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A362564
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a(n) is the largest integer x such that n + 2^x is a square, or -1 if no such number exists.
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1
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3, 1, 0, 5, 2, -1, 1, 3, 4, -1, -1, 2, -1, 1, 0, 7, 9, -1, -1, 4, 2, -1, 1, 0, -1, -1, -1, 3, -1, -1, -1, 5, 8, 1, 0, 6, -1, -1, -1, -1, 7, -1, -1, -1, 2, -1, 1, 4, 5, -1, -1, -1, -1, -1, -1, 3, 6, -1, -1, 2, -1, 1, 0, 9, 10, -1, -1, 11, -1, -1, -1, -1, 3, -1, -1, -1, 2, -1, 1, 6, -1, -1, -1, 4, -1
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OFFSET
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1,1
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COMMENTS
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a(n) is the maximum integer solution x of the indefinite equation n + 2^x = r^2 where n is a constant and r is a positive integer, or -1 if there are no solutions.
See A247763 for the number of solutions and for further information, references and links about this problem.
If n == 0 (mod 4), we first try x = 0 or 1; when x >= 2, the result can be derived from the result for n/4 (see formula).
If n == 2 (mod 4), the only possible values of x is 1, as otherwise n + 2^x == 2 (mod 4), so nonsquare.
If n == 3 (mod 4), the only possible values of x are 0 and 1, as otherwise n + 2^x == 3 (mod 4), so nonsquare.
If n == 1 (mod 4), or n = 4*k + 1 (k >= 0): we suggest that r = 2*m + 1, 4*k + 1 + 2^x = (2*m + 1)^2, thus k + 2^(x - 2) = m*(m + 1); if k is odd, the only possible values of x is 2 because m*(m + 1) is even.
If n = k^2 (k >= 1), 2^x = (r + k)*(r - k), so 2k must be in the form 2^i - 2^j.
The problem can be solved via reduction to three Mordell curves: n + z^3 = y^2, n + 2z^3 = y^2 or equivalently 4n + (2z)^3 = (2y)^2, n + 4z^3 = y^2 or equivalently 16n + (4z)^3 = (4y)^2, where z := 2^floor(x/3). For a given n, each of these three curves is known to have only a finite number of integer points (y,z), proving that x cannot be unbounded. - Max Alekseyev, Apr 26 2023
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LINKS
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FORMULA
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a(4*k + 2) = 1 if k + 1 is a square, or -1 otherwise.
a(4*k + 3) = 1 if 4*k + 5 is a square, or 0 is k + 1 is a square and 4*k + 5 is a nonsquare, or -1 otherwise.
a(4*k + 4) = a(k) + 2 if a(k) >= 0, or 0 if 4*k + 1 is a square and a(k) = -1, or -1 otherwise.
a(8*k + 5) = 2 if 8*k + 9 is a square, or -1 otherwise.
a((2^i - 2^j)^2) = i + j + 2 for i,j >= 0.
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EXAMPLE
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See COMMENTS section for further proof.
For n = 1, 1 + 2^3 = 9 = 3^2;
for n = 4, 4 + 2^5 = 36 = 6^2;
for n = 7, 7 + 2^1 = 9 = 3^2;
for n = 9, 9 + 2^4 = 25 = 5^2.
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PROG
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(Sage) def a362564(n): return max((3*v-2*k for k in range(3) for z, _, _ in EllipticCurve([0, 4^k*n]).integral_points() if z>=1<<k and z==1<<(v:=valuation(z, 2))), default=-1) # Max Alekseyev, Apr 26 2023
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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