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Search: id:A138851
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| A138851 |
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Nearest integer to 1/(round(x)-x), where exp(pi sqrt(n))-744 = (12(x^2-1))^3. |
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+0
4
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| -4, -3, -2, 2, 3, 5, 12, -33, -7, -4, -2, 2, 3, 6, 8954018, -6, -3, 2, 3, 9, -12, -3, -2, 4, 18, -6, -2, 3, 14, -5, -2, 4, -21, -3, 3, 51, -3, 3, 2683620901418, -3, 4, -9, 2, 11, -3, 4, -5, 3, -10, 2, -17, 2, -14, 2, -7, 3, -4, 7, -2, -16, 3, -3, 31514540715033062, 3, -3, -12, 5, 2, -3, -9, 12, 4, 2, -2, -3, -4, -7, -10, -16, -19, -16
(list; graph; listen)
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OFFSET
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5,1
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COMMENT
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Records are attained at the larger Heegener numbers (A003173).
T. Piezas draws attention on the fact that the integers very close to exp(pi sqrt(n)) are of the form (12(k^2-1))^3+744. Here this closeness is expressed as the (rounded value) of the reciprocal of the (signed) distance of these k-values from the integers.
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LINKS
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T. Piezas, "More on e^(pi*sqrt(163))" on sci.math.research, April 13, 2008 and The Ramanujan Pages
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EXAMPLE
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We have e^(pi sqrt(19))-744 = (12(x^2-1))^3 with x = 2.9999998883... = 3 - 1/8954017.533..., therefore a(19) = 8954018.
In the same way, e^(pi sqrt(163))-744 = (12(x^2-1))^3 with x = 230.999999999999999999999999999890... = 231 - 1/9093255353570474976233448828.20..., thus a(163) = 9093255353570474976233448828.
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PROGRAM
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(PARI) default(realprecision, 200); A138851(n)={ n=frac( sqrt( sqrtn( exp( sqrt(n)*Pi )-744, 3)/12 + 1 )); round( 1/(round(n)-n)) }
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CROSSREFS
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Cf. A003173, A014708, A056581 and references therein.
Sequence in context: A117462 A155462 A109496 this_sequence A090342 A010307 A001178
Adjacent sequences: A138848 A138849 A138850 this_sequence A138852 A138853 A138854
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KEYWORD
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sign
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AUTHOR
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M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Apr 16 2008
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