A211434 Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w+2x+5y=0.

1, 1, 5, 9, 17, 25, 33, 45, 57, 73, 89, 105, 125, 145, 169, 193, 217, 245, 273, 305, 337, 369, 405, 441, 481, 521, 561, 605, 649, 697, 745, 793, 845, 897, 953, 1009, 1065, 1125, 1185, 1249, 1313, 1377, 1445, 1513, 1585, 1657, 1729, 1805, 1881

Offset: 0

Author

Clark Kimberling, Apr 11 2012

Keywords

• nonn
• easy

Comments

For a guide to related sequences, see A211422.

Also, a(n) is the number of ordered pairs (w,x) with both terms in {-n,...,0,...,n} and w+2x divisible by 5. If (w,x) is such a pair it is easy to see that (-x,w), (-w,-x), and (x,-w) also are such pairs. If both w and x are nonzero these four pairs lie one in each quadrant. If one of w or x is zero, the other must be a multiple of 5. This means that a(n) equals 4*A211523(n) (the nonzero pairs) plus 4*floor(n/5) + 1 (pairs with w or x equal to zero). Since the sequences A211523(n), floor(n/5), and the constant sequence all satisfy the recurrence conjectured in the formula section, a(n) must also satisfy the recurrence, so this proves the conjecture. - Pontus von Brömssen, Jan 17 2020

Links

• Pontus von Brömssen, Table of n, a(n) for n = 0..1024
• Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,1,-2,1).

Crossrefs

Cf. A211422, A211523.

Programs

Magma

a:=[]; for n in [0..50] do m:=0; for i, j in [-n..n] do if (i+2*j) mod 5 eq 0  then m:=m+1; end if; end for; Append(~a, m); end for; a; // Marius A. Burtea, Jan 19 2020

Magma

R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1 - x + 4*x^2 + 4*x^4 - x^5 + x^6) / ((1 - x)^3*(1 + x + x^2 + x^3 + x^4)))); // Marius A. Burtea, Jan 19 2020

Mathematica

t[n_] := t[n] = Flatten[Table[w + 2 x + 5 y, {w, -n, n}, {x, -n, n}, {y, -n, n}]]
c[n_] := Count[t[n], 0]
t = Table[c[n], {n, 0, 70}]  (* A211434 *)
(t - 1)/4                    (* A011858 *)

Formula

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

G.f.: (1 - x + 4*x^2 + 4*x^4 - x^5 + x^6) / ((1 - x)^3*(1 + x + x^2 + x^3 + x^4)).

a(n) = 2*a(n-1) - a(n-2) + a(n-5) - 2*a(n-6) + a(n-7) for n>6.

(End)

a(n) = (4*n*(n+1) + c(n))/5, where c(n) is 5 if n is 0 or 4 (mod 5), -3 if n is 1 or 3 (mod 5), and 1 if n is 2 (mod 5). - Pontus von Brömssen, Jan 17 2020