|
| |
|
|
A138851
|
|
Nearest integer to 1/(round(x)-x), where exp(pi sqrt(n))-744 = (12(x^2-1))^3.
|
|
4
| |
|
|
-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; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 5,1
|
|
|
COMMENTS
| 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.
|
|
|
LINKS
| T. Piezas, "More on e^(pi*sqrt(163))" on sci.math.research, April 13, 2008 and The Ramanujan Pages
|
|
|
EXAMPLE
| 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.
|
|
|
PROG
| (PARI) default(realprecision, 200); A138851(n)={ n=frac( sqrt( sqrtn( exp( sqrt(n)*Pi )-744, 3)/12 + 1 )); round( 1/(round(n)-n)) }
|
|
|
CROSSREFS
| Cf. A003173, A014708, A056581 and references therein.
Sequence in context: A117462 A155462 A109496 * A181061 A090342 A010307
Adjacent sequences: A138848 A138849 A138850 * A138852 A138853 A138854
|
|
|
KEYWORD
| sign
|
|
|
AUTHOR
| M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Apr 16 2008
|
| |
|
|