The highest number is a 4, which would make it ideal for repeating yourself and it starts with a 2.

Perform a Run Length Encoding with the numbers, starting with 0
0011011001111001011001100111011100110100011011010111010101101111011011100110101100111001011110100011011101101010
This gives a nice 112 character long string, which can be nicely grouped into pieces of 8
00110110 01111001 01100110 01110111 00110100 01101101 01110101 01101111 01101110 01101011 00111001 01111010 00110111 01101010
Convert binary to ASCII for the final solve
6yfw4muonk9z7j

Convert each character to it’s Morse code
..- -.-- ..- -..- --.- -... --.- ---.. ...-- -.... --- .--- .--. -.... ----. .-. --.. -... --... .-. --.- ....- -.-. -... --.- -..- --- .-.. .--- ....- --.. -- 
Converting . to 0 and - to 1, we get something that we can once again divide nicely into 8 groups
00110110 01100111 01100011 01111000 00111000 01110111 01101000 01111001 01100100 01100001 01101000 01101010 00110110 01111010 00111000 01110011
Converting binary to ASCII, we get our solve
6gcx8whydahj6z8s

Sofar we had two binary solves, one involving RLE. Might it be that it is needed again?

Obviously we cannot run alphabet characters through a RLE cipher, but we can convert them to their alphabet index first.
Convert each character to it’s alphabet index, where a=0 and z=25:
22 11 1 11 22 13 21 21 11 21 15 21 3 1 21 4 13 5 22 11 11 3 21 22 4 11 3 11 3 21 12 11 3 12 22 21 11 13 12 3 21 22 21 15 21 3 12 4 1
This looks promising, we are starting with a 2 again and no number higher than 5.
Let us group up the numbers
221111122132121112115213121413522111132122411311321121131222211113123212221152131241
and then perform a Run Length Encoding with them
00110101011001000110110101101000001101110110111101110000011001010111001001100001011101000110100101110110011001010111011000110110011010000011011101100001
What do you know, another long string that can be grouped into 8 groups
00110101 01100100 01101101 01101000 00110111 01101111 01110000 01100101 01110010 01100001 01110100 01101001 01110110 01100101 01110110 00110110 01101000 00110111 01100001
Finally convert binary to ASCII for the solve
5dmh7operativev6h7a

Look at all the information in the picture

Taking the obvious first, we see a long string of 1 and 0, with a single 2 in there.
Read out the entire string, starting from the single 2
200110001001101000011011000110010001101100011010100110011001100000011000000110000001100000011000000110000
Reading the 2 as a hint to binary and seeing that the remaining character’s length is divisible by 8, let us group them
00110001 00110100 00110110 00110010 00110110 00110101 00110011 00110000 00110000 00110000 00110000 00110000 00110000
Converting Binary to ASCII we get
1462653000000
Recognize that it look very much like a Unix timestamp, we use an Timestamp Epoch convert and get the UTC date and time of
2016-05-07T20:30:00+00:00

So now we have a date and time, but that doesn’t mean much, lets have a look at the other clues.
The filename of image was lccn-n96059244.jpg
Googling a bit, we see that lccn stands for Library of Congress Control Number.
Taking that knowledge, we search their database and find that n96059244 points to Tim Hawkinson, which corresponds with what looks to be in the image.
Over at the University of California, San Diego, there is a Stuart collection which includes the Tim Hawkinson Bear.
So it seems that something might happen at or around the bear on the 7th May 2016 at 20:30, which was confirmed later in the post “A Mission in San Diego”