New answers tagged

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lst = [] print("ENTER ZERO NUMBER FOR EXIT !!!!!!!!!!!!") print("ENTER LIST ELEMENTS :: ") while True: n = int(input()) if n == 0 : print("!!!!!!!!!!! EXIT !!!!!!!!!!!!") break else : lst.append(n) oddTup = [] evenTup = [] for j in lst : if j % 2 == 0 : evenTup. append(j) else : ...


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350. Husk, 166 bytes, A000774 *Π¹→ṁ\ḣ " Explanation " " *Π¹→ṁ\ḣ " → 1 plus " ṁ\ sum of reciprocals of " ḣ the range [1..N] " * times " Π¹ the factorial of N Try it online! Next sequence! It was surprisingly tricky to get the explanation to come out to the correct length.


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Barrel, 26 bytes Disclaimer: The language is newer than the question, but I didn't even think of golfing this until after I'd created the language. I did update the language after I originally wrote the answer, and changed my answer, but that was because I was fixing the interpreter and made some changes to the spec to make the language work better. I wasn't ...


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349. Kotlin, 774 bytes, A000607 val primes = mutableListOf<Int> (2) val prev = mutableListOf<Double>(1.0) fun sequence(n: Int): Double { if (n < prev.size) return prev.get(n) var sum = 0.0 for (k in (primes.last() + 1)..n) { var prime = true for (p in primes) if (k % p == 0) { ...


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05AB1E, 32 bytes ∞ʒDg6-sηε2.£}¦D€SP.ι3Ý3*>è2ä€ËP* Try it online! Link includes a header and footer to increase efficiency by making it start looking from \$1,000,000\$, but works (extremely slowly) without them. The program will not terminate after finding all \$900\$ Meeker numbers. ∞ʒDg6-sηε2.£}¦D€SP.ι3Ý3*>è2ä€ËP* # full program ʒ ...


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Jelly, 20 bytes DḌ,PƊƝFḊm3⁼2/Ạ ȷ6Ç#Ṫ A monadic Link accepting \$n\$ that yields the \$n^\text{th}\$ Meeker number Try it online! Very slow so will probably time out for \$n \gt 36\$. How? ȷ6Ç#Ṫ - Main Link: integer, n ȷ6 - 10^6 # - find the first n integers starting at 10^6 for which: Ç - call Link 1 Ṫ - tail DḌ,PƊƝFḊm3⁼2/Ạ - Link 1: ...


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Python - 11 bytes print -1 0 1 I would say "Good luck improving this.", but past experience has taught me never to assume: though, this won't be easily beat (for i in range(3), %ding the spaces and %sing the 1s are all overweight, so you can eliminate those.) (Edit 1: -4 due to caird, as is expected)


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Grok, 13 bytes }q {p1+YzP9>!


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Jelly, 3 bytes -r1 Try it online! 1r- Try it online! -ŒR Try it online! 1ŒR Try it online! Ø-Ż Try it online! Ø+Ż Try it online! -rN Try it online! 1rN Try it online! 2Ż’ Try it online! 3Ḷ’ Try it online! and a bonus one: M, 2 bytes -R Try it online!


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dotcomma, 67 bytes [[[[[[[[[[.,][.].,][.].,][.].,][.].,][.].,][.].,][.].,][.].,][.].,] Try it online! A simple approach which nests [[...][.].,], which adds one to and then pushes each number from one to ten. The "root" of this is [.,] which pushes one. Explanation: In dotcomma, dots and commas are the only operators. Their behaviors are extremely ...


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Perl 5, 58 bytes /(.)(.)(.(.))(.)/,$1*$2==$3&&$5*$4==$'&&say for 1e6..1e7-1 Try it online! Outputs the entire list of Meeker numbers.


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Scala, 103 bytes Stream from 1 map "".+filter{x=>x.size==7&&x.map(_-48).sliding(4,3).forall{x=>x(0)*x(1)==10*x(2)+x(3)}} Try it in Scastie! A pretty straightforward answer. You can treat it like a function giving the nth Meeker number (0-indexed) or as a list of all Meeker numbers.


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R, 101 91 bytes `[`=`for`;a[1:9,b[0:9,e[0:9,show(paste0(a,b,if(a*b<10)0,d<-a*b,e,if((f=d%%10*e)<10)0,f))]]] Try it online! Inspired by and taking a similar approach to Manish Kundu's answer. This is a horrible piece of R hackery that redefines square brackets '[]', normally used for indexing, to be a short alias for the for looping function. The ...


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R, 76 66 bytes Edit: -10 bytes thanks to Giuseppe f=function(b,a,n)`if`(nchar(b)>n,substr(b,n,n),f(paste0(b,a),b,n)) Try it online! Recursive function: input the starting strings b and a (note reversed order), and the 1-based index to output. Could be a bit shorter (53 bytes) if inputs are vectors of characters instead of strings. R, 48 bytes function(...


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Pyth, 58 bytes A(/%Q100T+1/Q100)A(*GH*%*GHT%QT)s[+T/QT*"0"<GTG%QT*"0"<HTH Try it online! Same as my Python solution. A assigns the two provided elements to the variables G and H respectively. Q is the input (zero-indexed).


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C (clang), 107 99 bytes Saved 8 bytes thanks to the man himself Arnauld!!! a;b;p;q;f(){for(a=0;9/++a;)for(b=0;b<100;)printf("%d%d%02d%d%02d ",a,p=b/10,p*=a,q=b++%10,p%10*q);} Try it online! Prints them all.


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JavaScript (SpiderMonkey), 65 bytes for(i=99;i<999;print((e-i+'0'-a*b+e)*~99+a*b%10*e))[a,b,e]=++i+'' Try it online! For example, when i == 267: i=267; [a,b,e]=i+''; print([a, b, e]); // ["2", "6", "7"] print(e-i); // -260 print(e-i+'0'); // "-2600" print(a*b); // 12 print(e-i+'0'-a*b); // -2612 print(e-i+'0'-a*b+...


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Python 2, 121 115 98 bytes def f(n,t=10):p,q=n%100/t,1+n/100;r=p*q;s=n%t*(r%t);print`t+n/t`+'0'*(r<t)+`r`+`n%t`+'0'*(s<t)+`s` Try it online! On noticing the digits, I observed a pattern and got this constant time solution. Takes n as input (0-indexed) and prints n'th meeker number. For example, the first 2 digits follow the pattern: 10, 11, 12, ... ...


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Charcoal, 33 bytes F…χ¹⁰⁰Fχ⟦⪫⟦ι﹪%02dΠικ﹪%02d×﹪Πιχκ⟧ω Try it online! Link is to verbose version of code. Prints all 900 Meeker numbers. Explanation: F…χ¹⁰⁰ Loop over all possible first pairs of digits. Fχ Loop over all possible fifth digits. ⟦⪫⟦...⟧ω Print the following items on one line and then move to the next line. ι The first two digits. ﹪%02dΠι ...


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Haskell, 82 80 76 bytes -6 bytes thanks to kops, for rearranging things and proposing a better interpretation of the rules :P [n|n<-show<$>[1..],[a,b,c,d,e,f,g]<-[read.pure<$>n],a*b==10*c+d,d*e==10*f+g] Try it online! The list of all Meeker numbers, represented as strings. Haskell, 80 bytes [n>>=show|n@[a,b,c,d,e,f,g]<-mapM([0.....


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APL (Dyalog Extended), 45 bytes (SBCS) Full program. Prints all meeker numbers. Requires 0-based indexing (⎕IO←0). 1e6+⍸{(×⌿i⊇⍵)≡10⊥⍵[2+i←0 1,⍪3 4]}¨10⊤¨1e6…1e7 Try it online! (limited to upper bound of 2 000 000 ― above code works offline) 1e6…1e7 numbers 1 000 000 through 10 000 000 10⊤¨ base-10 representation of each (splits digits of numbers into lists ...


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Java (JDK), 70 bytes (a,b,n)->{for(var t=a;b.length()<=n;a=b,b=t)t=b+a;return b.charAt(n);} Try it online!


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Husk, 27 bytes föΛo§=oΠ←od→C2§e←→X4dfo=7LN Try it online! returns the list of all meeker numbers. It's horribly inefficient, so here's a version which starts from 1e7, and shows the first n values. EDIT: corrected the answer after Dominic Van Essen found a bug.


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JavaScript (Node.js), 86 bytes _=>[...Array(1e7).keys()].filter(x=>(s=x+'')[0]*s[1]==s[2]+s[3]&&s[3]*s[4]==s[5]+s[6]) Try it online! I feel like this is too long. Explanation Basically, this gets all numbers from 0 to 9999999, coerces each to a string and checks character-wise. One of Javascript's quirks is that multiplied strings are coerced ...


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R, 96 92 84 bytes function(A,B,I,`!`=function(k)"if"(k,"if"(k>1,paste0(!k-1,!k-2),B),A))substr(!I,I,I) Try it online! Takes I 1-indexed. -8 bytes thanks to @Dominic


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Arn -al, 9 bytes I'm not particularly sure why this works... w®¦€•3=⁺■ Try it! 1-indexed, outputs the nth term Explained Unpacked: 1 1 1{#+}-> Exploits Arn's weird precedence bugs to save a byte. A more "normal" version would have {_ _+} as the block or something like {#_+} (as # is a suffix) [ ... ] Implied by `-a` flag 1 First ...


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Vyxal, d, 5 bytes 5vτ½⌈ Try it Online! A port of the Jelly answer which is a port of short husk answer. Explained 5vτ½⌈ 5vτ # convert each digit of the input to base 5 ½ # halve each item in that list (halving vectorises all the way down) ⌈ # ceiling each item in that list # -d deep sums the list and implicitly outputs


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Charcoal, 42 bytes ≔⁻NLζθ≔⁰εW¬‹θ⁰«F¬&ε⊗ε≧⁻⁺Lζ∧﹪ε²Lηθ≦⊕廧⁺ηζθ Try it online! Link is to verbose version of code. Takes the index as the first input. Explanation: I wanted to avoid building up a humunguous string but the code is still slow because I don't know a good way of calculating Fibbinary numbers. ≔⁻NLζθ Subtract the length of B from n. ≔⁰ε Start ...


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Charcoal, 13 bytes IΣ⭆S⭆↨Iι⁵L↨λ³ Try it online! Link is to verbose version of code. Explanation: S Convert input to a string ⭆ Map over characters and join ι Current character I Cast to integer ↨ ⁵ Convert to base 5 ⭆ Map over base 5 digits and join λ Current ...


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Retina 0.8.2, 20 15 bytes . $*1, 1{5}|11? Try it online! Link includes test cases. Explanation: . $*1, Convert each digit to unary separately. 1{5}|11? Count the number of 5s, 2s and 1s needed to make each digit.


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Python 3 + Z3, N=5 Not 100% sure if this is correct. Basically a standard SAT/SMT coding of the Hamiltonian cycle problem, with additional clauses to block intersecting diagonals and collinear segments. Each time a solution is found all of its distinct rotations/flips are blocked. I've run it up to N=6 (2120 solutions found), which took 30 minutes. Edit: I ...


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Python 2, 37 bytes f=lambda n:n and n/5%2-n%5/-2+f(n/10) Try it online! Uses a formula rather than a lookup table for each digit. n/5%2 counts the five-cent coin, and subtracting -n%5/-2 is equivalent to adding (n%5+1)/2 for the one- and two-cent coins. n%10 0 1 2 3 4 5 6 7 8 9 ----------------------------- n/5%2 0 0 0 0 0 1 1 1 1 1 0-n%5/-2 0 1 ...


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Wolfram Language (Mathematica), 39 bytes ⌈.4#⌉-⌊#/5⌋&@*IntegerDigits/*Tr Try it online!


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05AB1E, 6 bytes Takes inputs [A, B] and n. λèì}sè Try it online! Commented: λè } # get the nth element of the sequence generated by: ì # prepending the current string to the last string # that starts with [A, B] s # swap implicit input n to the top of the stack è # index with n into the nth element of the sequence


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PowerShell, 63 bytes param($x,$y,$n)$a=$x,$y 1..$n|%{$a+=$a[$_]+$a[$_-1]} $a[-1][$n] Try it online! -11 bytes thanks to julian


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Haskell, 15 bytes a#b=b++b#(a++b) Try it online! Takes strings a and b as input, returns the whole infinite Fibonacci word, as is usually allowed in sequence challenges. How? Not much to say. This answer relies on the identity $$ F(a,b)=b+F(b,a+b), $$ where \$F(a,b)\$ is the infinite Fibonacci word with starting words \$a\$ and \$b\$.


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Husk, 5 bytes A rip-off of Delfad0r's beautiful Haskell answer. Go upvote that! S+S₀+ Try it online! Returns the entire infinite string/list. S+S₀+ a b is (S+) ((S₀) (+a)) b, which expands to (+b) ((S₀ (+a)) b) (where ₀ is a self-reference to the main function) and then to (+b) (₀ b (+a b)), which is basically b + F(b, a + b). Safer answer, 8 bytes !₁ S+S₁+ ...


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Jelly, 6 5 bytes ⁹;¡⁵ị Try it online! Uses 1 indexing -1 byte thanks to Jonathan Allan, noticing that we could avoid ⁵ becoming the right argument to ;¡ by forcing ;¡ into a nilad-dyad pair with ⁹! Dyadic ¡ is essentially Jelly's generalised Fibonacci operator How it works ⁹;¡⁵ị - Main link. Takes A on the left, B on the right and I as the third argument ⁹ ...


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Python 3, 36 bytes f=lambda a,b,i:b[i:i+1]or f(b,b+a,i) Try it online! -8 bytes thanks to dingledooper Not a particularly creative approach. In fact, basically this is just what l4m2 did but Python will error when accessing out of bounds instead of returning undefined. Using b[i:i+1] returns b[i] (for strings) if it's in range, but doesn't error and ...


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JavaScript (Node.js), 26 bytes n=>g=a=>b=>b[n]||g(b)(b+a) Try it online! Thank tsh for -1 Byte


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Python 2, 48 bytes f=lambda x:x and int("0112212233"[x%10])+f(x/10) Try it online! 49 bytes in Python 3 because you'd need // for floor division.


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Husk, 11 9 8 bytes Edit: -1 byte thanks to caird coinheringaahing ṁo⌈½ṁB5d Try it online! d # get the digits ṁB5 # convert them all to base-5 # (this gives a 1 for each 5-denomination coin needed, # as well as the leftover for each digit. # We'll need 2 more coins for those with leftover 3 or 4, # ...


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05AB1E, 11 bytes S<•δ¬Èº•sèO Try it online! Same approach as my Jelly answer. How it works S<•δ¬Èº•sèO - Program. Input: n S - Cast n to digits < - Decrement •δ¬Èº• - Compressed integer: 1122122330 sè - Using n's digits, index into the digits of 1122122330 O - Sum Kudos to Kevin Cruijssen's excellent ...


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R, 54 51 47 45 bytes Edit: converted to console input instead of a full function to try not to fall behind Robin Ryder's answer d=utf8ToInt(scan(,''))-48;sum(d>0,d>5,d%%5>2) Try it online!


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JavaScript (ES7),  36  35 bytes Similar to other answers. Using a Black Magic formula instead of a lookup table. f=n=>n&&(n%10)**29%3571%4+f(n/10|0) Try it online! Here is a script that looks for \$(p,m)\$ pairs such that \$(n^p\bmod m)\bmod 4=a_n\$ for all \$n\in[0..9]\$. It's worth noting that this code takes IEEE-754 precision errors into ...


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C (gcc), 39 bytes f(n){n=n?f(n/10)+""[n%10]:0;} Try it online! JavaScript (Node.js), 37 bytes f=n=>n&&+"0112212233"[n%10]+f(n/10|0) Try it online!


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Haskell, 55…41 40 bytes -1 byte thanks to xnor, for using a string instead of the hard-coded list. a=0:tail[i+read[j]|i<-a,j<-"0112212233"] Try it online! a is the infinite sequence. How? It's not hard to find the recursive formula $$ a(n)=a\left(\left\lfloor\frac{n}{10}\right\rfloor\right)+a(n \operatorname{mod} 10) $$ with the base cases ...


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Husk, 23 bytes L◄Lfo=⁰Σ↑o≤⁰LṖƒ(+İ€m*10 Try it online! Extremely slow past 11. Explanation L◄Lfo=⁰Σ↑o≤⁰LṖƒ(+İ€m*10 ƒ( create an infinite list using: İ€ currency denomination builtin: [1,1/2,1/5,...200] + plus m*10 the input mapped to *10 this gives [1,1/...


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R, 64 52 50 45 bytes sum(c(1,2,1:3)[.6*utf8ToInt(scan(,""))-28.4]) Try it online! Same strategy as Delfad0r's Haskell answer, which is nicely explained. First, scan(,"") reads in input as a string. Then, utf8ToInt(...)-48 takes a string of digits and converts it to a vector of integer digits. This works out shorter than taking input as ...


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Jelly, 7 bytes Db5FHĊS Try it online! Steals Ports Dominic Van Essen's Husk answer, be sure to upvote that! How it works Db5FHĊS - Main link. Takes an integer n on the left D - Convert to digits b5 - Convert each digit to base 5 F - Flatten H - Halve each Ċ - Ceiling of each S - SUm Jelly, 11 bytes Dị“FȮŀO’D¤S Try it ...


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