A211422 Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w^2 + x*y = 0.
1, 9, 17, 25, 41, 49, 57, 65, 81, 105, 113, 121, 137, 145, 153, 161, 193, 201, 225, 233, 249, 257, 265, 273, 289, 329, 337, 361, 377, 385, 393, 401, 433, 441, 449, 457, 505, 513, 521, 529, 545, 553, 561, 569, 585, 609, 617, 625, 657, 713, 753, 761
Offset: 0
Keywords
A211433 Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w+2x+4y=0.
1, 1, 7, 11, 23, 27, 45, 53, 77, 85, 115, 127, 163, 175, 217, 233, 281, 297, 351, 371, 431, 451, 517, 541, 613, 637, 715, 743, 827, 855, 945, 977, 1073, 1105, 1207, 1243, 1351, 1387, 1501, 1541, 1661, 1701, 1827, 1871, 2003, 2047, 2185, 2233
Offset: 0
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 4. If (w,x) is such a pair it is easy to see that (-w,x), (-w,-x), and (w,-x) also are such pairs. The number of pairs with both w and x positive is given by A211521(n). If w=0, x must be even, and if x=0, w must be divisible by 4. This means that a(n) = 4*A211521(n) + 2*floor(n/2) + 2*floor(n/4) + 1. Since the sequences A211521(n), floor(n/2), floor(n/4), 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 19 2020
Links
- Pontus von Brömssen, Table of n, a(n) for n = 0..1024
- Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1,-1,1).
Programs
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Magma
a:=[]; for n in [0..50] do m:=0; for i, j in [-n..n] do if (i+2*j) mod 4 eq 0 then m:=m+1; end if; end for; Append(~a, m); end for; a; // Marius A. Burtea, Jan 19 2020
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Magma
R
:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1 + 5*x^2 + 4*x^3 + 5*x^4 + x^6) / ((1 - x)^3*(1 + x)^2*(1 + x^2)))); // Marius A. Burtea, Jan 19 2020 -
Mathematica
t[n_] := t[n] = Flatten[Table[w + 2 x + 4 y, {w, -n, n}, {x, -n, n}, {y, -n, n}]] c[n_] := Count[t[n], 0] t = Table[c[n], {n, 0, 70}] (* A211433 *) (t - 1)/2 (* integers *)
Formula
Conjectures from Colin Barker, May 15 2017: (Start)
G.f.: (1 + 5*x^2 + 4*x^3 + 5*x^4 + x^6) / ((1 - x)^3*(1 + x)^2*(1 + x^2)).
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-4) - a(n-5) - a(n-6) + a(n-7) for n>6.
(End)
Comments
Examples
Links
Crossrefs
Programs
Mathematica
t[n_] := t[n] = Flatten[Table[w^2 + x*y, {w, -n, n}, {x, -n, n}, {y, -n, n}]] c[n_] := Count[t[n], 0] t = Table[c[n], {n, 0, 70}] (* A211422 *) (t - 1)/8 (* A120486 *)