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A159915
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a(n) = floor((n+1)/4)*floor(n/2).
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1
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0, 0, 0, 1, 2, 2, 3, 6, 8, 8, 10, 15, 18, 18, 21, 28, 32, 32, 36, 45, 50, 50, 55, 66, 72, 72, 78, 91, 98, 98, 105, 120, 128, 128, 136, 153, 162, 162, 171, 190, 200, 200, 210, 231, 242, 242, 253, 276, 288, 288, 300, 325, 338, 338, 351, 378, 392, 392, 406, 435, 450, 450
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OFFSET
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0,5
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COMMENTS
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Half the number of (n-2)-element subsets of {1,...,n} with odd sum of the elements.
This is half the antepenultimate column of A159916, cf. formula.
The number of subsets of {1,...,n} with n-2 elements, adding up to an odd integer, is always even (cf. examples), so we divide it by 2.
We prefer to include a(0)=a(1)=a(2)=0, even if it might seem more natural to start only at n=2 or n=3.
From the rational g.f. it can be seen that the sequence is a linear recurrence with constant coefficients (3,-5,7,-7,5,-3,1) of order 7.
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LINKS
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FORMULA
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G.f.: x^3(1 - x + x^2)/(1 - 3*x + 5*x^2 - 7*x^3 + 7*x^4 - 5*x^5 + 3*x^6 - x^7).
a(n) = 3*a(n-1) - 5*a(n-2) + 7*a(n-3) - 7*a(n-4) + 5*a(n-5) - 3*a(n-6) + a(n-7) for n > 7.
For n > 2, a(n) = A159916(n*(n-1)/2 + n - 2)/2 = T(n,n-2)/2 as defined there.
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EXAMPLE
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a(0)=a(1)=0 since there are no subsets with -2 or -1 elements.
a(2)=0 since the sum of the elements of a 0-element subset is zero.
a(3)=1 since for n=3 we have two singleton subsets of {1,2,3}, {1} and {3}, with odd sum of elements.
a(4)=2 since for n=4 we have four 2-element subsets of {1,2,3,4} with odd sum: {1,2}, {2,3}, {1,4}, {3,4}.
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PROG
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(PARI) A159915(n)=polcoeff((1-x+x^2)/(1-3*x+5*x^2-7*x^3+7*x^4-5*x^5+3*x^6-x^7)+O(x^(n-2)), n-3)
a(n, t=[0, 0, 0, 1, 2, 2, 3], c=[1, -3, 5, -7, 7, -5, 3]~)=while(n-->5, t=concat(vecextract(t, "^1"), t*c)); t[n+2] /* Note: a(n+1, [0, 0, 0, 0, 1, 2, 2]) gives the same result as a(n) */
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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