Bernoulli Numbers
Mathematica, 40 28 23 22 bytes
Using the famous formula n*ζ(1−n)=−Bn, where ζ is the Riemann Zeta function.
If[#>0,-Zeta[1-#]#,1]&
The same length:
B@0=1;B@n_=-Zeta[1-n]n
Original solution, 40 bytes, using the generating function of Bernoulli numbers.
#!SeriesCoefficient[t/(1-E^-t),{t,0,#}]&
Julia, 58 bytes
B(m)=m<1?1:1-sum(k->big(binomial(m,k))*B(k)/(m-k+1),0:m-1)
This creates a recursive function B that accepts an integer and returns a BigFloat (i.e. high precision floating point).
Ungolfed:
function B(m::Integer)
m == 0 && return 1
return 1 - sum(k -> big(binomial(m, k)) * B(k) / (m-k+1), 0:m-1)
end
Minkolang 0.14, 97 bytes
I actually tried doing it recursively first, but my interpreter, as currently designed, actually can't do it. If you try to recurse from within a for loop, it starts a new recursion. So I went for the tabulating approach...which had precision problems. So I did the whole thing with fractions. With no built-in support for fractions. [sigh]
n1+[xxi$z0z2%1+F0c0=$1&$d4Mdm:1R:r$dz1Az0A]$:N.
11z[i0azi6M*i1azi-1+*d0c*1c2c*-1c3c*4$X]f
z1=d1+f
Try it here. Bonus: the array has all fractions for every previous Bernoulli number!
Explanation (in a bit)
n1+ Take number from input (N) and add 1
[ Open for loop that runs N+1 times (starts at zero)
xx Dump the top two values of the stack
i$z Store the loop counter in the register (m)
0 Push 0
z2%1+ Push 1 if N is even, 2 if odd
F Gosub; pops y,x then goes to codebox(x,y), to be returned to
0c Copy the first item on the stack
, 1 if equal to 0, 0 otherwise
$1& Jump 11 spaces if top of stack is not 0
(If top of stack is not 0, then...)
$d Duplicate whole stack
4M Pop b,a and push GCD(a,b)
dm Duplicate and merge (a,b,c,c -> a,c,b,c)
: Divide
1R Rotate 1 item to the right (0G works too)
: Divide
r Reverse stack
(In both cases...)
$d Duplicate whole stack
z1A Store denominator of B_m in array
z0A Store numerator of B_m in array
] Close for loop
$: Divide (float division)
N. Output as number and stop.
11 Push two 1s (a, b)
z[ Open a for loop that repeats m times
i0a Retrieve numerator of B_k (p)
zi Push m, k
6M Pop k,m and push mCk (binomial) (x)
* p*x (c)
i1a Retrieve denominator of B_k (q)
zi-1+ m-k+1 (y)
* q*y (d)
d Duplicate top of stack
0c* a*d
1c2c* b*c
- a*d-b*c
1c3c* b*d
4$X Dump the bottom four items of stack
]f Jump back to F
z m
1= 0 (if m is 1) or 1 (otherwise)
d1+ Duplicate and add 1 (1 or 2)
f Jump back to F
The third line is responsible for 1/2 if m is 1 and 0/1 if m is an odd number greater than 1. The second line calculates B_m with the summation formula given in the question, and does so entirely with numerators and denominators. Otherwise it'd be a lot shorter. The first half of the first line does some bookkeeping and chooses whether to execute the second or third line, and the second half divides the numerator and denominator by their GCD (if applicable) and stores those values. And outputs the answer at the end.