|
| |
|
|
A309989
|
|
Digits of one of the two 17-adic integers sqrt(-1).
|
|
8
|
|
|
|
4, 2, 10, 5, 12, 16, 12, 8, 13, 3, 14, 0, 6, 1, 0, 15, 1, 8, 14, 5, 7, 16, 14, 1, 5, 13, 9, 6, 5, 12, 16, 15, 9, 16, 14, 12, 16, 1, 3, 6, 4, 10, 15, 5, 16, 12, 2, 1, 5, 4, 0, 15, 2, 11, 14, 9, 5, 1, 11, 16, 15, 7, 5, 6, 14, 3, 12, 0, 0, 11, 12, 13, 9, 5, 4, 16, 13
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,1
|
|
|
COMMENTS
|
This square root of -1 in the 17-adic field ends with digit 4. The other, A309990, ends with digit 13 (D when written as a 17-adic number).
|
|
|
LINKS
|
Seiichi Manyama, Table of n, a(n) for n = 0..10000
Wikipedia, p-adic number
|
|
|
FORMULA
|
a(n) = (A286877(n+1) - A286877(n))/17^n.
For n > 0, a(n) = 16 - A309990(n).
|
|
|
EXAMPLE
|
The solution to x^2 == -1 (mod 17^4) such that x == 4 (mod 17) is x == 27493 (mod 17^4), and 27493 is written as 5A24 in heptadecimal, so the first four terms are 4, 2, 10 and 5.
|
|
|
PROG
|
(PARI) a(n) = truncate(sqrt(-1+O(17^(n+1))))\17^n
|
|
|
CROSSREFS
|
Cf. A286877, A286878.
Digits of p-adic square roots:
A318962, A318963 (2-adic, sqrt(-7));
A271223, A271224 (3-adic, sqrt(-2));
A269591, A269592 (5-adic, sqrt(-4));
A210850, A210851 (5-adic, sqrt(-1));
A290794, A290795 (7-adic, sqrt(-6));
A290798, A290799 (7-adic, sqrt(-5));
A290796, A290797 (7-adic, sqrt(-3));
A051277, A290558 (7-adic, sqrt(2));
A321074, A321075 (11-adic, sqrt(3));
A321078, A321079 (11-adic, sqrt(5));
A322091, A322092 (13-adic, sqrt(-3));
A286838, A286839 (13-adic, sqrt(-1));
A322087, A322088 (13-adic, sqrt(3));
this sequence, A309990 (17-adic, sqrt(-1)).
Sequence in context: A138569 A191725 A198326 * A283938 A283943 A283942
Adjacent sequences: A309986 A309987 A309988 * A309990 A309991 A309992
|
|
|
KEYWORD
|
nonn,base
|
|
|
AUTHOR
|
Jianing Song, Aug 26 2019
|
|
|
STATUS
|
approved
|
| |
|
|
