|
|
A304023
|
|
a(n) is the smallest integer with n digits in base 3/2 expressed in base 3/2.
|
|
3
|
|
|
0, 20, 210, 2100, 21010, 210110, 2101100, 21011000, 210110000, 2101100010, 21011000110, 210110001100, 2101100011010, 21011000110100, 210110001101000, 2101100011010010, 21011000110100110, 210110001101001100, 2101100011010011010, 21011000110100110100, 210110001101001101010
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Excluding 0, every term starts with 2 and has exactly one 2.
The last digit is always zero.
Removing the last digit produces the sequence A303500 of the smallest even integers in base 3/2.
The value of this sequence in base 10 is A070885.
When subtracting 1 from the value of this sequence we get A304025.
The largest integer with a given number of digits in base 3/2 can be produced directly from this sequence by replacing 21 at the beginning and 0 at the end with 2, and by shifting the rest up by 1, see sequence A304024.
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
The number 5 in base 3/2 is 22, and the number 6 is 210. Therefore, 210 is the smallest three-digit integer.
|
|
MAPLE
|
b:= proc(n) b(n):= `if`(n=1, 1, 3*ceil(b(n-1)/2)) end:
g:= proc(n) g(n):= `if`(n<2, 0, irem(n, 3, 'q')+g(2*q)*10) end:
a:= n-> g(b(n)):
|
|
PROG
|
(Python)
def f(n): return 0 if n < 1 else f(n//3*2)*10 + n%3
def a(n):
k = 0
while len(str(f(k))) != n: k += 1
return f(k)
(PARI) f(n) = if( n<1, 0, f(n\3 * 2) * 10 + n%3);
a(n) = {my(k=0); while(#Str(f(k)) != n, k++); f(k); } \\ Michel Marcus, Jun 19 2018
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|