A211480 Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w + 2x + 3y = 1.

0, 3, 8, 16, 27, 40, 56, 75, 96, 120, 147, 176, 208, 243, 280, 320, 363, 408, 456, 507, 560, 616, 675, 736, 800, 867, 936, 1008, 1083, 1160, 1240, 1323, 1408, 1496, 1587, 1680, 1776, 1875, 1976, 2080, 2187, 2296, 2408, 2523, 2640, 2760, 2883

Offset: 0

Author

Clark Kimberling, Apr 12 2012

Keywords

• nonn

Comments

For a guide to related sequences, see A211422.

For n>2, this is the number of 1's in the partitions of 4n-4 into 4 parts. - Wesley Ivan Hurt, Mar 13 2014

List of triples: [4*k*(3*k-1), 4*k*(3*k+1), 3*(2*k+1)^2], respectively A014642, 8*A005449, 3*A016754. - Luce ETIENNE, May 31 2017

Conjecture: Number of partitions of 4n+2 into 3 parts. - George Beck, Mar 23 2023

Links

• Robert Israel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A211422.

Cf. A005449, A014642, A016754.

Programs

Maple

f:= proc(n) local x,r,ymin,ymax;
  r:= 0:
  for x from -n to n do
    ymin:= max(-n, ceil((-n+1-2*x)/3));
    ymax:= min(n, floor((n+1-2*x)/3));
    if ymin <= ymax then r:= r + ymax-ymin+1 fi
  od;
  r
end proc:
map(f, [$0..50]); # Robert Israel, Jun 09 2023

Mathematica

t[n_] := t[n] = Flatten[Table[w + 2 x + 3 y - 1, {w, -n, n}, {x, -n, n}, {y, -n, n}]] c[n_] := Count[t[n], 0] t = Table[c[n], {n, 0, 70}] (* A211480 *)
b[0] := 0; b[n_] := Sum[((4 n - 2 - i)*Floor[(4 n - 2 - i)/2] - i (4 n - 2 - i) + (i + 2) (Floor[(4 n - 2 - i)/2] - i)) (Floor[(Sign[(Floor[(4 n - 2 - i)/2] - i)] + 2)/2])/(4 n), {i, 0, 2 n}]; Table[b[n - 1] + 2 (n - 1), {n, 50}] (* Wesley Ivan Hurt, Mar 13 2014 *)

Formula

Conjectures from Colin Barker, May 15 2017: (Start)

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

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5) for n>=5.

(End)

Conjecture: a(n) = (8*floor(n/3)*(2*n-3*floor(n/3)-1)+3*(1-(-1)^(n+2-floor((n+2)/3))))/2 = floor((2*n-1)^2/3). - Luce ETIENNE, May 25 2017

Extensions

Offset corrected by Robert Israel, Jun 09 2023