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Why does converting from np.float16 to np.float32 modify the value?

When converting a number from half to single floating representation I see a change in the numeric value.
Here I have 65500 stored as a half precision float, but upgrading to single precision changes the underlying value to 65504, which is many floating point increments away from the target.
In this specific case, why does this happen?
(Pdb) np.asarray(65500,dtype=np.float16).astype(np.float32)
array(65504., dtype=float32)
As a side note, I also observe
(Pdb) int(np.finfo(np.float16).max)
65504

The error is not "many floating point increments away" [corrected to match OP's improved wording]. Read the standard IEEE 754-2008. It specifies 10 bits for the mantissa, or 1024 distinct values. Your value is on the close order of 2^16, so you have an increment of 2^6, or 64.
The format also gives 1 bit for the sign and 5 for the characteristic (exponent).
65500 is stored as something equivalent to + 2^6 * 1023.5. This translates directly to 65504 when you convert to float32. You lost the precision when you converted your larger number to 10 bits of precision. When you convert in either direction, the result is always constrained by the less-precise type.

Python: Float to Decimal conversion and subsequent shifting may give wrong result

When I convert a float to decimal.Decimal in Python and afterwards call Decimal.shift it may give me completely wrong and unexpected results depending on the float. Why is this the case?
Converting 123.5 and shifting it:
from decimal import Decimal
a = Decimal(123.5)
print(a.shift(1)) # Gives expected result
The code above prints the expected result of 1235.0.
If I instead convert and shift 123.4:
from decimal import Decimal
a = Decimal(123.4)
print(a.shift(1)) # Gives UNexpected result
it gives me 3.418860808014869689941406250E-18 (approx. 0) which is completely unexpected and wrong.
Why is this the case?
Note:
I understand the floating-point imprecision because of the representation of floats in memory. However, I can't explain why this should give me such a completely wrong result.
Edit:
Yes, in general it would be best to not convert the floats to decimals but convert strings to decimals instead. However, this is not the point of my question. I want to understand why the shifting after float conversion gives such a completely wrong result. So if I print(Decimal(123.4)) it gives 123.40000000000000568434188608080148696899414062 so after shifting I would expect it to be 1234.0000000000000568434188608080148696899414062 and not nearly zero.

You need to change the Decimal constructor input to use strings instead of floats.
a = Decimal('123.5')
print(a.shift(1))
a = Decimal('123.4')
print(a.shift(1))
or
a = Decimal(str(123.5))
print(a.shift(1))
a = Decimal(str(123.4))
print(a.shift(1))
The output will be as expected.
>>> 1235.0
>>> 1234.0
Decimal instances can be constructed from integers, strings, floats, or tuples. Construction from an integer or a float performs an exact conversion of the value of that integer or float.
For floats, Decimal calls Decimal.from_float()
Note that Decimal.from_float(0.1) is not the same as Decimal('0.1'). Since 0.1 is not exactly representable in binary floating point, the value is stored as the nearest representable value which is 0x1.999999999999ap-4. The exact equivalent of the value in decimal is 0.1000000000000000055511151231257827021181583404541015625.
Internally, the Python decimal library converts a float into two integers representing the numerator and denominator of a fraction that yields the float.
n, d = abs(123.4).as_integer_ratio()
It then calculates the bit length of the denominator, which is the number of bits required to represent the number in binary.
k = d.bit_length() - 1
And then from there the bit length k is used to record the coefficient of the decimal number by multiplying the numerator * 5 to the power of the bit length of the denominator.
coeff = str(n*5**k)
The resulting values are used to create a new Decimal object with constructor arguments of sign, coefficient, and exponent using this values.
For the float 123.5 these values are
>>> 1 1235 -1
and for the float 123.4 these values are
1 123400000000000005684341886080801486968994140625 -45
So far, nothing is amiss.
However when you call shift, the Decimal library has to calculate how much to pad the number with zeroes based on the shift you've specified. To do this internally it takes the precision subtracted by length of the coefficient.
amount_to_pad = context.prec - len(coeff)
The default precision is only 28 and with a float like 123.4 the coefficient becomes much longer than the default precision as noted above. This creates a negative amount to pad with zeroes and makes the number very tiny as you noted.
A way around this is to increase the precision to the length of the exponent + the length of the number you started with (45 + 4).
from decimal import Decimal, getcontext
getcontext().prec = 49
a = Decimal(123.4)
print(a)
print(a.shift(1))
>>> 123.400000000000005684341886080801486968994140625
>>> 1234.000000000000056843418860808014869689941406250
The documentation for shift hints that the precision is important for this calculation:
The second operand must be an integer in the range -precision through precision.
However it does not explain this caveat for floats that don't play nice with memory limitations.
I would expect this to raise some kind of error and prompt you to change your input or increase the precision, but at least you know!
#MarkDickinson noted in a comment above that you can view this Python bug tracker for more information: https://bugs.python.org/issue7233

How many digits can float8, float16, float32, float64, and float128 contain?

Numpy's dtype documentation only shows "x bits exponent, y bits mantissa" for each float type, but I couldn't translate that to exactly how many digits before/after the decimal point. Is there any simple formula/table to look this up in?

This is not as simple as usually expected. For accuracy of mantissa, there generally are two values:
Given a value in decimal representation, how many decimal digits can be guaranteedly preserved if converted from decimal to a selected binary format and back (with default rounding).
Given a value in binary format, how many decimal digits are needed if value is converted to decimal format and back to original binary format (again, with default rounding) to get the original value unchanged.
In both cases, decimal representation is treated as independent of used exponent, without leading and trailing zeros (for example, all of 0.0123e4, 1.23e2, 1.2300e2, 123, 123.0, 123000.000e-3 are 3 digits).
For 32-bit binary float, these two sizes are 6 and 9 decimal digits, respectively. In C <float.h>, these are FLT_DIG and FLT_DECIMAL_DIG. (This is weird that 32-bit float keeps 7 decimal digits for total most of all numbers, but there are exceptions.)
In C++, look at std::numeric_limits<float>::digits10 and std::numeric_limits<float>::max_digits10, respectively.
For 64-bit binary float, these are 15 and 17 (DBL_DIG and DBL_DECIMAL_DIG, respectively; and std::numeric_limits<double>::{digits10, max_digits10}).
General formulas for them (thx2 #MarkDickinson)
${format}_DIG (digits10): floor((p-1)*log10(2))
${format}_DECIMAL_DIG (max_digits10): ceil(1+p*log10(2))
where p is number of digits in mantissa (including hidden one for normalized IEEE754 case).
Also, comments with some mathematical explanation at C++ numeric limits page:
The standard 32-bit IEEE 754 floating-point type has a 24 bit fractional part (23 bits written, one implied), which may suggest that it can represent 7 digit decimals (24 * std::log10(2) is 7.22), but relative rounding errors are non-uniform and some floating-point values with 7 decimal digits do not survive conversion to 32-bit float and back: the smallest positive example is 8.589973e9, which becomes 8.589974e9 after the roundtrip. These rounding errors cannot exceed one bit in the representation, and digits10 is calculated as (24-1)*std::log10(2), which is 6.92. Rounding down results in the value 6.
Look for values for 16- and 128-bit floats in comments (but see below for what is 128-bit float in real).
For exponent, this is simpler because each of the border values (minimum normalized, minimum denormalized, maximum represented) are exact and can be easily obtained and printed.
#PaulPanzer suggested numpy.finfo. It gives first of these values ({format}_DIG); maybe it is the thing you search:
>>> numpy.finfo(numpy.float16).precision
3
>>> numpy.finfo(numpy.float32).precision
6
>>> numpy.finfo(numpy.float64).precision
15
>>> numpy.finfo(numpy.float128).precision
18
but, on most systems (my one was Ubuntu 18.04 on x86-84) the value is confusing for float128; it is really for 80-bit x86 "extended" float with 64 bits significand; real IEEE754 float128 has 112 significand bits and so real value shall be around 33, but numpy presents another type under this name. See here for details: in general, float128 is a delusion in numpy.
UPD3: you mentioned float8 - there is no such type in IEEE754 set. One could imagine such type for some utterly specific purposes, but its range will bee too narrow for any universal usage.

To keep it simple, generally
Data-Type | Precision
----------------------
float16 | 3
float32 | 7
float64 | 15
float128 | 18

Python: Why does have 2^-n work for n>52 and not 1+2^-n-1?

I'm pretty new to python, and I've made a table which calculates T=1+2^-n-1 and C=2^n, which both give the same values from n=40 to n=52, but for n=52 to n=61 I get 0.0 for T, whereas C gives me progressively smaller decimals each time - why is this?
I think I understand why T becomes 0.0, because of python using binary floating point and because of the machine epsilon value - but I'm slightly confused as to why C doesn't also become 0.0.
import numpy as np
import math
t=np.zeros(21)
c=np.zeros(21)
for n in range(40,61):
m=n-40
t[m]=1+2**(-n)-1
c[m]=2**(-n)
print (n,t[m],c[m])

The "floating" in floating point means that values are represented by storing a fixed number of leading digits and a scale factor, rather than assuming a fixed scale (which would be fixed point).
2**-53 only takes one (binary) digit to represent (not including the scale), but 1+2**-53 would take 54 to represent exactly. Python floats only have 53 binary digits of precision; 2**-53 can be represented exactly, but 1+2**-53 gets rounded to exactly 1, and subtracting 1 from that gives exactly 0. Thus, we have
>>> 2**-53
1.1102230246251565e-16
>>> 1+(2**-53)-1
0.0
Postscript: you might wonder why 2**-53 displays as a value not equal to the exact mathematical value when I said it was exact. That's due to the float->string conversion logic, which only keeps enough decimal digits to reconstruct the original float (instead of printing a bunch of digits at the end that are usually just noise).

The difference between both is indeed due to floating-point representation. Indeed, if you perform 1 + X where X is a very very small number, then the floating-point representation sets its exponent value to 0 and the precision is ensured by the mantissa, which is 52-bit on a 64-bit computer. Therefore, 1 + 2^(-X) if X > 52 is equal to 1. However, even 2^-100 can be represented in double-precision floating-point, so you can see C decrease for a larger number of samples.

Safest way to convert float to integer in python?

Python's math module contain handy functions like floor & ceil. These functions take a floating point number and return the nearest integer below or above it. However these functions return the answer as a floating point number. For example:
import math
f=math.floor(2.3)
Now f returns:
2.0
What is the safest way to get an integer out of this float, without running the risk of rounding errors (for example if the float is the equivalent of 1.99999) or perhaps I should use another function altogether?

All integers that can be represented by floating point numbers have an exact representation. So you can safely use int on the result. Inexact representations occur only if you are trying to represent a rational number with a denominator that is not a power of two.
That this works is not trivial at all! It's a property of the IEEE floating point representation that int∘floor = ⌊⋅⌋ if the magnitude of the numbers in question is small enough, but different representations are possible where int(floor(2.3)) might be 1.
To quote from Wikipedia,
Any integer with absolute value less than or equal to 224 can be exactly represented in the single precision format, and any integer with absolute value less than or equal to 253 can be exactly represented in the double precision format.

Use int(your non integer number) will nail it.
print int(2.3) # "2"
print int(math.sqrt(5)) # "2"

You could use the round function. If you use no second parameter (# of significant digits) then I think you will get the behavior you want.
IDLE output.
>>> round(2.99999999999)
3
>>> round(2.6)
3
>>> round(2.5)
3
>>> round(2.4)
2

Combining two of the previous results, we have:
int(round(some_float))
This converts a float to an integer fairly dependably.

That this works is not trivial at all! It's a property of the IEEE floating point representation that int∘floor = ⌊⋅⌋ if the magnitude of the numbers in question is small enough, but different representations are possible where int(floor(2.3)) might be 1.
This post explains why it works in that range.
In a double, you can represent 32bit integers without any problems. There cannot be any rounding issues. More precisely, doubles can represent all integers between and including 253 and -253.
Short explanation: A double can store up to 53 binary digits. When you require more, the number is padded with zeroes on the right.
It follows that 53 ones is the largest number that can be stored without padding. Naturally, all (integer) numbers requiring less digits can be stored accurately.
Adding one to 111(omitted)111 (53 ones) yields 100...000, (53 zeroes). As we know, we can store 53 digits, that makes the rightmost zero padding.
This is where 253 comes from.
More detail: We need to consider how IEEE-754 floating point works.
1 bit 11 / 8 52 / 23 # bits double/single precision
[ sign | exponent | mantissa ]
The number is then calculated as follows (excluding special cases that are irrelevant here):
-1sign × 1.mantissa ×2exponent - bias
where bias = 2exponent - 1 - 1, i.e. 1023 and 127 for double/single precision respectively.
Knowing that multiplying by 2X simply shifts all bits X places to the left, it's easy to see that any integer must have all bits in the mantissa that end up right of the decimal point to zero.
Any integer except zero has the following form in binary:
1x...x where the x-es represent the bits to the right of the MSB (most significant bit).
Because we excluded zero, there will always be a MSB that is one—which is why it's not stored. To store the integer, we must bring it into the aforementioned form: -1sign × 1.mantissa ×2exponent - bias.
That's saying the same as shifting the bits over the decimal point until there's only the MSB towards the left of the MSB. All the bits right of the decimal point are then stored in the mantissa.
From this, we can see that we can store at most 52 binary digits apart from the MSB.
It follows that the highest number where all bits are explicitly stored is
111(omitted)111. that's 53 ones (52 + implicit 1) in the case of doubles.
For this, we need to set the exponent, such that the decimal point will be shifted 52 places. If we were to increase the exponent by one, we cannot know the digit right to the left after the decimal point.
111(omitted)111x.
By convention, it's 0. Setting the entire mantissa to zero, we receive the following number:
100(omitted)00x. = 100(omitted)000.
That's a 1 followed by 53 zeroes, 52 stored and 1 added due to the exponent.
It represents 253, which marks the boundary (both negative and positive) between which we can accurately represent all integers. If we wanted to add one to 253, we would have to set the implicit zero (denoted by the x) to one, but that's impossible.

If you need to convert a string float to an int you can use this method.
Example: '38.0' to 38
In order to convert this to an int you can cast it as a float then an int. This will also work for float strings or integer strings.
>>> int(float('38.0'))
38
>>> int(float('38'))
38
Note: This will strip any numbers after the decimal.
>>> int(float('38.2'))
38

math.floor will always return an integer number and thus int(math.floor(some_float)) will never introduce rounding errors.
The rounding error might already be introduced in math.floor(some_large_float), though, or even when storing a large number in a float in the first place. (Large numbers may lose precision when stored in floats.)

Another code sample to convert a real/float to an integer using variables.
"vel" is a real/float number and converted to the next highest INTEGER, "newvel".
import arcpy.math, os, sys, arcpy.da
.
.
with arcpy.da.SearchCursor(densifybkp,[floseg,vel,Length]) as cursor:
for row in cursor:
curvel = float(row[1])
newvel = int(math.ceil(curvel))

Since you're asking for the 'safest' way, I'll provide another answer other than the top answer.
An easy way to make sure you don't lose any precision is to check if the values would be equal after you convert them.
if int(some_value) == some_value:
some_value = int(some_value)
If the float is 1.0 for example, 1.0 is equal to 1. So the conversion to int will execute. And if the float is 1.1, int(1.1) equates to 1, and 1.1 != 1. So the value will remain a float and you won't lose any precision.