A271937 a(n) = (7/4)*n^2 + (5/2)*n + (7 + (-1)^n)/8.

1, 5, 13, 24, 39, 57, 79, 104, 133, 165, 201, 240, 283, 329, 379, 432, 489, 549, 613, 680, 751, 825, 903, 984, 1069, 1157, 1249, 1344, 1443, 1545, 1651, 1760, 1873, 1989, 2109, 2232, 2359, 2489, 2623, 2760, 2901, 3045, 3193, 3344, 3499, 3657, 3819, 3984, 4153

Offset: 0

Author

Vincenzo Librandi, Apr 20 2016

Keywords

• nonn
• easy

Comments

Let P be a polygon with vertices (0,0), (0,2), (1,1) and (0,3/2). The number of integer points in nP is counted by this quasi-polynomial (nP is the n-fold dilation of P). See Wikipedia in Links section.

From Bob Selcoe, Sep 10 2016: (Start)

a(n) = the number of partitions in reverse lexicographic order starting with n 3's followed by n 2's; i.e., the number of partitions summing to 5n such that no part > 3 and the number of 3's digits <= the number of 2's digits.

First differences are A047346(n+1); second differences are 4 when n is even and 3 when n is odd (i.e., A010702(n+1)); third differences are 1 when n is even and -1 when n is odd. (End)

Examples

			a(1) = 5; the 5 partitions are: {3,2}; {3,1,1}; {2,2,1}; {2,1,1,1}; {1,1,1,1,1}.
a(3) = 24: floor(8/2) + floor(11/2) + floor(14/2) + floor(17/2) = 4+5+7+8 = 24.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Wikipedia, Examples of Ehrhart Quasi-Polynomials.
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

First bisection (after 1) is A168235.

Second bisection is A135703 (without 0).

Cf. A006578, A047346.

Programs

Magma

[(7/4)*n^2+(5/2)*n+(7+(-1)^n)/8: n in [0..50]];

Mathematica

Table[(7/4) n^2 + (5/2) n + (7 + (-1)^n)/8, {n, 0, 50}]
LinearRecurrence[{2,0,-2,1},{1,5,13,24},50] (* Harvey P. Dale, Mar 23 2025 *)

PARI

Vec((1+3*x+3*x^2)/((1-x)^3*(1+x)) + O(x^99)) \\ Altug Alkan, Sep 10 2016

Formula

O.g.f.: (1 + 3*x + 3*x^2)/((1 - x)^3*(1 + x)).

E.g.f.: (7 + 34*x + 14*x^2)*exp(x)/8 + exp(-x)/8.

a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).

a(2*k) = k*(7*k + 5) + 1, a(2*k+1) = (k + 1)*(7*k + 5).

From Bob Selcoe, Sep 10 2016 (Start):

a(n) = (n+1)^2 + A006578(n).

a(n) = a(n-1) + A047346(n+1).

a(n) = Sum_{j=0..n} floor((2n+3j+2)/2).

(End)

Extensions

Edited and extended by Bruno Berselli, Apr 20 2016