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0, 1, 2, 9, 28, 65, 126, 217, 344, 513, 730, 1001, 1332, 1729, 2198, 2745, 3376, 4097, 4914, 5833, 6860, 8001, 9262, 10649, 12168, 13825, 15626, 17577, 19684, 21953, 24390, 27001, 29792, 32769, 35938, 39305, 42876, 46657, 50654, 54873, 59320
(list; graph; refs; listen; history; internal format)
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OFFSET
| -1,3
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COMMENTS
| Sequence allows us to find X values of the equation: 1!*X^4 + 2!*(X - 1)^3 + 3!*(X - 2)^2 + (4^2)*(X -3) + 5^2 = Y^3. To prove that 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) must be a cube, so X = n^3 + 1 and Y = n(n^3 + 2). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Dec 04 2007
Number of units of a(n) belongs to a periodic sequence: 0, 1, 2, 9, 8, 5, 6, 7, 4, 3.We conclude that a(n) and a(n+10) have the same number of units. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 05 2009]
Where records occur in the (real) sequence Ceiling[k^(1/3)] - k^(1/3), k=1,2,3,... [From John W. Layman (layman(AT)math.vt.edu), Sep 07 2010]
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LINKS
| Nathaniel Johnston, Table of n, a(n) for n = -1..10000
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MATHEMATICA
| Table[n^3+1, {n, -1, 5!}] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), May 27 2010]
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CROSSREFS
| Sequence in context: A155472 A100293 A202679 * A121643 A183376 A131066
Adjacent sequences: A001090 A001091 A001092 * A001094 A001095 A001096
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KEYWORD
| nonn
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Ray Wills [ rwills(AT)vmprofs.estec.esa.nl ]
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EXTENSIONS
| More terms from James A. Sellers (sellersj(AT)math.psu.edu), Sep 19 2000
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