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%I #42 Aug 29 2020 02:14:54
%S 2,9,16,29,42,61,80,105,130,161,192,229,266,309,352,401,450,505,560,
%T 621,682,749,816,889,962,1041,1120,1205,1290,1381,1472,1569,1666,1769,
%U 1872,1981,2090,2205,2320,2441,2562,2689,2816,2949,3082,3221,3360,3505,3650
%N Number of length three 0..n arrays with the sum of two elements equal to twice the third.
%C Number of length three 0..n vectors that contain their arithmetic mean. - _Hywel Normington_, Aug 15 2020
%H David A. Corneth, <a href="/A248434/b248434.txt">Table of n, a(n) for n = 1..10000</a> (first 210 terms from R. H. Hardin)
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).
%F Empirical: a(n) = 2*a(n-1) -2*a(n-3) +a(n-4).
%F Empirical for n mod 2 = 0: a(n) = (3/2)*n^2 + n + 1.
%F Empirical for n mod 2 = 1: a(n) = (3/2)*n^2 + n - (1/2).
%F From _Hywel Normington_, Aug 21 2020: (Start)
%F a(n) = a(n-1) + 1 + 6*floor(n/2)
%F a(n) = A319127(n+1) + n + 1 = 6*floor((n+1)/2)*floor(n/2) + n + 1.
%F (End)
%F From _Colin Barker_, Aug 28 2020: (Start)
%F G.f.: x*(2 + 5*x - 2*x^2 + x^3) / ((1 - x)^3*(1 + x)).
%F a(n) = (1 + 3*(-1)^n + 4*n + 6*n^2) / 4 for n>0.
%F (End)
%e Some solutions for n=6:
%e ..2....3....6....1....3....4....3....1....6....2....4....0....4....5....4....3
%e ..6....1....2....0....2....3....3....2....5....3....0....1....3....6....4....5
%e ..4....5....4....2....1....5....3....3....4....1....2....2....2....4....4....4
%o (PARI) a(n) = {my(res = 2); if(n % 2 == 0, res+=(1 + 6*floor(n/2))); n = (n-1)>>1; res+=6*n^2 + 8*n; res} \\ _David A. Corneth_, Aug 26 2020
%o (PARI) first(n) = {my(res = vector(n), inc = 7); res[1] = 2; for(i = 2, n, res[i] = res[i-1] + inc; inc += 6 * (i%2 == 1)); res} \\ _David A. Corneth_, Aug 26 2020
%Y Row 1 of A248433.
%Y Cf. A168328, A319127, A319127. First differences A168301.
%K nonn
%O 1,1
%A _R. H. Hardin_, Oct 06 2014
%E Name simplified by _Andrew Howroyd_, Aug 14 2020
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