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A211434
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Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w+2x+5y=0.
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2
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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
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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 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
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LINKS
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FORMULA
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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
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MATHEMATICA
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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 *)
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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 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<x>:=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
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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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