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Numbers n for which the symmetric representation of sigma(n) has at least 3 parts, all having the same area.
3

%I #22 Jan 15 2015 13:25:42

%S 15,5950

%N Numbers n for which the symmetric representation of sigma(n) has at least 3 parts, all having the same area.

%C a(3) > 36000000.

%C Also intersection of A241558 and A241559 (minimum = maximum) minus the union of A238443 and A239929 (number of parts <= 2).

%e The parts of the symmetric representations of sigma(15) and sigma(5950) are {8, 8, 8} and {4464, 4464, 4464}, respectively, so a(1) = 15 and a(2) = 5950.

%e From _Omar E. Pol_, Dec 09 2014: (Start)

%e Illustration of the symmetric representation of sigma(15) = 8 + 8 + 8 = 24 in the first quadrant:

%e .

%e . _ _ _ _ _ _ _ _ 8

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

%e . |

%e . |_ _

%e . |_ |_ 8

%e . | |_

%e . |_ _ |

%e . |_|_ _ _ 8

%e . | |

%e . | |

%e . | |

%e . | |

%e . | |

%e . | |

%e . | |

%e . |_|

%e .

%e The three parts have the same area.

%e (End)

%t (* T[], row[], cD[] & tD[] are defined in A239663 *)

%t a251820[n_] := Module[{pT = T[n, 1], cT, cL, cW = 0, cR = 0, sects = {}, j = 1, r = row[n], test = True}, While[test && j <= r, cT = T[n, j+1]; cL = pT - cT; cW += (-1)^(j+1) * tD[n, j]; If[cW == 0 && cR != 0, AppendTo[sects, cR]; cR = 0; If[Min[sects] != Max[sects], test = False], cR += cL * cW]; pT = cT; j++]; If[cW != 0, AppendTo[sects, 2 * cR - cW]]; Min[sects] == Max[sects] && Length[sects] > 1]

%t Select[Range[50000], a251820] (* data *)

%Y Cf. A000203, A237593, A237270, A238443, A239663, A241558, A241559.

%K nonn,more,hard,bref

%O 1,1

%A _Hartmut F. W. Hoft_, Dec 09 2014