Output the Euler Numbers

Question

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

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

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

• user45941: @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.

Answer 1

Mathematica, 6 bytes

EulerE

-cough-

• hyperneutrino: When I saw the title, I immediately thought "Surely Mathematica must have a builtin for this".
• Luis Mendo: That applies to pretty much every title... even detecting goats in images
• Greg Martin: GoatImageQ is underappreciated
• Magic Octopus Urn: Can you explain this? Edit: this was a joke.

Answer 2

J, 10 bytes

(1%6&o.)t:

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

• orlp: Does J do symbolic analysis to get the generating function? It doesn't run into floating point errors even for n=30.
• miles: @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.

Answer 3

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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• Jeppe Stig Nielsen: The built-in function is eulerfrac.

Answer 4

Maple, 5 bytes

euler

Hurray for builtins?

• user36219: Love it when mathematica is too verbose for a maths problem...

Answer 5

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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Answer 6

Mathematica, without built-in, 18 bytes

Using @rahnema1's formula:

2Im@PolyLog[-#,I]&

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21 bytes:

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

Answer 7

JavaScript (Node.js), (削除) 70 (削除ここまで) (削除) 46 (削除ここまで) 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ドル\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ドル\leq i\leq n\$). As a result, \$F(n,i)=0\$ where \$i > n\$. This completes the proof of the validity of the algorithm.

enter image description here

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

Answer 8

Python 2.7, 46 bytes

Using scipy.

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

Answer 9

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
}

Answer 10

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 [1]

Answer 11

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
[email protected]_2/:
_1.@v:-1
v:-1_1-.@
_5.@v:-1
v:-1_"="-.@
_"}$#"*+.@v:-1
8**-.@v:-1_"'PO"
"0h;"3_"A+y^"9*+**[email protected]*8+*:*

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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:>10p00円:>20p\>010g20g2*-00g1>-:30pv>\:
_12ドル 0g2%2*-*10g20g110g20g-240pv^1g03:_^*
>-#1:_!>\#<:#*_40ドルg:1-40p!#v_*\>00円
@.$_^#`g00:<|!`g01::+1\+*/\<
+4%1-*/+2円+^>$10ドルg::2>\#<1#*-#2:#\_$*1円

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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.

Answer 12

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]]

Answer 13

APL(NARS), 42 chars, 84 bytes

E←{0≥w←⍵:1⋄1-+/{(⍵!w)×ばつ(2*w-1+⍵)×ばつ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)×ばつ(2*w-1+⍵)×ばつ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):

 ×ばつ+/{+/⍵{⍺÷⍨(0J2*-⍺)×ばつ(⍵!⍺)×ばつ( ̄1*⍵)×ばつ(×ばつ⍵)*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.

Answer 14

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

• FlipTack: 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).

Answer 15

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: [1]
 { 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
}

Answer 16

Wolfram Language (Mathematica), (削除) 47 (削除ここまで) 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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