Counting +1 primes
JavaScript ES6, 141 bytes 143 147 160
Saves 13 bytes, thanks to @Naouak
n=>[...t=n.toString(2)].map((l,i)=>t.slice(0,v=i+1)+1+t.slice(v)).filter((l,i,a)=>a.indexOf(l)==i&&(p=(n,c)=>n%c&&c>n-2||p(n,++c))('0b'+l,2))
Similar method to my TeaScript answer, uses RegExp (you heard me right) to check for primes.
Ungolfed
n=>
[...t = n.toString(2)] // To binary
.map((l,i)=> // Make cycles
t.slice(0, v = i+1)
+ 1
+ t.slice(v)
).filter((l,i,a)=>
a.indexOf(l) == i && // Remove Duplicates
(p=(n,c)=> // Prime checking
n % c &&
c > n - 2 ||
p(n,++c)
)('0b'+l,2)
).length
Pyth, 20 bytes
s/LPd{mij\1c.BQ]d2hQ
Test Suite
s/LPd{mij\1c.BQ]d2hQ
Q = eval(input())
m hQ For insertion position in [0 ... Q]
.BQ Convert Q to binary string
c ]d Chop at insertion position
j\1 Join on '1'
i 2 Convert to integer
{ Deduplicate
/LPd Map each number to the number of times it occurs in its
prime factorization, e.g. whether or not it is prime.
s Sum and print.
TeaScript, 22 bytes
x÷¿®x÷E(i¬,1)¤©d¡F(¥)n
TeaScript is starting to look like APL... The special characters are converted to longer, commonly repeated sequences
Online interpreter be sure to check "Inputs are numbers."
Explanation && Ungolfed
xT(2)s``m(#P(xT(2)E(i+1,1),2))d()F($P)n
xT(2) // Take input, convert to binary
s``m(# // Loop over input
P( // Convert to decimal...
xT(2) // Input to binary
E(i+1,1) // Inset 1 into (above) at current index in loop
,2)
)d() // Remove duplicates
F($P) // Filter items that aren't prime
n // Grab length.