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A211433
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Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w+2x+4y=0.
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2
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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FORMULA
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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)
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MATHEMATICA
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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 *)
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PROG
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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
(Magma) R<x>:=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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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