Django pagination (get the page number corresponding to the object)

Question

I have paginate I'm trying to get the index page from an object page (e.g. pagination in reverse order)

get_paginated_posts returns a paginator for the model Post:

class PostManager(models.Manager):
    def get_paginated_posts(self, request=None):
        if request and request.user.has_perm('blog.change_post'):
            posts = super(PostManager, self).filter(is_update=False)
        else:        
            posts = super(PostManager, self).filter(publish=True, is_update=False)
        return Paginator(posts, POSTS_PER_PAGE)
    .
    .

This is my model

class Post(models.Model):
    .
    .
    .
    def get_page(self, request=None):
        paginator = Post.objects.get_paginated_posts(request)
        for i in range(1, paginator.num_pages+1):
            if self in paginator.page(i).object_list:                
                return i
            pass
        return False 

I am concerned about the call Post.objects.get_paginated_poststo the get_page function.
Is it correct to call a class Postfrom an instance? Is there any other better way to make this possible?
Why can't I call super(Post, self).objects.get_paginated_poststo do the same?
I understand that it self.objects.get_paginated_postsdoes not work due to the lack of access for the object to its manager.

solvable

Final code proposed by Tomas Elendt:

class PostManager(models.Manager):
    def get_paginated_posts(self, user=None):
        if user and user.has_perm('blog.change_post'):
            posts = super(PostManager, self).filter(is_update=False)
        else:        
            posts = super(PostManager, self).filter(publish=True, is_update=False)
        return Paginator(posts, POSTS_PER_PAGE)

class Post(models.Model):
    .
    def get_page(self, request=None):
        return self._default_manager.filter(is_update = False, time__gt=self.time).count()/POSTS_PER_PAGE +1 
        #Just a one line now :P 

Answer 1

, . , - !

, Post (, Paginator). , , -. POSTS_PER_PAGE, .

IMHO PostManager Post . , , , user_id ( ).

EDIT:

from django.db import models
from django.contrib.auth.models import User

POSTS_PER_PAGE = 10

class Post(models.Model):
    """
    >>> from datetime import datetime, timedelta
    >>> from django.db import connection
    >>> from django.conf import settings
    >>>
    >>> user = User.objects.create_user("test", "[e-mail]")
    >>> for i in xrange(100):
    ...     p = Post.objects.create(author=user,
    ...                             pub_date=datetime.now() - timedelta(hours=i))
    >>> post = Post.objects.all()[68]
    >>> settings.DEBUG = True    # monkey-patching settings - ugly
    >>> connection.queries = []  # cleaning previous queries
    >>> post.get_page()
    7
    >>> len(connection.queries)  # print number of queries of `get_page` call
    1
    """
    pub_date = models.DateTimeField(auto_now_add=True)
    author = models.ForeignKey(User)
    class Meta:
        ordering = ["-pub_date"]

    def get_page(self):
        return self._default_manager.filter(author__id=self.author_id).filter(
            pub_date__gt=self.pub_date).count() / POSTS_PER_PAGE + 1