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A183928
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Number of nondecreasing arrangements of n numbers in -2..2 with sum zero and sum of squares less than 2n.
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1
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1, 2, 2, 5, 8, 10, 15, 21, 26, 35, 44, 53, 67, 81, 94, 114, 134, 153, 179, 206, 232, 266, 300, 334, 377, 420, 462, 515, 568, 620, 683, 747, 810, 885, 960, 1035, 1123, 1211, 1298, 1400, 1502, 1603, 1719, 1836, 1952, 2084, 2216, 2348, 2497, 2646, 2794, 2961, 3128
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OFFSET
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1,2
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COMMENTS
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Let x_i be the number of numbers in such an arrangement with |i| = 2. Then x_1 + 4*x_2 < 2*n and x_1 is even, x_1 + x_2 != 1.
A pair (x_1, x_2) has a fixed number of arrangements corresponding to it, independent of the number of zeros in the arrangement, hence independent of n. For example, for (x_1, x_2) = (2, 1), we could find the arrangements (-1, -1, 2) and (-2, 1, 1) and from there on find some solutions counted in n = 4 (see example below) by putting an extra 0 to each arrangement to find (-1, -1, 0, 2) and (-2, 0, 1, 1).
For some n, we might find that (x_1, x_2) is feasible by the restrictions above. If all those 1's and 2's are signed positive, their sums become S = x_1 + 2 * x_2. We then change signs of 1's and 2's such that the sum of all terms is 0. If we change the sign of 1 from + to -, the total is reduced by 2 * 1 = 2. Similarily, if we change the sign of 2 from + to -, we reduce the total by 2*2 = 4. Let s_i be the number of i's whose signs we change from + to -, x_i - s_i, s_i >= 0. So the sum S must be even. We get the new sum 0 = (x_1 - 2*s_1) + (2 * x_2 - 2*s_2) = x_1 + 2 * x_2 - 2 * (s_1 + s_2) = S - 2 * (s_1 + s_2). I.e., s_1 + s_2 = S/2, which is a Diophantine equation. Note that the solution to this equation can be used for computation of various values of a(n). See the example for an application.
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LINKS
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FORMULA
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Empirical: a(n) = a(n-1)+2*a(n-3)-a(n-4)-a(n-5)-a(n-6)-a(n-7)+2*a(n-8)+a(n-10)-a(n-11).
Empirical g.f.: x*(1 + x + x^3 + x^5) / ((1 - x)^4*(1 + x)*(1 + x^2)*(1 + x + x^2)^2). - Colin Barker, Oct 21 2017
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EXAMPLE
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All solutions for n=4:
-1 0 -2 -1 -1
0 0 0 -1 -1
0 0 1 0 1
1 0 1 2 1
Suppose we try to find a(6) and see that (0, 0, 1, 1, 2, 2) has sum of squares <= 2*6 = 12. This corresponds to x_1 = x_2 = 2 and gives S = x_1 + 2*x_2 = 2 + 2*2 = 6. We change signs so that s_1 + 2*s_2 = S/2 = 6/2 = 3, with s_1, s_2 in {0, 1, 2}. We see that (s_1, s_2) in {(1, 1)} so the only solution for this is (0, 0, -1, 1, -2, 2) which is (-2, -1, 0, 0, 1, 2) when sorted. - David A. Corneth, Oct 24 2017
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PROG
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(PARI) a(n) = {my(res = 0); for(x_2 = 0, (2*n - 1) \ 4, forstep(x_1 = 0, min(2*n - 4 * x_2 - 1, n), 2, if(x_1 + x_2 != 1 && x_1 + x_2 <= n && !(x_1 == 0 && x_2%2 == 1), res += (2 * (min((x_1 - 2*(x_2%2))\4, x_2\2) + 1) - !(x_2%2))))); res} \\ David A. Corneth, Oct 24 2017
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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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