New answers tagged sequence
1
Keg, 11 bytes (SBCS)
Now I realized that Keg is pretty hard to use...
¿Ï_^'(Ï_')"
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0
GolfScript, 12 bytes
~),(;{),(;}%
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0
W j f, 3 bytes
It's actually pretty trivial in W ...
M_M
Explanation
a % Take an input, say 3
M % Identity map every item in numeric value 3
% Since it is prohibited, a range is generated,
% namely [1 2 3]
a M % For every item in [1 2 3]:
% Note that single items are 1, 2, and 3, *not* the whole list
a % Generate a ...
2
Mathematica 46 42 bytes
f=2^(#/2)HypergeometricU[-#/2,.5,-.5]I^-#&
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Thanks to mabel for helping me shave 4 bytes by using what I think is an anonymous function.
As seen in https://www.wolframalpha.com/input/?i=OEIS+A000085 and http://mathworld.wolfram.com/ConfluentHypergeometricFunctionoftheSecondKind.html
2
Python, 226 bytes
Assumes the lists O and V are defined, as well as the time T, either an int or a float.
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s=lambda x:int(x/abs(x));r=range;f=lambda O,V,T:"".join(map(lambda t:t[1],sorted(sum([[((i-o)/v,"%+d"%(s(v)*c))for i in r((v>0)-(v<0)*(o==0),int((o+T*v)//1)+(v>0),s(v))]for c,o,v in zip(r(len(O)),O,V)],[]),key=lambda t:t[0])))
...
1
Elm, 45 bytes
f n = if n<2 then 1 else (n-1)*f(n-2)+f(n-1)
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Implementation of the recursive function.
1
GolfScript, 16 bytes
This is inspired by the Fibonacci GolfScript program.
~,1.@{)*1$+\}/\;
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Explanation
~, # Generate exclusive range from input to 0
1. # Make 2 copies of 1
@ # Make the each target on the top
{ }/ # Foreach over the range
) # Increment the current item
...
1
PHP, 48 bytes
function f($n){return$n<2?1:f(--$n)+$n*f($n-1);}
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Implementation of the recursive function.
Original version: 54 bytes (port of @Noodle9 answer):
for($i=$j=1;++$n<$argn;$j=$k){$k=$i;$i+=$n*$j;}echo$i;
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2
Red, 53 bytes
f: func[n][either 1 > n: n - 1[1][n *(f n - 1)+ f n]]
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1
Japt, 28 26 bytes
_pZÌ+(ZÊÉ *ZgJÑ}g´U1õ ï)gJ
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pZÌ+(ZÊÉ *ZgJÑ make sequence with next element from previous sequence
_ ... }g´U repeat input -1 times
1õ ï) starting with [[1,1]]
gJ implicit output last element of resulting sequence
0
Rogex, 27 bytes
90B300d00901e00100701a20f00
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Explained
90B 300 d00 # Set the loop value to 11
901 # Set the buffer to 1
e00 # While the buffer isn't equal to 11
100 701 # Print the buffer then increment it
a20 # Print a space
f00 # End loop
Rogex answer number 2! Also, my 101st answer! (´• ヮ •`)
3
C (gcc), 40 bytes
long f(long n){n=n<2?1:f(--n)+n*f(n-1);}
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Merely copied from @Arnauld answer
1
Retina 0.8.2, 80 bytes
.+
$*_;;_;
{`^_(_*;)(_*;_+)
$1_$2,$2
+`(,_*)_(;_+;(_*))|,
$3$1$2
^;_*;(_*).*
$.1
Try it online! Explanation:
.+
$*_;;_;
Initialise the work area with n, i=0, and T=[1] (note that the list indices are reversed, so that the first element of T is T[i] and the last is T[0]) converted to unary.
{`
Loop n times (actually until the ...
2
J, 25 bytes
($:+]*$:@<:)@<:`1:@.(<&2)
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$: is J's recursion operator, so this is a fairly straightforward impl of the recursive def.
I tried a couple other approaches (including using ^: power of operator) but wasn't able to get shorter than this.
Am curious if anyone can improve it...
1
Keg, 21 bytes
:1≤[_1|;:@Tƒ^:;@Tƒ*+]
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Simply implements the formula in a recursive manner. The footer is mainly for testing purposes
2
naz, 50 bytes
1x1f1a1o2x1v0m9a1a1o1v0x1f1f1f1f1f1f1f1f1f8s1o1s1o
Explanation (with 0x commands removed)
1x1f # Function 1
1a1o # Add 1 to the register and output
2x1v # Store the new value in variable 1
0m9a1a1o # Output a newline
1v # Read variable 1 into the register
...
3
Haskell, 29 bytes
f n|n<2=1|m<-n-1=f m+m*f(n-2)
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Implements the recursive formula.
20
Python, 33 bytes
f=lambda n:n<2or~-n*f(n-2)+f(n-1)
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Uses the recursive formula.
1
Charcoal, 20 bytes
⊞υ¹ＦＮ⊞υ⁺§υι×ι§υ⊖ιＩ⊟υ
Try it online! Link is to verbose version of code. Uses the recurrence relation. Explanation:
⊞υ¹
T(0)=1
ＦＮ
Loop n times.
⊞υ⁺§υι×ι§υ⊖ι
T(i+1)=T(i)+iT(i-1)
Ｉ⊟υ
Output T(n).
8
Jelly, 6 bytes
Œ!ỤƑ€S
A monadic Link accepting a non-negative integer which yields a positive integer.
Try it online! Or see the first 10 (n=10 is too slow)
How?
Builds all permutations of [1...n] (or if n=0 just [[]]) and counts the number which are involutions.
Œ!ỤƑ€S - Link: integer, n e.g. 0 3
Œ! - all permutations [[]] ...
0
Lua, 96 bytes
n=io.read()i=0 while i do for j=0,n..i do if (n..i)+0==j*j then print(i)return end end i=i+1 end
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7
TI-BASIC, 28 22 bytes
∑((Ans nCr (2K))(2K)!/(2^KK!),K,0,int(.5Ans
Input is n in Ans.
Output is the nth telephone number.
I was going to use the first formula mentioned in the challenge, but testing it resulted in ERROR: OVERFLOWs even on smaller numbers because TI-BASIC handles n!! as two successive factorials instead of a double factorial.
At least it ...
12
Oasis, 7 6 5 bytes
àn*+1
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# to compute the nth telephone number f(n):
à # push the telephone numbers f(n-1) and f(n-2)
n # push n
* # multiply: n * f(n-2)
+ # add: n * f(n-2) + f(n-1)
1 # base case: f(0) = 1
2
Perl 6, 26 bytes
{{(1,1,++$×*+*...*)[$_]}}
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5
JavaScript (ES6), 25 bytes
0-indexed. Uses the recurrence relation.
f=n=>n<2||f(--n)+n*f(n-1)
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9
cQuents, 10 bytes
=1:Z+Y($-2
1-indexed.
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Explanation
=1 first term in sequence is 1
: given n, output nth term in sequence (1-indexed)
each term is
Z+ (n-1) term +
Y (n-2) term *
($-2 (index - 2)
3
Python 3, 81 \$\cdots\$ 57 56 bytes
def f(n):
a=b=i=1
while i<n:a,b=a+i*b,a;i+=1
return a
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Uses 0 based indexing and implements the recursive formula.
4
05AB1E, 12 8 bytes
-4 bytes by porting Stephen's cQuent answer
1λèsN<*+
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3
Brachylog, 8 bytes
;.c~^₂∧ℕ
Sadly enough it takes 3 bytes to check if a number is square and an extra ℕ is needed if the input is a square itself.
Explanation
;.c~^₂∧ℕ # Find a value
;.c # that concatenate to the input
~^₂ # gives a square number
∧ℕ # and make sure that value is a number
# (otherwise ...
1
05AB1E, 8 7 bytes
PLIδÖʒO
Inspired by @jimmy23013's CJam answer, so make sure to upvote him as well!
Input as a pair of integers \$[m,n]\$ and output as a list of [0,1] and [1,0] pairs for \$V\$ and \$H\$ respectively.
Try it online or verify all test cases.
Explanation:
P # Take the product of the (implicit) input-pair: m*n
L # Pop and ...
0
GolfScript, 24 bytes
I don't feel right to have a Burlesque answer without a GolfScript answer.(Please don't mind if this is a little bit slow, although they seem to produce the right result.)
-1{).2$\+~.),{.*}%?)!}do
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Explanation
-1 # The counter for the current number (it is >= 0)
{ # Do at ...
0
PHP, 50 bytes
<?=array_sum(str_split(decbin($argn&-2^$argn*2)));
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Unfortunately, really don't think I can get it any more golfed than this!
The idea is to count transitions 1->0 or 0->1.
2
Pyth, 18 16 15 11 10 bytes
Uses the floating point square root test, like most of the other answers. As such, the precision is eventually insufficient, and the smallest square it fails with is 4503599761588225 – with an input of 45035997, it returns 61588224 instead of 61588225 as it should. The smallest input it fails with is 20256971, returning 228498166 ...
3
Pyth, 35 33 30 23 21 bytes
This is a program that uses only integer arithmetic. Since Python has arbitrary-precision integers, this is guaranteed to work for arbitrarily large numbers, whereas the square-root test used by most of the other answers fails with numbers greater than 252, having false positives with numbers very close to a perfect square.
...
4
Python 2, 47 49 bytes
f=lambda p,i=0:int(`p`+`i`)**.5%1and f(p,i+1)or i
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5
C++ (clang), 194 \$\cdots\$ 128 129 bytes
#import<string>
#define z std::to_string
int f(int n){for(int j=-1;;)for(long i=0,m=stol(z(n)+z(++j));i++<m;)if(i*i==m)return j;}
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Added a byte to fix overflow error kindly pointed out by Deadcode.
2
Japt, 10 bytes
_siU ¬%1}f
Saved a byte thanks to @AZTECCO
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4
C, 123 115 bytes (thanks, Jo King!)
n,k,m=10,x,p;main(){scanf("%d",&n);for(;k<m||(m*=10);++k)for(p=0;x=n*m+k,p<=x;++p)if(p*p==x)return printf("%d",k);}
I noticed this version can fail on not-very-large inputs due to integer overflow. So I, did a separate version, with less overflow.
C, less overflow. 117 bytes
n,k,m=10,x,p;main(){scanf("%d",&...
1
Charcoal, 16 bytes
≔⁰ηＷ﹪₂Ｉ⁺θη¹≦⊕ηＩη
Try it online! Link is to verbose version of code. Explanation:
≔⁰η
Initialise the result to zero.
Ｗ﹪₂Ｉ⁺θη¹
Repeat until the concatenation is a square.
≦⊕η
Increment the result.
Ｉη
Output the result.
36 bytes for an efficient solution:
≔×χＮη
Try it online! Link is to verbose version of code. Explanation:
...
1
cQuents, 13 bytes
#|1:#bA~N
?$$
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Explanation
#|1 output nth element in sequence (remove to output indefinitely)
: mode: sequence, output nth element in sequence
# conditional: count N up, when true, add to sequence
b call second line, passing
A~N input concatenated to ...
3
Perl 5 -pal, 30 bytes
$_=0;$_++while"@F$_"**.5=~/\./
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1
Perl 5, 35 40 bytes
$k=0;L:$_="$n$k"**.5;!/\./ or$k++,goto L
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2
PHP, 53 47 44 bytes
for($k=-1;fmod(sqrt($argn.++$k),1););echo$k;
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0
Haskell, 93 88 bytes
[j|i<-(show.(^2))<$>[0..],i/=n,n`isPrefixOf`i,j<-[drop(length n)i],j=="0"||j!!0/='0']!!0
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4
R, 45 bytes
n=scan();while((paste0(n,+F):1)^.5%%1)F=F+1;F
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Throws a warning for each iteration noting that it only uses the first element of the vector as a loop condition. This will run out of space for large inputs since it constructs a vector of length \$n'k\$ at each iteration; this version will work for larger inputs and should be a bit ...
2
Japt, 10 bytes
@s+X ¬v1}a
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1
Jelly, 8 bytes
0ṭVƲɗ1#
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A monadic link taking as its argument an integer and returning an integer.
Explanation
0 ɗ1# | Starting with zero, find 1 integer where the following is true as a dyad, using the original argument as the right argument
ṭ | - Tag onto the end
V | - Evaluate after converting to string and ...
0
PHP, 83 86 bytes
g:substr($k=$i*++$i,0,$l=strlen($n=$argn))==$n&&("$k")[$l]&&die(substr($k,$l));goto g;
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-9 bytes thanks to @640KB and bugs fixed (+12). Only fails for case 10 currently.
Ungolfed
<?php
$number=$argv[1];
$len = strlen($number);
$i=0;
do {
$k = $i*$i;
if (substr($k,0,$len)==$number && ...
3
Ruby -pl, 43 bytes
i=1;i+=1until"#{i*i}"[/^#$_(?!0.|$)/]
$_=$'
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Tests each square until it finds one that starts with the input and is followed by a number with no leading zero. I thought going in this direction rather than the more natural one would be shorter, but now I'm doubting that it is.
1
JavaScript (Node.js), 38 Bytes
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f=(a,b=[0])=>(a+b)**.5%1?f(a,[++b]):+b
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