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Search: id:A135300
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| A135300 |
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Sequence allows us to find X positive values of the equation: 1!*X^4 - 2!*(X + 1)^3 + 3!*(X + 2)^2 - (4^2)*(X + 3) + 5^2 = Y^3. |
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+0
2
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| 1, 7, 26, 63, 124, 215, 342, 511, 728, 999, 1330, 1727, 2196, 2743, 3374, 4095, 4912, 5831, 6858, 7999, 9260, 10647, 12166, 13823, 15624, 17575, 19682, 21951, 24388, 26999, 29790, 32767, 35936, 39303, 42874
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OFFSET
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1,2
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COMMENT
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To prove that X = 1 or X = n^3 - 1: Y^3 = 1!*X^4 - 2!*(X + 1)^3 + 3!*(X + 2)^2 - (4^2)*(X + 3) + 5^2 = X^4 - 2*(X + 1)^3 + 6*(X + 2)^2 - 16(X + 3) + 25 = X^4 - 2*X^3 + 2X - 1 = (X + 1)(X^3 - 3*X^2 + 3X - 1) = (X + 1)*(X - 1)^3 it means: X = 1 or (X + 1) must be a cube, so (X, Y) = (1, 0) or (X, Y) = (n^3 - 1, n(n^3 - 2)) with n>=2.
Apart from the first term, the same as A068601. R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 29 2008
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FORMULA
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a(1) = 1 and a(n)= n^3 - 1 with n>=2.
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CROSSREFS
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Cf. A001093.
Adjacent sequences: A135297 A135298 A135299 this_sequence A135301 A135302 A135303
Sequence in context: A049453 A046433 A128972 this_sequence A024001 A068601 A006325
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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), Dec 04 2007
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