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A115878
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a(n) is the number of positive solutions of the Diophantine equation x^2 = y(y+n).
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10
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0, 0, 1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 4, 2, 1, 2, 1, 1, 4, 1, 1, 4, 2, 1, 3, 1, 1, 4, 1, 3, 4, 1, 4, 2, 1, 1, 4, 4, 1, 4, 1, 1, 7, 1, 1, 7, 2, 2, 4, 1, 1, 3, 4, 4, 4, 1, 1, 4, 1, 1, 7, 4, 4, 4, 1, 1, 4, 4, 1, 7, 1, 1, 7, 1, 4, 4, 1, 7, 4, 1, 1, 4, 4, 1, 4, 4, 1, 7, 4, 1, 4, 1, 4, 10, 1, 2, 7, 2, 1, 4
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OFFSET
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1,9
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COMMENTS
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For odd n, a(n) can be computed from the prime signature. - David A. Corneth, Oct 07 2018
Number of r X s rectangles with integer side lengths such that r + s = n, r < s and (s-r) | (s*r). - Wesley Ivan Hurt, Apr 24 2020
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LINKS
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FORMULA
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a((2k+1) * 2^m) = floor(tau((2k + 1) ^ 2) / 2) for m <= 2.
a((2k+1) * 2^m) = (2m - 3) * a(2k+1) + (m-2) for m > 2. (End)
a(n) = Sum_{i=1..floor((n-1)/2)} (1 - ceiling(i*(n-i)/(n-2*i)) + floor(i*(n-i)/(n-2*i))). - Wesley Ivan Hurt, Apr 24 2020
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EXAMPLE
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a(15) = 4 since there are 4 solutions (x,y) to x^2 = y(y+15), namely (4,1), (10,5), (18, 12) and (56, 49).
Note how each x is obtained from each such divisor pair n2/d and d of n2 as (n2/d - d)/4, when their difference is a positive multiple of four, thus in case of n2 = 15^2 = 225 we get (225/1 - 1)/4 = 56, (225/3 - 3)/4 = 18, (225/5 - 5) = 10 and (225/9 - 9)/4 = 4. - Antti Karttunen, Oct 06 2018
a(96) = 10. We compute P, the largest power of 2 dividing n = 96. Then compute min(P, 4) and divide n by it. This gives 96/4 = 24. Then find the number of divisors of 24^2, which is 21. Dividing by 2 rounding down to the nearest integer gives 10, the value of a(96). - David A. Corneth, Oct 06 2018
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MATHEMATICA
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a[n_] := Sum[Boole[d^2 < n^2 && Mod[n^2/d-d, 4] == 0], {d, Divisors[n^2]}];
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PROG
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(PARI) A115878(n) = { my(n2 = n^2); sumdiv(n2, d, ((d*d)<n2)&&(0==(((n2/d)-d)%4))); }; \\ Antti Karttunen, Oct 06 2018
(PARI) a(n) = my(v=min(2, valuation(n, 2))); numdiv((n>>v)^2)>>1 \\ David A. Corneth, Oct 06 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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