|
|
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;
text;
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
|
|
|
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
|
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|