetry hence the number of middle divisors of 2 is equal to 2 - 1 = 1.

%e For n = 6 there are three vertices on the axis of symmetry hence the number of middle divisors of 6 is equal to 3 - 1 = 2.

%e For n = 10 there is only one vertex on the axis of symmetry hence the number of middle divisors of 10 is equal to 1 - 1 = 0.

%t a[n_] := 1 + DivisorSum[n, 1 &, n/2 <= #^2 < 2*n &]; Array[a, 100] (* _Amiram Eldar_, Oct 17 2021 *)

%o (PARI)

%o A067742(n) = sumdiv(n, d, my(d2 = d^2); n / 2 < d2 && d2 <= n << 1); \\ From A067742

%o A348406(n) = (1 + A067742(n));

%Y Companion of A348364.

%Y Cf. A067742, A071090, A071540, A071562, A071563, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A240542, A279387, A281007, A299761, A303297, A340833, A340847, A346868, A347950, A348327.

%K nonn

%O 1,1

%A _Omar E. Pol_, Oct 17 2021