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

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, 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, 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

Offset: 1

Author

Omar E. Pol, Nov 05 2015

Keywords

• nonn

Comments

The diagram of the asymmetric representation of sigma in an octant has been obtained according to the following way: A196020 --> A236104 --> A235791 --> A237591.

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.

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.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..5000 (computed from the b-file of A237271 provided by Michel Marcus)

Crossrefs

Cf. A000203, A196020, A235791, A236104, A237270, A237271, A237591, A237593, A244050, A244971, A245092, A262619.

Formula

a(n) = (A237271(n) + 1)/2, if A237271(n) is odd.

a(n) = A237271(n)/2, if A237271(n) is even.