A033183 a(n) = number of pairs (p,q) such that 4*p + 9*q = n.

1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 3, 3, 2

Offset: 0

Author

Michel Tixier (tixier(AT)dyadel.net)

Keywords

• nonn

Comments

From Reinhard Zumkeller, Nov 07 2009: (Start)

In other words: number of partitions into 4 or 9;

a(n) <= A078134(n); a(A078135(n)) = 0;

a(A167632(n)) = n and a(m) < n for m < A167632(n). (End)

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, -1).

Crossrefs

Cf. A033182.

Programs

Mathematica

CoefficientList[Series[1/((1-x^4)(1-x^9)),{x,0,80}],x] (* or  *) LinearRecurrence[{0,0,0,1,0,0,0,0,1,0,0,0,-1}, {1,0,0,0,1,0,0,0,1,1,0,0,1}, 80] (* Harvey P. Dale, Oct 13 2012 *)

Formula

a(n) = [ 7 n/9 ]+1+[ -3 n/4 ].

G.f.: 1/((1-x^4)*(1-x^9)). - Vladeta Jovovic, Nov 12 2004

a(n) = a(n-4) + a(n-9) - a(n-13). - R. J. Mathar, Dec 04 2011