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Search: id:A132411
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| A132411 |
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Sequence allows us to find X values of the equation: X^3 - (X + 1)^2 + X + 2 = Y^2. |
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+0
4
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| 0, 1, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763
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OFFSET
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0,3
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COMMENT
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To prove that X = 1 or X = n^2 - 1: Y^2 = X^3 - (X + 1)^2 + X + 2 = X^3 - X^2 - X + 1 = (X + 1)(X^2 - 2X + 1) = (X + 1)*(X - 1)^2 it means: X = 1 or (X + 1) must be a perfect square, so X = 1 or X = n^2 - 1 with n>=1. which gives: (X, Y) = (0, 1) or (X, Y) = (1, 0) or (X, Y) = (n^2 - 1, n*(n^2 - 2))with n>=2.
An equivalent technique of integer factorization would work for example for the equation X^3+3*X^2-9*X+5=(X+5)(X-1)^2=Y^2, looking for perfect squares of the form X+5=n^2. Another example is X^3+X^2-5*X+3=(X+3)*(X-1)^2=Y^2 with solutions generated from perfect squares of the form X+3=n^2. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 20 2007
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FORMULA
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a(o) = 0, a(1) = 1 and a(n) = n^2 - 1 with n>=2.
G.f.: x+x^2*(-3+x)/(-1+x)^3 . - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 20 2007
Starting (1, 3, 8, 15, 24,...) = binomial transfor of [1, 2, 3, -1, 1, -1,...] - Gary W. Adamson (qntmpkt(AT)yahoo.com), May 12 2008
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EXAMPLE
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0^3 - 1^2 + 2 = 1^2, 1^3 - 2^2 + 3 = 0^2, 3^3 - 4^2 + 5 = 4^2.
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CROSSREFS
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Cf. A028560, A005563.
Sequence in context: A083656 A013648 A005563 this_sequence A147998 A131386 A164003
Adjacent sequences: A132408 A132409 A132410 this_sequence A132412 A132413 A132414
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KEYWORD
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nonn
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AUTHOR
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Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 12 2007
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