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A262618 Number of parts in the asymmetric representation of sigma(n) in an octant.

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%I A262618 #24 Dec 17 2017 03:08:27
%S A262618 1,1,1,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,2,1,2,1,1,1,1,1,2,1,
%T A262618 2,1,1,1,2,1,1,1,1,1,2,1,1,1,2,2,2,1,1,1,2,1,2,1,1,1,1,1,3,1,2,1,1,1,
%U A262618 2,2,1,1,1,1,2,1,2,1,1,1,3,1,1,1,2,1,2,1,1,1,2,1,2,1,2,1,1,2,3,1,1,1,1,1,2,1
%N A262618 Number of parts in the asymmetric representation of sigma(n) in an octant.
%C A262618 The diagram of the asymmetric representation of sigma in an octant has been obtained according to the following way: A196020 --> A236104 --> A235791 --> A237591.
%C A262618 Consider that the hypotenuse of the first triangle of the diagram has length 2, so the area of the triangle is equal to 1 and the sum of the areas of all parts added at n-th stage equals sigma(n), the sum of the divisors of n.
%C A262618 a(n) is also the number of terraces at n-th level (starting from the top) in an octant of the step pyramids described in A245092 and A244050.
%H A262618 Antti Karttunen, <a href="/A262618/b262618.txt">Table of n, a(n) for n = 1..5000</a> (computed from the b-file of A237271 provided by Michel Marcus)
%F A262618 a(n) = (A237271(n) + 1)/2, if A237271(n) is odd.
%F A262618 a(n) = A237271(n)/2, if A237271(n) is even.
%Y A262618 Cf. A000203, A196020, A235791, A236104, A237270, A237271, A237591, A237593, A244050, A244971, A245092, A262619.
%K A262618 nonn
%O A262618 1,9
%A A262618 _Omar E. Pol_, Nov 05 2015