A130695 Number of ways to write n as (a+1)(b+1)(c+1) - abc with a, b, c nonnegative integers.

1, 3, 3, 6, 3, 9, 4, 9, 6, 12, 3, 15, 6, 12, 9, 15, 3, 21, 7, 15, 9, 18, 6, 24, 9, 15, 9, 24, 6, 30, 6, 15, 15, 24, 9, 30, 7, 21, 12, 30, 3, 33, 15, 21, 15, 24, 6, 39, 12, 27, 12, 27, 9, 42, 12, 21, 15, 36, 6, 45, 13, 18, 21, 36, 12, 39, 6, 33, 15, 45, 9, 42, 12, 24, 24, 30, 9, 57, 18, 30

Offset: 1

Author

Jan Kristian Haugland, Jul 10 2007

Keywords

• nonn

Examples

			a(7) = 4 because 7 = 7*1*1-6*0*0 = 1*7*1-0*6*0 = 1*1*7-0*0*6 = 2*2*2-1*1*1.
G.f. = x + 3*x^2 + 3*x^3 + 6*x^4 + 3*x^5 + 9*x^6 + 4*x^7 + 9*x^8 + 6*x^9 + ...
		

Links

• _Jan Kristian Haugland_, Jul 10 2007, Table of n, a(n) for n = 1..200

Crossrefs

Cf. A238872.

Programs

Mathematica

f[{a_,b_,c_}]:=(a+1)(b+1)(c+1)-a*b*c; nn=80;Take[Transpose[Sort[Tally[f/@ Tuples[Range[0,nn],3]],#1[[1]]<#2[[1]]&]] [[2]],nn] (* Harvey P. Dale, Mar 05 2012 *)
a[ n_] := Length @ FindInstance[ {x >= 0, y >= 0, z >= 0, x y + y z + z x + x + y + z + 1 == n}, {x, y, z}, Integers, 10^9]; (* Michael Somos, Jul 04 2015 *)
a[ n_] := (2 + (-1)^n) Length @ FindInstance[ {1 <= y <= n, 1 <= x <= y, 1 <= z <= y, y^2 - (x^2 - x + z^2 - z) / 2 == n}, {x, y, z}, Integers, 10^9]; (* Michael Somos, Jul 04 2015 *)

Formula

a(2*n) = A238872(2*n) / 3 if n>0. a(2*n + 1) = A238872(2*n + 1). - Michael Somos, Jul 04 2015