Smarandache prime-digital sequence: Difference between revisions
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func main() { |
func main() { |
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fmt.Println("The first 25 terms of the Smarandache prime-digital sequence are:") |
fmt.Println("The first 25 terms of the Smarandache prime-digital sequence are:") |
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⚫ | |||
fmt.Println(" 1. 2") |
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⚫ | |||
n = listSPDSPrimes(n.AddOne(), 4, 25, false) |
n = listSPDSPrimes(n.AddOne(), 4, 25, false) |
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fmt.Println("\nHigher terms:") |
fmt.Println("\nHigher terms:") |
Revision as of 15:05, 4 July 2019
The Smarandache prime-digital sequence (SPDS for brevity) is the sequence of primes whose digits are themselves prime.
For example 257 is an element of this sequence because it is prime itself and its digits: 2, 5 and 7 are also prime.
- Task
- Show the first 25 SPDS primes.
- Show the hundredth SPDS prime.
- See also
- OEIS A019546: Primes whose digits are primes.
- https://www.scribd.com/document/214851583/On-the-Smarandache-prime-digital-subsequence-sequences
F#
This task uses Extensible Prime Generator (F#) <lang fsharp> // Generate Smarandache prime-digital sequence. Nigel Galloway: May 31st., 2019 let rec spds g=seq{yield! g; yield! (spds (Seq.collect(fun g->[g*10+2;g*10+3;g*10+5;g*10+7]) g))}|>Seq.filter(isPrime) spds [2;3;5;7] |> Seq.take 25 |> Seq.iter(printfn "%d") printfn "\n\n100th item of this sequence is %d" (spds [2;3;5;7] |> Seq.item 99) printfn "1000th item of this sequence is %d" (spds [2;3;5;7] |> Seq.item 999) </lang>
- Output:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273 100th item of this sequence is 33223 1000th item of this sequence is 3273527
Factor
Naive
<lang factor>USING: combinators.short-circuit io lists lists.lazy math math.parser math.primes prettyprint sequences ; IN: rosetta-code.smarandache-naive
- smarandache? ( n -- ? )
{ [ number>string string>digits [ prime? ] all? ] [ prime? ] } 1&& ;
- smarandache ( -- list ) 1 lfrom [ smarandache? ] lfilter ;
- smarandache-demo ( -- )
"First 25 members of the Smarandache prime-digital sequence:" print 25 smarandache ltake list>array . "100th member: " write smarandache 99 [ cdr ] times car . ;
MAIN: smarandache-demo</lang>
- Output:
First 25 members of the Smarandache prime-digital sequence: { 2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273 } 100th member: 33223
Optimized
<lang factor>USING: combinators generalizations io kernel math math.functions math.primes prettyprint sequences ; IN: rosetta-code.smarandache
! Observations: ! * For 2-digit numbers and higher, only 3 and 7 are viable in ! the ones place. ! * Only 2, 3, 5, and 7 are viable anywhere else. ! * It is possible to use this information to drastically ! reduce the amount of numbers to check for primality. ! * For instance, by these rules we can tell that the next ! potential Smarandache prime digital after 777 is 2223.
- next-one ( n -- n' ) 3 = 7 3 ? ; inline
- next-ten ( n -- n' )
{ { 2 [ 3 ] } { 3 [ 5 ] } { 5 [ 7 ] } [ drop 2 ] } case ;
- inc ( seq quot: ( n -- n' ) -- seq' )
[ 0 ] 2dip [ change-nth ] curry keep ; inline
- inc1 ( seq -- seq' ) [ next-one ] inc ;
- inc10 ( seq -- seq' ) [ next-ten ] inc ;
- inc-all ( seq -- seq' )
inc1 [ zero? not [ next-ten ] when ] V{ } map-index-as ;
- carry ( seq -- seq' )
dup [ 7 = not ] find drop { { 0 [ inc1 ] } { f [ inc-all 2 suffix! ] } [ cut [ inc-all ] [ inc10 ] bi* append! ] } case ;
- digits>integer ( seq -- n ) [ 10 swap ^ * ] map-index sum ;
- next-smarandache ( seq -- seq' )
[ digits>integer prime? ] [ carry dup ] do until ;
- .sm ( seq -- ) <reversed> [ pprint ] each nl ;
- first25 ( -- )
2 3 5 7 [ . ] 4 napply V{ 7 } clone 21 [ next-smarandache dup .sm ] times drop ;
- nth-smarandache ( n -- )
4 - V{ 7 } clone swap [ next-smarandache ] times .sm ;
- smarandache-demo ( -- )
"First 25 members of the Smarandache prime-digital sequence:" print first25 nl { 100 1000 10000 100000 } [ dup pprint "th member: " write nth-smarandache ] each ;
MAIN: smarandache-demo</lang>
- Output:
First 25 members of the Smarandache prime-digital sequence: 2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273 100th member: 33223 1000th member: 3273527 10000th member: 273322727 100000th member: 23325232253
Go
Basic
<lang go>package main
import (
"fmt" "math/big"
)
var b = new(big.Int)
func isSPDSPrime(n uint64) bool {
nn := n for nn > 0 { r := nn % 10 if r != 2 && r != 3 && r != 5 && r != 7 { return false } nn /= 10 } b.SetUint64(n) if b.ProbablyPrime(0) { // 100% accurate up to 2 ^ 64 return true } return false
}
func listSPDSPrimes(startFrom, countFrom, countTo uint64, printOne bool) uint64 {
count := countFrom for n := startFrom; ; n += 2 { if isSPDSPrime(n) { count++ if !printOne { fmt.Printf("%2d. %d\n", count, n) } if count == countTo { if printOne { fmt.Println(n) } return n } } }
}
func main() {
fmt.Println("The first 25 terms of the Smarandache prime-digital sequence are:") fmt.Println(" 1. 2") n := listSPDSPrimes(3, 1, 25, false) fmt.Println("\nHigher terms:") indices := []uint64{25, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000} for i := 1; i < len(indices); i++ { fmt.Printf("%6d. ", indices[i]) n = listSPDSPrimes(n+2, indices[i-1], indices[i], true) }
}</lang>
- Output:
The first 25 terms of the Smarandache prime-digital sequence are: 1. 2 2. 3 3. 5 4. 7 5. 23 6. 37 7. 53 8. 73 9. 223 10. 227 11. 233 12. 257 13. 277 14. 337 15. 353 16. 373 17. 523 18. 557 19. 577 20. 727 21. 733 22. 757 23. 773 24. 2237 25. 2273 Higher terms: 100. 33223 200. 223337 500. 723337 1000. 3273527 2000. 22332337 5000. 55373333 10000. 273322727 20000. 727535273 50000. 3725522753 100000. 23325232253
Optimized
This version is inspired by the optimizations used in the Factor and Phix entries which are expressed here as a kind of base-4 arithmetic using a digits set of {2, 3, 5, 7} where leading '2's are significant.
This is more than 30 times faster than the above version (runs in about 12.5 seconds on my Celeron @1.6GHx) and could be quickened up further (to around 4 seconds) by using a wrapper for GMP rather than Go's native big.Int type. <lang go>package main
import (
"fmt" "math/big"
)
type B2357 []byte
var bi = new(big.Int)
func isSPDSPrime(b B2357) bool {
bi.SetString(string(b), 10) return bi.ProbablyPrime(0) // 100% accurate up to 2 ^ 64
}
func listSPDSPrimes(startFrom B2357, countFrom, countTo uint64, printOne bool) B2357 {
count := countFrom n := startFrom for { if isSPDSPrime(n) { count++ if !printOne { fmt.Printf("%2d. %s\n", count, string(n)) } if count == countTo { if printOne { fmt.Println(string(n)) } return n } } if printOne { n = n.AddTwo() } else { n = n.AddOne() } }
}
func incDigit(digit byte) byte {
switch digit { case '2': return '3' case '3': return '5' case '5': return '7' default: return '9' // say }
}
func (b B2357) AddOne() B2357 {
le := len(b) b[le-1] = incDigit(b[le-1]) for i := le - 1; i >= 0; i-- { if b[i] < '9' { break } else if i > 0 { b[i] = '2' b[i-1] = incDigit(b[i-1]) } else { b[0] = '2' nb := make(B2357, le+1) copy(nb[1:], b) nb[0] = '2' return nb } } return b
}
func (b B2357) AddTwo() B2357 {
return b.AddOne().AddOne()
}
func main() {
fmt.Println("The first 25 terms of the Smarandache prime-digital sequence are:") n := listSPDSPrimes(B2357{'2'}, 0, 4, false) n = listSPDSPrimes(n.AddOne(), 4, 25, false) fmt.Println("\nHigher terms:") indices := []uint64{25, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000} for i := 1; i < len(indices); i++ { fmt.Printf("%6d. ", indices[i]) n = listSPDSPrimes(n.AddTwo(), indices[i-1], indices[i], true) }
}</lang>
- Output:
Same as before.
Julia
The prime single digits are 2, 3, 5, and 7. Except for 2 and 5, any number ending in 2 or 5 is not prime. So we start with [2, 3, 5, 7] and then add numbers that end in 3 or 7 and that only contain 2, 3, 5, and 7. This can be done via permutations of combinations with repetition. <lang julia> using Combinatorics, Primes
combodigits(len) = sort!(unique(map(y -> join(y, ""), with_replacement_combinations("2357", len))))
function getprimes(N, maxdigits=9)
ret = [2, 3, 5, 7] perms = Int[] for i in 1:maxdigits-1, combo in combodigits(i), perm in permutations(combo) n = parse(Int64, String(perm)) * 10 push!(perms, n + 3, n + 7) end for perm in sort!(perms) if isprime(perm) && !(perm in ret) push!(ret, perm) if length(ret) >= N return ret end end end
end
const v = getprimes(10000) println("The first 25 Smarandache primes are: ", v[1:25]) println("The 100th Smarandache prime is: ", v[100]) println("The 10000th Smarandache prime is: ", v[10000])
</lang>
- Output:
The first 25 Smarandache primes are: [2, 3, 5, 7, 23, 37, 53, 73, 223, 227, 233, 257, 277, 337, 353, 373, 523, 557, 577, 727, 733, 757, 773, 2237, 2273] The 100th Smarandache prime is: 33223 The 10000th Smarandache prime is: 273322727
Perl 6
<lang perl6>use Lingua::EN::Numbers; use ntheory:from<Perl5> <:all>;
- Implemented as a lazy, extendable list
my $spds = grep { .&is_prime }, flat [2,3,5,7], [23,27,33,37,53,57,73,77], -> $p
{ state $o++; my $oom = 10**(1+$o); [ flat (2,3,5,7).map: -> $l { (|$p).map: $l*$oom+* } ] } … *;
say 'Smarandache prime-digitals:'; printf "%22s: %s\n", ordinal(1+$_).tclc, comma $spds[$_] for flat ^25, 99, 999, 9999, 99999;</lang>
- Output:
Smarandache prime-digitals: First: 2 Second: 3 Third: 5 Fourth: 7 Fifth: 23 Sixth: 37 Seventh: 53 Eighth: 73 Ninth: 223 Tenth: 227 Eleventh: 233 Twelfth: 257 Thirteenth: 277 Fourteenth: 337 Fifteenth: 353 Sixteenth: 373 Seventeenth: 523 Eighteenth: 557 Nineteenth: 577 Twentieth: 727 Twenty-first: 733 Twenty-second: 757 Twenty-third: 773 Twenty-fourth: 2,237 Twenty-fifth: 2,273 One hundredth: 33,223 One thousandth: 3,273,527 Ten thousandth: 273,322,727 One hundred thousandth: 23,325,232,253
Phix
Optimised. As noted on the Factor entry, candidates>10 must end in 3 or 7 (since they would not be prime if they ended in 2 or 5), which we efficiently achieve by alternately adding {4,-4}. Digits to the left of that must all be 2/3/5/7, so we add {1,2,2,-5}*10^k to cycle round those digits. Otherwise it is exactly like counting by adding 1 to each digit and carrying 1 left when we do a 9->0.
I had planned to effectively merge a list of potential candidates with a list of all prime numbers, but because of the massive gaps (eg between 777,777,777 and 2,222,222,223) it proved much faster to test each candidate for primality individually. Timings below show just how much this improves things. <lang Phix>atom t0 = time() sequence spds = {2,3,5,7} atom nxt_candidate = 23 sequence adj = {{4,-4},sq_mul({1,2,2,-5},10)},
adjn = {1,1}
include mpfr.e mpz zprime = mpz_init() randstate state = gmp_randinit_mt()
procedure populate_spds(integer n)
while length(spds)<n do mpz_set_d(zprime,nxt_candidate) if mpz_probable_prime_p(zprime,state) then spds &= nxt_candidate end if for i=1 to length(adjn) do sequence adjs = adj[i] integer adx = adjn[i] nxt_candidate += adjs[adx] adx += 1 if adx<=length(adjs) then adjn[i] = adx exit end if adjn[i] = 1 if i=length(adjn) then -- (this is eg 777, by now 223 carry 1, -> 2223) adj = append(adj,sq_mul(adj[$],10)) adjn = append(adjn, 1) nxt_candidate += adj[$][2] exit end if end for end while
end procedure
populate_spds(25) printf(1,"spds[1..25]:%v\n",{spds[1..25]}) for n=2 to 5 do
integer p = power(10,n) populate_spds(p) printf(1,"spds[%d]:%d\n",{p,spds[p]})
end for ?elapsed(time()-t0)</lang>
- Output:
spds[1..25]:{2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273} spds[100]:33223 spds[1000]:3273527 spds[10000]:273322727 spds[100000]:23325232253 "3.6s"
For comparison, on the same machine:
Factor (as optimised) took 45s to calculate the 100,000th number.
Go took 1 min 50 secs to calculate the 100,000th number.
Julia crashed when the limit was changed to 100,000, however it took 11s just to calculate the 10,000th number anyway.
Perl 6 was by far the slowest of all I tried, taking 1 min 15s just to calculate the 10,000th number.
REXX
The prime number generator has been simplified and very little optimization was included. <lang rexx>/*REXX program lists a sequence of SPDS (Smarandache prime-digital sequence) primes.*/ parse arg n q /*get optional number of primes to find*/ if n== | n=="," then n= 25 /*Not specified? Then use the default.*/ if q= then q= 100 /* " " " " " " */ say '═══listing the first' n "SPDS primes═══" call spds n
do i=1 for words(q)+1; y=word(q, i); if y== | y=="," then iterate say say '═══listing the last of ' y "SPDS primes═══" call spds -y end /*i*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ spds: parse arg x 1 ox; x= abs(x) /*obtain the limit to be used for list.*/
c= 0 /*C number of SPDS primes found so far*/ #= 0 /*# number of primes found so far*/ do j=1 by 2 while c<x; z= j /*start: 1st even prime, then use odd. */ if z==1 then z= 2 /*handle the even prime (special case) */ /* [↓] divide by the primes. ___ */ do k=2 to # while k*k<=z /*divide Z with all primes ≤ √ Z */ if z//@.k==0 then iterate j /*÷ by prev. prime? ¬prime ___ */ end /*j*/ /* [↑] only divide up to √ Z */ #= # + 1; @.#= z /*bump the prime count; assign prime #*/ if verify(z, 2357)>0 then iterate j /*Digits ¬prime? Then skip this prime.*/ c= c + 1 /*bump the number of SPDS primes found.*/ if ox<0 then iterate /*don't display it, display the last #.*/ say right(z, 21) /*maybe display this prime ──► terminal*/ end /*j*/ /* [↑] only display N number of primes*/ if ox<0 then say right(z, 21) /*display one (the last) SPDS prime. */ return</lang>
- output when using the default inputs:
═══listing the first 25 SPDS primes═══ 2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273 ═══listing the last of 100 SPDS primes═══ 33223 ═══listing the last of 1000 SPDS primes═══ 3273527
Ring
<lang ring>
- Project: Calmo primes
load "stdlib.ring" limit = 25 max = 300000 num = 0 see "working..." + nl see "wait for done..." + nl see "First 25 Calmo primes are:" + nl for n = 1 to max
if isprime(n) res = calmo(n) if res = 1 num = num + 1 if num < limit + 1 see "" + num + ". " + n + nl ok if num = 100 see "The hundredth Calmo prime is:" + nl see "" + num + ". " + n + nl exit ok ok ok
next see "done..." + nl
func calmo(p)
sp = string(p) for n = 1 to len(sp) if not isprime(sp[n]) return 0 ok next return 1
</lang>
- Output:
working... wait for done... First 25 Calmo primes are: 1. 2 2. 3 3. 5 4. 7 5. 23 6. 37 7. 53 8. 73 9. 223 10. 227 11. 233 12. 257 13. 277 14. 337 15. 353 16. 373 17. 523 18. 557 19. 577 20. 727 21. 733 22. 757 23. 773 24. 2237 25. 2273 The hundredth Calmo prime is: 100. 33223 done...
Sidef
<lang ruby>func is_prime_digital(n) {
n.is_prime && n.digits.all { .is_prime }
}
say is_prime_digital.first(25).join(',') say is_prime_digital.nth(100)</lang>
- Output:
2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273 33223
zkl
GNU Multiple Precision Arithmetic Library
Using GMP ( probabilistic primes), because it is easy and fast to generate primes.
Extensible prime generator#zkl could be used instead. <lang zkl>var [const] BI=Import("zklBigNum"); // libGMP
spds:=Walker.zero().tweak(fcn(ps){
var [const] nps=T(0,0,1,1,0,1,0,1,0,0); // 2,3,5,7 p:=ps.nextPrime().toInt(); if(p.split().filter( fcn(n){ 0==nps[n] }) ) return(Void.Skip); p // 733 --> (7,3,3) --> () --> good, 29 --> (2,9) --> (9) --> bad
}.fp(BI(1)));</lang> Or <lang zkl>spds:=Walker.zero().tweak(fcn(ps){
var [const] nps="014689".inCommon; p:=ps.nextPrime().toInt(); if(nps(p.toString())) return(Void.Skip); p // 733 --> "" --> good, 29 --> "9" --> bad
}.fp(BI(1)));</lang> <lang zkl>println("The first 25 terms of the Smarandache prime-digital sequence are:"); spds.walk(25).concat(",").println();
println("The hundredth term of the sequence is: ",spds.drop(100-25).value); println("1000th item of this sequence is : ",spds.drop(1_000-spds.n).value);</lang>
- Output:
The first 25 terms of the Smarandache prime-digital sequence are: 2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273 The hundredth term of the sequence is: 33223 1000th item of this sequence is : 3273527