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%I #68 Sep 08 2022 08:46:02
%S 0,0,2,3,7,9,15,18,26,30,40,45,57,63,77,84,100,108,126,135,155,165,
%T 187,198,222,234,260,273,301,315,345,360,392,408,442,459,495,513,551,
%U 570,610,630,672,693,737,759,805,828,876,900,950,975,1027,1053
%N Number of ordered triples (w,x,y) with all terms in {1,...,n} and 2w = 2n - 2x + y.
%C For a guide to related sequences, see A211422.
%C a(n) = sum of natural numbers in interval (floor((n+1)/2),n]. - _Jaroslav Krizek_, Mar 05 2014
%C For n > 0, 2*a(n-1) is the sum of the largest parts of the partitions of 2n into two distinct even parts. - _Wesley Ivan Hurt_, Dec 19 2017
%C From _Paul Curtz_, Oct 23 2018: (Start)
%C Consider the 51 first nonnegative numbers in the following boustrophedon distribution:
%C 35--36--37--38--39--40--41--42--43--44--45
%C 34--33--32--31--30--29--28--27--26--46
%C 12--13--14--15--16--17--18--25--47
%C 11--10---9---8---7--19--24--48
%C 1---2---3---6--20--23--49
%C 0---4---5--21--22--50
%C a(n+1) is the union of the main vertical (0,2, 9,15, 30,40, ... ) and of the shifted main antidiagonal (3,7, 18,26, 45,57, ... ). (End)
%C Sum of the shortest side lengths of all integer-sided triangles with perimeter 3(n+1) whose sides lengths are in arithmetic progression (For example, when n=4 there are two triangles with perimeter 3(4+1) = 15 whose side lengths are in arithmetic progression: [3,5,7] and [4,5,6]; thus a(4) = 3+4 = 7). - _Wesley Ivan Hurt_, Nov 01 2020
%H Vincenzo Librandi, <a href="/A211539/b211539.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).
%F a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
%F From _Jaroslav Krizek_, Mar 05 2014: (Start)
%F a(n) = T(n+1) - T(floor((n+1)/2)) - (n+1), where T(k) = A000217(k).
%F a(n) = Sum_{k=floor((n+1)/2)+1..n} k.
%F a(n) = a(n-1) + n for even n; a(n) = a(n-1) + (n-1)/2 for odd n. (End)
%F From _Ralf Stephan_, Mar 10 2014: (Start)
%F a(n) = (1/16) * (6n^2 + 2n - 3 + (2n+3)*(-1)^n ).
%F G.f.: (x^3+2x^2)/((1+x)^2*(1-x)^3). (End)
%F From _Paul Curtz_, Oct 22 2018: (Start)
%F a(2n) = A005449(n), a(2n+1) = A045943(n).
%F a(2n) + a(2n+1) = A045944(n).
%F a(3n) = 3*(0, 1, 5, 10, 19, 28, 42, ...).
%F a(n+1) = a(n) + A065423(n+2).
%F a(-n) = A211538(n+2). (End)
%F E.g.f.: (3*x*(1 + x)*cosh(x) + (-3 + 5*x + 3*x^2)*sinh(x))/8. - _Stefano Spezia_, Nov 02 2020
%e G.f. = 2*x^2 + 3*x^3 + 7*x^4 + 9*x^5 + 15*x^6 + 18*x^7 + ... - _Michael Somos_, Nov 14 2018
%p a:=n->add(k,k=floor((n+1)/2)+1..n): seq(a(n),n=0..55); # _Muniru A Asiru_, Oct 26 2018
%t t[n_] := t[n] = Flatten[Table[2 w + 2 x - y - 2 n, {w, 1, n}, {x, 1, n}, {y, 1, n}]]
%t c[n_] := Count[t[n], 0]
%t t = Table[c[n], {n, 0, 70}] (* A211539 *)
%t FindLinearRecurrence[t]
%t CoefficientList[Series[(x^3 + 2 x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 60}], x] (* _Vincenzo Librandi_, Mar 12 2014 *)
%o (PARI) a(n)=(1/16)*(6*n^2+2*n-3+(2*n+3)*(-1)^n) \\ _Ralf Stephan_, Mar 10 2014
%o (Magma) I:=[0,0,2,3,7]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..60]]; // _Vincenzo Librandi_, Mar 12 2014
%o (GAP) a:=[0];; for n in [2..55] do if n mod 2 = 0 then Add(a,a[n-1]+n); else Add(a,a[n-1]+(n-1)/2); fi; od; Concatenation([0],a); # _Muniru A Asiru_, Oct 26 2018
%Y Cf. A211422.
%Y Cf. A000217, A005449, A045943, A045944, A065423, A211538.
%K nonn,easy
%O 0,3
%A _Clark Kimberling_, Apr 15 2012
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