# Output the Euler Numbers

Given a non-negative integer $$\n ,\$$ output the $$\n^{\text{th}}\$$ Euler number (OEIS A122045).

All odd-indexed Euler numbers are $$\0 .\$$ The even-indexed Euler numbers can be computed with the following formula ($$\i \equiv \sqrt{-1}\$$ refers to the imaginary unit): $$E_{2n} = i \sum_{k=1}^{2n+1}{ \sum_{j=0}^{k}{ \left(\begin{array}{c}k \\ j \end{array}\right) \frac{{\left(-1\right)}^{j} {\left(k-2j\right)}^{2n+1}}{2^k i^k k} } } \,.$$

## Rules

• $$\n\$$ will be a non-negative integer such that the $$\n^{\text{th}}\$$ Euler number is within the representable range of integers for your language.

## Test Cases

0 -> 1
1 -> 0
2 -> -1
3 -> 0
6 -> -61
10 -> -50521
20 -> 370371188237525

• @donbright You're missing a set of parentheses: wolframalpha.com/input/… - with that, the two summands are both -i/2, which yield -i when added. Multiply that by the i outside of the summation, and you get 1.
– user45941
Jan 22, 2017 at 21:01

# Mathematica, 6 bytes

EulerE


-cough-

• When I saw the title, I immediately thought "Surely Mathematica must have a builtin for this". Jan 21, 2017 at 2:48
• That applies to pretty much every title... even detecting goats in images Jan 21, 2017 at 16:19
• GoatImageQ is underappreciated Jan 21, 2017 at 19:34
• Can you explain this? Edit: this was a joke. Jan 24, 2017 at 21:30

# J, 10 bytes

(1%6&o.)t:


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Uses the definition for the exponential generating function sech(x).

• Does J do symbolic analysis to get the generating function? It doesn't run into floating point errors even for n=30.
– orlp
Jan 21, 2017 at 0:49
• @orlp I'm not sure what it does internally, but J knows the Taylor series for a subset of verbs. Any function you can define using a combination of those verbs will be valid for t. or t: where are g.f. and e.g.f. A curious note is that tan(x) is not supported but sin(x)/cos(x) is. Jan 21, 2017 at 0:57

# PARI/GP, 9 bytes

eulerfrac

This built-in was added in version 2.13.0, after this challenge was asked.

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# PARI/GP, 24 bytes

n->2*imag(polylog(-n,I))

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# PARI/GP, 32 bytes

n->n!*Vec(1/cosh(x+O(x^n++)))[n]

This is the original answer.

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• The built-in function is eulerfrac. Dec 19, 2020 at 12:16

# Maple, 5 bytes

euler


Hurray for builtins?

• Love it when mathematica is too verbose for a maths problem... Jan 21, 2017 at 7:48

# Maxima, 5 bytes / 42 bytes

Maxima has a built in:

euler


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The following solution does not require the built in from above, and uses the formula that originally defined the euler numbers. We are basically looking for the n-th coefficient of the series expansion of 1/cosh(t) = sech(t) (up to the n!)

f(n):=coeff(taylor(sech(x),x,0,n)*n!,x,n);


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# Mathematica, without built-in, 18 bytes

Using @rahnema1's formula:

2Im@PolyLog[-#,I]&


21 bytes:

Sech@x~D~{x,#}/.x->0&


# JavaScript (Node.js), 7046 44 bytes

F=(a,b=a)=>a?F(--a,b)*~b+F(a,b-=2)*(a-b):+!b


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Surprised to find no JavaScript answer yet, so I'll have a try.

The code consists of only basic mathematics, but the mathematics behind the code requires calculus. The recursion formula is derived from the expansion of the derivatives of $$\\mathrm{sech}(x)\$$ of different orders.

### Explanation

Here I'll use some convenient notation. Let $$\T^n:=\mathrm{tanh}^n(t)\$$ and $$\S^n:=\mathrm{sech}^n(t)\$$. Then we have

$$\frac{d^nS}{dt^n}=\sum_{i=0}^nF(n,i)T^{n-i}S^{i+1}$$

Since $$\\frac{dT}{dt}=S^2\$$ and $$\\frac{dS}{dt}=-TS\$$, we can deduce that

\begin{equation} \begin{aligned} \frac{d}{dt}(T^aS^b)&=aT^{a-1}(S^2)(S^b)+bS^{b-1}(-TS)(T^a) \\ &=aT^{a-1}S^{b+2}-bT^{a+1}S^b \end{aligned} \end{equation}

Let $$\b=i+1\$$ and $$\a=n-i\$$, we can rewrite the relation above as

\begin{equation} \begin{aligned} \frac{d}{dt}(T^{n-i}S^{i+1})&=(n-i)T^{n-i-1}S^{i+3}-(i+1)T^{n-i+1}S^{i+1}\\ &=(n-i)T^{(n+1)-(i+2)}S^{(i+2)+1}-(i+1)T^{(n+1)-i}S^{i+1} \end{aligned} \end{equation}

That is, $$\F(n,i)\$$ contributes to both $$\F(n+1,i+2)\$$ and $$\F(n+1,i)\$$. As a result, we can write $$\F(n,i)\$$ in terms of $$\F(n-1,i-2)\$$ and $$\F(n-1,i)\$$:

$$F(n,i)=(n-i+1)F(n-1,i-2)-(i+1)F(n-1,i)$$

with initial condition $$\F(0,0)=1\$$ and $$\F(0,i)=0\$$ where $$\i\neq 0\$$.

The related part of the code a?F(--a,b)*~b+F(a,b-=2)*(a-b):+!b is exactly calculating using the above recurrence formula. Here's the breakdown:

F(--a,b)                 // F(n-1, i)                   [ a = n-1, b = i   ]
*~b                      // *-(i+1)                     [ a = n-1, b = i   ]
+F(a,b-=2)               // +F(n-1, i-2)                [ a = n-1, b = i-2 ]
*(a-b)                   // *((n-1)-(i-2))              [ a = n-1, b = i-2 ]
// which is equivalent to *(n-i+1)


Since $$\T(0)=0\$$ and $$\S(0)=1\$$, $$\E_n\$$ equals the coefficient of $$\S^{n+1}\$$ in the expansion of $$\\frac{d^nS}{dt^n}\$$, which is $$\F(n,n)\$$.

For branches that $$\F(0,0)\$$ can never be reached, the recurrences always end at 0, so $$\F(n,i)=0\$$ where $$\i<0\$$ or $$\i\$$ is odd. The latter one, particularly, implies that $$\E_n=0\$$ for all odd $$\n\$$s. For even $$\i\$$s strictly larger than $$\n\$$, the recurrence may eventually allow $$\0\leq i\leq n\$$ to happen at some point, but before that step it must reach a point where $$\i=n+1\$$, and the recurrence formula shows that the value must be 0 at that point (since the first term is multiplied by $$\n-i+1=n-(n+1)+1=0\$$, and the second term is farther from the "triangle" of $$\0\leq i\leq n\$$). As a result, $$\F(n,i)=0\$$ where $$\i > n\$$. This completes the proof of the validity of the algorithm. ### Extensions

The code can be modified to calculate three more related sequences:

Tangent Numbers (46 bytes)

F=(a,b=a)=>a?F(--a,b)*++b+F(a,b-=3)*(a-b):+!~b


Secant Numbers (45 bytes)

F=(a,b=a)=>a?F(--a,b)*++b+F(a,b-=3)*(a-b):+!b


Euler Zigzag Numbers (48 bytes)

F=(a,b=a)=>a?F(--a,b)*++b+F(a,b-=3)*(a-b):!b+!~b


# Python 2.7, 46 bytes

Using scipy.

from scipy.special import*
lambda n:euler(n)[n]


# Perl 6, 78 bytes

{(->*@E {1-sum @E».&{$_*2**(@E-1-$++)*[*](@E-$++^..@E)/[*] 1..$++}}...*)[$_]}  Uses the iterative formula from here: $$E_n = 1 - \sum_{k=0}^{n-1} \left[ E_k \cdot 2^{(n-1-k)} \cdot \binom{n}{k} \right]$$ ### How it works The general structure is a lambda in which an infinite sequence is generated, by an expression that is called repeatedly and gets all previous values of the sequence in the variable @E, and then that sequence is indexed with the lambda argument: { ( -> *@E { } ... * )[$_] }


The expression called for each step of the sequence, is:

1 - sum @E».&{              # 1 - ∑
$_ # Eₙ * 2**(@E - 1 -$++)     # 2ⁿ⁻ˡ⁻ᵏ
* [*](@E - $++ ^.. @E) # (n-k-1)·...·(n-1)·n / [*] 1..$++            # 1·2·...·k
}


# Maxima, 29 bytes

z(n):=2*imagpart(li[-n](%i));


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Two times imaginary part of polylogarithm function of order -n with argument i 

# Befunge, 115 bytes

This just supports a hardcoded set of the first 16 Euler numbers (i.e. E0 to E15). Anything beyond that wouldn't fit in a 32-bit Befunge value anyway.

&:2%v
v@.0_2/:
_1.@v:-1
v:-1_1-.@
_5.@v:-1
v:-1_"="-.@
_"}$#"*+.@v:-1 8**-.@v:-1_"'PO" "0h;"3_"A+y^"9*+**.@.-*8+*:*  Try it online! I've also done a full implementation of the formula provided in the challenge, but it's nearly twice the size, and it's still limited to the first 16 values on TIO, even though that's a 64-bit interpreter. <v0p00+1& v>1:>10p\00:>20p\>010g20g2*-00g1>-:30pv>\: _$12 0g2%2*-*10g20g110g20g-240pv^1g03:_^*
>-#1:_!>\#<:#*_$40g:1-40p!#v_*\>0\0 @.$_^#g00:<|!g01::+1\+*/\<
+4%1-*/+\2+^>10g::2>\#<1#*-#2:#\_\$*\1


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The problem with this algorithm is that the intermediate values in the series overflow much sooner than the total does. On a 32-bit interpreter it can only handle the first 10 values (i.e. E0 to E9). Interpreters that use bignums should do much better though - PyFunge and Befungee could both handle at least up to E30.

# Python2, (sympy rational), 153 bytes

from sympy import *
t=n+2
print n,re(Add(*map(lambda (k,j):I**(k-2*j-1)*(k-2*j)**(n+1)*binomial(k,j)/(k*2**k),[(c/t+1,c%t) for c in range(0,t**2-t)])))


This is very suboptimal but it's trying to use basic sympy functions and avoid floating point. Thanks @Mego for setting me straight on the original formula listed above. I tried to use something like @xnor's "combine two loops" from Tips for golfing in Python

• You can do import* (remove the space in between) to save a byte. Also, you need to take the number as an input somehow (snippets which assume the input to be in a variable are not allowed). Jan 25, 2017 at 21:20

# Axiom, 5 bytes

euler


for OEIS A122045; this is 57 bytes

g(n:NNI):INT==factorial(n)*coefficient(taylor(sech(x)),n)


test code and results

(102) -> [[i,g(i)] for i in [0,1,2,3,6,10,20]]
(102)
[[0,1],[1,0],[2,- 1],[3,0],[6,- 61],[10,- 50521],[20,370371188237525]]

(103) -> [[i,euler(i)] for i in [0,1,2,3,6,10,20]]
(103)
[[0,1],[1,0],[2,- 1],[3,0],[6,- 61],[10,- 50521],[20,370371188237525]]


## CJam (34 bytes)

{1a{_W%_,,.*0+(+W%\_,,:~.*.+}@*W=}


Online demo which prints E(0) to E(19). This is an anonymous block (function).

The implementation borrows Shieru Akasoto's recurrence and rewrites it in a more CJam friendly style, manipulating entire rows at a time.

## Dissection

{           e# Define a block
1a        e#   Start with row 0: 
{         e#   Loop...
_W%     e#     Take a copy and reverse it
_,,.*   e#     Multiply each element by its position
0+(+    e#     Pop the 0 from the start and add two 0s to the end
W%      e#     Reverse again, giving [0 0 (i-1)a_0 (i-2)a_1 ... a_{i-2}]
\       e#     Go back to the other copy
_,,:~.* e#     Multiply each element by -1 ... -i
.+      e#     Add the two arrays
}         e#
@*        e#   Bring the input to the top to control the loop count
W=        e#   Take the last element
}


# APL(NARS), 42 chars, 84 bytes

E←{0≥w←⍵:1⋄1-+/{(⍵!w)×(2*w-1+⍵)×E⍵}¨¯1+⍳⍵}


Follow the formula showed from "smls", test:

  E 0
1
E 1
0
E 3
0
E 6
¯61
E 10
¯50521


the last case return one big rational as result because i enter 20x (the big rational 20/1) and not 20 as i think 20.0 float 64 bit...

  E 20x
370371188237525


It would be more fast if one return 0 soon but would be a little more long (50 chars):

  E←{0≥w←⍵:1⋄0≠2∣w:0⋄1-+/{(⍵!w)×(2*w-1+⍵)×E⍵}¨¯1+⍳⍵}
E 30x
¯441543893249023104553682821


it would be more fast if it is used the definition on question (and would be a little more long 75 chars):

  f←{0≥⍵:1⋄0≠2∣⍵:0⋄0J1×+/{+/⍵{⍺÷⍨(0J2*-⍺)×(⍵!⍺)×(¯1*⍵)×(⍺-2×⍵)*n}¨0..⍵}¨⍳n←1+⍵}
f 0
1
f 1
0
f 3
0
f 6
¯61J0
f 10
¯50521J¯8.890242766E¯9
f 10x
¯50521J0
f 20x
370371188237525J0
f 30x
¯441543893249023104553682821J0
f 40x
14851150718114980017877156781405826684425J0
f 400x
290652112822334583927483864434329346014178100708615375725038705263971249271772421890927613982905400870578615922728
107805634246727371465484012302031163270328101126797841939707163099497536820702479746686714267778811263343861
344990648676537202541289333151841575657340742634189439612727396128265918519683720901279100496205972446809988
880945212776281115581267184426274778988681851866851641727953206090552901049158520028722201942987653512716826
524150450130141785716436856286094614730637618087804268356432570627536028770886829651448516666994497921751407
121752827492669601130599340120509192817404674513170334607613808215971646794552204048850269569900253391449524
735072587185797183507854751762384660697046224773187826603393443429017928197076520780169871299768968112010396
81980247383801787585348828625J0


The result above it is one complex number that has only the real part.

# Wolfram Language (Mathematica), 47 46 bytes

Without using any special functions:

e@0=1;e@n_:=-n!Sum[e[n-k]/k!/(n-k)!,{k,2,n,2}]


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