I'm working on modernizing Rosetta Code's infrastructure. Starting with communications. Please accept this time-limited open invite to RC's Slack.. --Michael Mol (talk) 20:59, 30 May 2020 (UTC)

# Consecutive primes with ascending or descending dfferences

Consecutive primes with ascending or descending dfferences
You are encouraged to solve this task according to the task description, using any language you may know.

Find and display here on this page, the longest sequence of consecutive prime numbers where the differences between the primes are strictly ascending.

Do the same for sequences of primes where the differences are strictly descending.

In both cases, show the sequence for primes   <   1,000,000.

If there are multiple sequences of the same length, only the first need be shown.

## ALGOL 68

`BEGIN # find sequences of primes where the gaps between the elements #      # are strictly ascending/descending                            #    # reurns a list of primes up to n #    PROC prime list = ( INT n )[]INT:         BEGIN            # sieve the primes to n #            INT no = 0, yes = 1;            [ 1 : n ]INT p;            p[ 1 ] := no; p[ 2 ] := yes;            FOR i FROM 3 BY 2 TO n DO p[ i ] := yes OD;            FOR i FROM 4 BY 2 TO n DO p[ i ] := no  OD;            FOR i FROM 3 BY 2 TO ENTIER sqrt( n ) DO                IF p[ i ] = yes THEN FOR s FROM i * i BY i + i TO n DO p[ s ] := no OD FI            OD;            # replace the sieve with a list #            INT p pos := 0;            FOR i TO n DO IF p[ i ] = yes THEN p[ p pos +:= 1 ] := i FI OD;            p[ 1 : p pos ]         END # prime list # ;    # shos the results of a search #    PROC show sequence = ( []INT primes, STRING seq name, INT seq start, seq length )VOID:         BEGIN            print( ( "    The longest sequence of primes with "                   , seq name                   , " differences contains "                   , whole( seq length, 0 )                   , " primes"                   , newline                   , "        First such sequence (differences in brackets):"                   , newline                   , "            "                   )                 );            print( ( whole( primes[ seq start ], 0 ) ) );            FOR p FROM seq start + 1 TO seq start + ( seq length - 1 ) DO                print( ( " (", whole( ABS( primes[ p ] - primes[ p - 1 ] ), 0 ), ") ", whole( primes[ p ], 0 ) ) )            OD;            print( ( newline ) )         END # show seuence # ;    # find the longest sequence of primes where the successive differences are ascending/descending #    PROC find sequence = ( []INT primes, BOOL ascending, REF INT seq start, seq length )VOID:         BEGIN            seq start     := seq length := 0;            INT start diff = IF ascending THEN 0 ELSE UPB primes + 1 FI;            FOR p FROM LWB primes TO UPB primes DO                INT prev diff := start diff;                INT length    := 1;                FOR s FROM p + 1 TO UPB primes                WHILE INT diff = ABS ( primes[ s ] - primes[ s - 1 ] );                      IF ascending THEN diff > prev diff ELSE diff < prev diff FI                DO                    length   +:= 1;                    prev diff := diff                OD;                IF length > seq length THEN                    # found a longer sequence #                    seq length := length;                    seq start  := p                FI            OD         END # find sequence #;    INT max number   = 1 000 000;    []INT primes     = prime list( max number );    INT asc length  := 0;    INT asc start   := 0;    INT desc length := 0;    INT desc start  := 0;    find sequence( primes, TRUE,   asc start,  asc length );    find sequence( primes, FALSE, desc start, desc length );    # show the sequences #    print( ( "For primes up to ", whole( max number, 0 ), newline ) );    show sequence( primes, "ascending",   asc start,  asc length );    show sequence( primes, "descending", desc start, desc length )END`
Output:
```For primes up to 1000000
The longest sequence of primes with ascending differences contains 8 primes
First such sequence (differences in brackets):
128981 (2) 128983 (4) 128987 (6) 128993 (8) 129001 (10) 129011 (12) 129023 (14) 129037
The longest sequence of primes with descending differences contains 8 primes
First such sequence (differences in brackets):
322171 (22) 322193 (20) 322213 (16) 322229 (8) 322237 (6) 322243 (4) 322247 (2) 322249
```

## C#

Extended the limit up to see what would happen.

`using System.Linq;using System.Collections.Generic;using TG = System.Tuple<int, int>;using static System.Console; class Program{    static void Main(string[] args)    {        const int mil = (int)1e6;        foreach (var amt in new int[] { 1, 2, 6, 12, 18 })        {            int lmt = mil * amt, lg = 0, ng, d, ld = 0;            var desc = new string[] { "A", "", "De" };            int[] mx = new int[] { 0, 0, 0 },                  bi = new int[] { 0, 0, 0 },                   c = new int[] { 2, 2, 2 };            WriteLine("For primes up to {0:n0}:", lmt);            var pr = PG.Primes(lmt).ToArray();            for (int i = 0; i < pr.Length; i++)            {                ng = pr[i].Item2; d = ng.CompareTo(lg) + 1;                if (ld == d)                    c[2 - d]++;                else                {                    if (c[d] > mx[d]) { mx[d] = c[d]; bi[d] = i - mx[d] - 1; }                    c[d] = 2;                }                ld = d; lg = ng;            }            for (int r = 0; r <= 2; r += 2)            {                Write("{0}scending, found run of {1} consecutive primes:\n  {2} ",                    desc[r], mx[r] + 1, pr[bi[r]++].Item1);                foreach (var itm in pr.Skip(bi[r]).Take(mx[r]))                    Write("({0}) {1} ", itm.Item2, itm.Item1); WriteLine(r == 0 ? "" : "\n");            }        }    }} class PG{    public static IEnumerable<TG> Primes(int lim)    {        bool[] flags = new bool[lim + 1];        int j = 3, lj = 2;        for (int d = 8, sq = 9; sq <= lim; j += 2, sq += d += 8)            if (!flags[j])            {                yield return new TG(j, j - lj);                lj = j;                for (int k = sq, i = j << 1; k <= lim; k += i) flags[k] = true;            }        for (; j <= lim; j += 2)            if (!flags[j])            {                yield return new TG(j, j - lj);                lj = j;            }    }}`
Output:
```For primes up to 1,000,000:
Ascending, found run of 8 consecutive primes:
128981 (2) 128983 (4) 128987 (6) 128993 (8) 129001 (10) 129011 (12) 129023 (14) 129037
Descending, found run of 8 consecutive primes:
322171 (22) 322193 (20) 322213 (16) 322229 (8) 322237 (6) 322243 (4) 322247 (2) 322249

For primes up to 2,000,000:
Ascending, found run of 9 consecutive primes:
1319407 (4) 1319411 (8) 1319419 (10) 1319429 (14) 1319443 (16) 1319459 (18) 1319477 (32) 1319509 (34) 1319543
Descending, found run of 8 consecutive primes:
322171 (22) 322193 (20) 322213 (16) 322229 (8) 322237 (6) 322243 (4) 322247 (2) 322249

For primes up to 6,000,000:
Ascending, found run of 9 consecutive primes:
1319407 (4) 1319411 (8) 1319419 (10) 1319429 (14) 1319443 (16) 1319459 (18) 1319477 (32) 1319509 (34) 1319543
Descending, found run of 9 consecutive primes:
5051309 (32) 5051341 (28) 5051369 (14) 5051383 (10) 5051393 (8) 5051401 (6) 5051407 (4) 5051411 (2) 5051413

For primes up to 12,000,000:
Ascending, found run of 9 consecutive primes:
1319407 (4) 1319411 (8) 1319419 (10) 1319429 (14) 1319443 (16) 1319459 (18) 1319477 (32) 1319509 (34) 1319543
Descending, found run of 10 consecutive primes:
11938793 (60) 11938853 (38) 11938891 (28) 11938919 (14) 11938933 (10) 11938943 (8) 11938951 (6) 11938957 (4) 11938961 (2) 11938963

For primes up to 18,000,000:
Ascending, found run of 10 consecutive primes:
17797517 (2) 17797519 (4) 17797523 (8) 17797531 (10) 17797541 (12) 17797553 (20) 17797573 (28) 17797601 (42) 17797643 (50) 17797693
Descending, found run of 10 consecutive primes:
11938793 (60) 11938853 (38) 11938891 (28) 11938919 (14) 11938933 (10) 11938943 (8) 11938951 (6) 11938957 (4) 11938961 (2) 11938963```

## C++

Library: Primesieve
`#include <cstdint>#include <iostream>#include <vector>#include <primesieve.hpp> void print_diffs(const std::vector<uint64_t>& vec) {    for (size_t i = 0, n = vec.size(); i != n; ++i) {        if (i != 0)            std::cout << " (" << vec[i] - vec[i - 1] << ") ";        std::cout << vec[i];    }    std::cout << '\n';} int main() {    std::cout.imbue(std::locale(""));    std::vector<uint64_t> asc, desc;    std::vector<std::vector<uint64_t>> max_asc, max_desc;    size_t max_asc_len = 0, max_desc_len = 0;    uint64_t prime;    const uint64_t limit = 1000000;    for (primesieve::iterator pi; (prime = pi.next_prime()) < limit; ) {        size_t alen = asc.size();        if (alen > 1 && prime - asc[alen - 1] <= asc[alen - 1] - asc[alen - 2])            asc.erase(asc.begin(), asc.end() - 1);        asc.push_back(prime);        if (asc.size() >= max_asc_len) {            if (asc.size() > max_asc_len) {                max_asc_len = asc.size();                max_asc.clear();            }            max_asc.push_back(asc);        }        size_t dlen = desc.size();        if (dlen > 1 && prime - desc[dlen - 1] >= desc[dlen - 1] - desc[dlen - 2])            desc.erase(desc.begin(), desc.end() - 1);        desc.push_back(prime);        if (desc.size() >= max_desc_len) {            if (desc.size() > max_desc_len) {                max_desc_len = desc.size();                max_desc.clear();            }            max_desc.push_back(desc);        }    }    std::cout << "Longest run(s) of ascending prime gaps up to " << limit << ":\n";    for (const auto& v : max_asc)        print_diffs(v);    std::cout << "\nLongest run(s) of descending prime gaps up to " << limit << ":\n";    for (const auto& v : max_desc)        print_diffs(v);    return 0;}`
Output:
```Longest run(s) of ascending prime gaps up to 1,000,000:
128,981 (2) 128,983 (4) 128,987 (6) 128,993 (8) 129,001 (10) 129,011 (12) 129,023 (14) 129,037
402,581 (2) 402,583 (4) 402,587 (6) 402,593 (8) 402,601 (12) 402,613 (18) 402,631 (60) 402,691
665,111 (2) 665,113 (4) 665,117 (6) 665,123 (8) 665,131 (10) 665,141 (12) 665,153 (24) 665,177

Longest run(s) of descending prime gaps up to 1,000,000:
322,171 (22) 322,193 (20) 322,213 (16) 322,229 (8) 322,237 (6) 322,243 (4) 322,247 (2) 322,249
752,207 (44) 752,251 (12) 752,263 (10) 752,273 (8) 752,281 (6) 752,287 (4) 752,291 (2) 752,293
```

This task uses Extensible Prime Generator (F#)

## F#

` // Longest ascending and decending sequences of difference between consecutive primes: Nigel Galloway. April 5th., 2021let fN g fW=primes32()|>Seq.takeWhile((>)g)|>Seq.pairwise|>Seq.fold(fun(n,i,g)el->let w=fW el in match w>n with true->(w,el::i,g) |_->(w,[el],if List.length i>List.length g then i else g))(0,[],[])for i in [1;2;6;12;18;100] do let _,_,g=fN(i*1000000)(fun(n,g)->g-n) in printfn "Longest ascending upto %d000000->%d:" i (g.Length+1); g|>List.rev|>List.iter(fun(n,g)->printf "%d (%d) %d " n (g-n) g); printfn ""                              let _,_,g=fN(i*1000000)(fun(n,g)->n-g) in printfn "Longest decending upto %d000000->%d:" i (g.Length+1); g|>List.rev|>List.iter(fun(n,g)->printf "%d (%d) %d " n (g-n) g); printfn "" `
Output:
```Longest ascending upto 1000000->8:
128981 (2) 128983 128983 (4) 128987 128987 (6) 128993 128993 (8) 129001 129001 (10) 129011 129011 (12) 129023 129023 (14) 129037
Longest decending upto 1000000->8:
322171 (22) 322193 322193 (20) 322213 322213 (16) 322229 322229 (8) 322237 322237 (6) 322243 322243 (4) 322247 322247 (2) 322249
Longest ascending upto 2000000->9:
1319407 (4) 1319411 1319411 (8) 1319419 1319419 (10) 1319429 1319429 (14) 1319443 1319443 (16) 1319459 1319459 (18) 1319477 1319477 (32) 1319509 1319509 (34) 1319543
Longest decending upto 2000000->8:
322171 (22) 322193 322193 (20) 322213 322213 (16) 322229 322229 (8) 322237 322237 (6) 322243 322243 (4) 322247 322247 (2) 322249
Longest ascending upto 6000000->9:
1319407 (4) 1319411 1319411 (8) 1319419 1319419 (10) 1319429 1319429 (14) 1319443 1319443 (16) 1319459 1319459 (18) 1319477 1319477 (32) 1319509 1319509 (34) 1319543
Longest decending upto 6000000->9:
5051309 (32) 5051341 5051341 (28) 5051369 5051369 (14) 5051383 5051383 (10) 5051393 5051393 (8) 5051401 5051401 (6) 5051407 5051407 (4) 5051411 5051411 (2) 5051413
Longest ascending upto 12000000->9:
1319407 (4) 1319411 1319411 (8) 1319419 1319419 (10) 1319429 1319429 (14) 1319443 1319443 (16) 1319459 1319459 (18) 1319477 1319477 (32) 1319509 1319509 (34) 1319543
Longest decending upto 12000000->10:
11938793 (60) 11938853 11938853 (38) 11938891 11938891 (28) 11938919 11938919 (14) 11938933 11938933 (10) 11938943 11938943 (8) 11938951 11938951 (6) 11938957 11938957 (4) 11938961 11938961 (2) 11938963
Longest ascending upto 18000000->10:
17797517 (2) 17797519 17797519 (4) 17797523 17797523 (8) 17797531 17797531 (10) 17797541 17797541 (12) 17797553 17797553 (20) 17797573 17797573 (28) 17797601 17797601 (42) 17797643 17797643 (50) 17797693
Longest decending upto 18000000->10:
11938793 (60) 11938853 11938853 (38) 11938891 11938891 (28) 11938919 11938919 (14) 11938933 11938933 (10) 11938943 11938943 (8) 11938951 11938951 (6) 11938957 11938957 (4) 11938961 11938961 (2) 11938963
Longest ascending upto 100000000->11:
94097537 (2) 94097539 94097539 (4) 94097543 94097543 (8) 94097551 94097551 (10) 94097561 94097561 (12) 94097573 94097573 (14) 94097587 94097587 (16) 94097603 94097603 (18) 94097621 94097621 (30) 94097651 94097651 (32) 94097683
Longest decending upto 100000000->10:
11938793 (60) 11938853 11938853 (38) 11938891 11938891 (28) 11938919 11938919 (14) 11938933 11938933 (10) 11938943 11938943 (8) 11938951 11938951 (6) 11938957 11938957 (4) 11938961 11938961 (2) 11938963
Real: 00:00:04.708
```

## Factor

Works with: Factor version 0.99 2021-02-05
`USING: arrays assocs formatting grouping io kernel literals mathmath.primes math.statistics sequences sequences.extrastools.memory.private ; << CONSTANT: limit 1,000,000 >> CONSTANT: primes \$[ limit primes-upto ] : run ( n quot -- seq quot )    [ primes ] [ <clumps> ] [ ] tri*    '[ differences _ monotonic? ] ; inline : max-run ( quot -- n )    1 swap '[ 1 + dup _ run find drop ] loop 1 - ; inline : runs ( quot -- seq )    [ max-run ] keep run filter ; inline : .run ( seq -- )    dup differences [ [ commas ] map ] [email protected]    [ "(" ")" surround ] map 2array round-robin " " join print ; : .runs ( quot -- )    [ runs ] keep [ < ] = "rising" "falling" ? limit commas    "Largest run(s) of %s gaps between primes less than %s:\n"    printf [ .run ] each ; inline [ < ] [ > ] [ .runs nl ] [email protected]`
Output:
```Largest run(s) of rising gaps between primes less than 1,000,000:
128,981 (2) 128,983 (4) 128,987 (6) 128,993 (8) 129,001 (10) 129,011 (12) 129,023 (14) 129,037
402,581 (2) 402,583 (4) 402,587 (6) 402,593 (8) 402,601 (12) 402,613 (18) 402,631 (60) 402,691
665,111 (2) 665,113 (4) 665,117 (6) 665,123 (8) 665,131 (10) 665,141 (12) 665,153 (24) 665,177

Largest run(s) of falling gaps between primes less than 1,000,000:
322,171 (22) 322,193 (20) 322,213 (16) 322,229 (8) 322,237 (6) 322,243 (4) 322,247 (2) 322,249
752,207 (44) 752,251 (12) 752,263 (10) 752,273 (8) 752,281 (6) 752,287 (4) 752,291 (2) 752,293
```

## FreeBASIC

Use any of the primality testing code on this site as an include; I won't reproduce it here.

`#define UPPER 1000000#include"isprime.bas" dim as uinteger champ = 0, record = 0, streak, i, j, n 'first generate all the primes below UPPERredim as uinteger prime(1 to 2)prime(1) = 2 : prime(2) = 3for i = 5 to UPPER step 2    if isprime(i) then        redim preserve prime(1 to ubound(prime) + 1)        prime(ubound(prime)) = i    end ifnext in = ubound(prime) 'now look for the longest streak of ascending primesfor i = 2 to n-1    j = i + 1    streak = 1    while j<=n andalso prime(j)-prime(j-1) > prime(j-1)-prime(j-2)        streak += 1        j+=1    wend    if streak > record then        champ = i-1        record = streak    end ifnext i print "The longest sequence of ascending primes (with their difference from the last one) is:"for i = champ+1 to champ+record    print prime(i-1);" (";prime(i)-prime(i-1);") ";next iprint prime(i-1) : print'now for the descending ones record = 0 : champ = 0for i = 2 to n-1    j = i + 1    streak = 1    while j<=n andalso prime(j)-prime(j-1) < prime(j-1)-prime(j-2)   'identical to above, but for the inequality sign        streak += 1        j+=1    wend    if streak > record then        champ = i-1        record = streak    end ifnext i print "The longest sequence of descending primes (with their difference from the last one) is:"for i = champ+1 to champ+record    print prime(i-1);" (";prime(i)-prime(i-1);") ";next iprint prime(i-1)`
Output:
```The longest sequence of ascending primes (with their difference from the last one) is:
128981 (2) 128983 (4) 128987 (6) 128993 (8) 129001 (10) 129011 (12) 129023 (14) 129037

The longest sequence of descending primes (with their difference from the last one) is:
322171 (22) 322193 (20) 322213 (16) 322229 (8) 322237 (6) 322243 (4) 322247 (2) 322249```

## Go

Translation of: Wren
`package main import (    "fmt"    "rcu") const LIMIT = 999999 var primes = rcu.Primes(LIMIT) func longestSeq(dir string) {    pd := 0    longSeqs := [][]int{{2}}    currSeq := []int{2}    for i := 1; i < len(primes); i++ {        d := primes[i] - primes[i-1]        if (dir == "ascending" && d <= pd) || (dir == "descending" && d >= pd) {            if len(currSeq) > len(longSeqs[0]) {                longSeqs = [][]int{currSeq}            } else if len(currSeq) == len(longSeqs[0]) {                longSeqs = append(longSeqs, currSeq)            }            currSeq = []int{primes[i-1], primes[i]}        } else {            currSeq = append(currSeq, primes[i])        }        pd = d    }    if len(currSeq) > len(longSeqs[0]) {        longSeqs = [][]int{currSeq}    } else if len(currSeq) == len(longSeqs[0]) {        longSeqs = append(longSeqs, currSeq)    }    fmt.Println("Longest run(s) of primes with", dir, "differences is", len(longSeqs[0]), ":")    for _, ls := range longSeqs {        var diffs []int        for i := 1; i < len(ls); i++ {            diffs = append(diffs, ls[i]-ls[i-1])        }        for i := 0; i < len(ls)-1; i++ {            fmt.Print(ls[i], " (", diffs[i], ") ")        }        fmt.Println(ls[len(ls)-1])    }    fmt.Println()} func main() {    fmt.Println("For primes < 1 million:\n")    for _, dir := range []string{"ascending", "descending"} {        longestSeq(dir)    }}`
Output:
```For primes < 1 million:

Longest run(s) of primes with ascending differences is 8 :
128981 (2) 128983 (4) 128987 (6) 128993 (8) 129001 (10) 129011 (12) 129023 (14) 129037
402581 (2) 402583 (4) 402587 (6) 402593 (8) 402601 (12) 402613 (18) 402631 (60) 402691
665111 (2) 665113 (4) 665117 (6) 665123 (8) 665131 (10) 665141 (12) 665153 (24) 665177

Longest run(s) of primes with descending differences is 8 :
322171 (22) 322193 (20) 322213 (16) 322229 (8) 322237 (6) 322243 (4) 322247 (2) 322249
752207 (44) 752251 (12) 752263 (10) 752273 (8) 752281 (6) 752287 (4) 752291 (2) 752293
```

## Julia

`using Primes function primediffseqs(maxnum = 1_000_000)    mprimes = primes(maxnum)    diffs = map(p -> mprimes[p[1] + 1] - p[2], enumerate(@view mprimes[begin:end-1]))    incstart, decstart, bestinclength, bestdeclength = 1, 1, 0, 0    for i in 1:length(diffs)-1        foundinc, founddec = false, false        for j in i+1:length(diffs)            if !foundinc && diffs[j] <= diffs[j - 1]                if (runlength = j - i) > bestinclength                    bestinclength, incstart = runlength, i                end                foundinc = true            end            if !founddec && diffs[j] >= diffs[j - 1]                if (runlength = j - i) > bestdeclength                    bestdeclength, decstart = runlength, i                end                founddec = true            end            foundinc && founddec && break        end    end    println("Ascending: ", mprimes[incstart:incstart+bestinclength], " Diffs: ", diffs[incstart:incstart+bestinclength-1])    println("Descending: ", mprimes[decstart:decstart+bestdeclength], " Diffs: ", diffs[decstart:decstart+bestdeclength-1])end primediffseqs() `
Output:
```Ascending: [128981, 128983, 128987, 128993, 129001, 129011, 129023, 129037] Diffs: [2, 4, 6, 8, 10, 12, 14]
Descending: [322171, 322193, 322213, 322229, 322237, 322243, 322247, 322249] Diffs: [22, 20, 16, 8, 6, 4, 2]
```

## Lua

This task uses `primegen` from: Extensible_prime_generator#Lua

`function findcps(primelist, fcmp)  local currlist = {primelist[1]}  local longlist, prevdiff = currlist, 0  for i = 2, #primelist do    local diff = primelist[i] - primelist[i-1]    if fcmp(diff, prevdiff) then      currlist[#currlist+1] = primelist[i]      if #currlist > #longlist then        longlist = currlist      end    else      currlist = {primelist[i-1], primelist[i]}    end    prevdiff = diff  end  return longlistend primegen:generate(nil, 1000000)cplist = findcps(primegen.primelist, function(a,b) return a>b end)print("ASC  ("..#cplist.."):  ["..table.concat(cplist, " ").."]")cplist = findcps(primegen.primelist, function(a,b) return a<b end)print("DESC ("..#cplist.."):  ["..table.concat(cplist, " ").."]")`
Output:
```ASC  (8):  [128981 128983 128987 128993 129001 129011 129023 129037]
DESC (8):  [322171 322193 322213 322229 322237 322243 322247 322249]```

## Mathematica/Wolfram Language

`prime = Prime[Range[PrimePi[10^6]]];s = Split[Differences[prime], Less];max = Max[Length /@ s];diffs = Select[s, Length/*EqualTo[max]];seqs = SequencePosition[Flatten[s], #, 1][[1]] & /@ diffs;Take[prime, # + {0, 1}] & /@ seqs s = Split[Differences[prime], Greater];max = Max[Length /@ s];diffs = Select[s, Length/*EqualTo[max]];seqs = SequencePosition[Flatten[s], #, 1][[1]] & /@ diffs;Take[prime, # + {0, 1}] & /@ seqs`
Output:
```128981	128983	128987	128993	129001	129011	129023	129037
402581	402583	402587	402593	402601	402613	402631	402691
665111	665113	665117	665123	665131	665141	665153	665177

322171	322193	322213	322229	322237	322243	322247	322249
752207	752251	752263	752273	752281	752287	752291	752293```

## Nim

`import math, strformat, sugar const N = 1_000_000 ##################################################################################################### Erathostenes sieve. var composite: array[2..N, bool]    # Initialized to false i.e. prime. for n in 2..int(sqrt(float(N))):  if not composite[n]:    for k in countup(n * n, N, n):      composite[k] = true let primes = collect(newSeq):               for n in 2..N:                 if not composite[n]: n  ##################################################################################################### Longest sequences. type Order {.pure.} = enum Ascending, Descending proc longestSeq(order: Order): seq[int] =  ## Return the longest sequence for the given order.   let ascending = order == Ascending  var    currseq: seq[int]    prevPrime = 2    diff = if ascending: 0 else: N   for prime in primes:    let nextDiff = prime - prevPrime    if nextDiff != diff and nextDiff > diff == ascending:      currseq.add prime    else:      if currseq.len > result.len:        result = move(currseq)      currseq = @[prevPrime, prime]    diff = nextDiff    prevPrime = prime   if currseq.len > result.len:    result = move(currseq)  proc `\$`(list: seq[int]): string =  ## Return the representation of a list of primes with interleaved differences.  var prevPrime: int  for i, prime in list:    if i != 0: result.add &" ({prime - prevPrime}) "    result.addInt prime    prevPrime = prime echo "For primes < 1000000.\n"echo "First longest sequence of consecutive primes with ascending differences:"echo longestSeq(Ascending)echo()echo "First longest sequence of consecutive primes with descending differences:"echo longestSeq(Descending)`
Output:
```For primes < 1000000.

First longest sequence of consecutive primes with ascending differences:
128981 (2) 128983 (4) 128987 (6) 128993 (8) 129001 (10) 129011 (12) 129023 (14) 129037

First longest sequence of consecutive primes with descending differences:
322171 (22) 322193 (20) 322213 (16) 322229 (8) 322237 (6) 322243 (4) 322247 (2) 322249```

## Pari/GP

Code is pretty reasonable, runs in ~70 ms at 1,000,000. Running under PARI (with gp2c translation) could take advantage of the diff structure of the prime table directly for small cases and avoid substantial overhead, gaining at least a factor of 2 in performance.

`showPrecPrimes(p, n)={  my(v=vector(n));  v[n]=p;  forstep(i=n-1,1,-1,    v[i]=precprime(v[i+1]-1)  );  for(i=1,n, print1(v[i]" "));}list(lim)={  my(p=3,asc,dec,ar,dr,arAt=3,drAt=3,last=2);  forprime(q=5,lim,    my(g=q-p);    if(g<last,      asc=0;      if(desc++>dr,        dr=desc;        drAt=q      )    ,g>last,      desc=0;      if(asc++>ar,        ar=asc;        arAt=q      )    ,      asc=desc=0    );    p=q;    last=g  );  print("Descending differences:");  showPrecPrimes(drAt, dr+2);  print("\nAscending differences:");  showPrecPrimes(arAt, ar+2);}list(10^6)`
Output:
```Descending differences:
322171 322193 322213 322229 322237 322243 322247 322249
Ascending differences:
128981 128983 128987 128993 129001 129011 129023 129037```

## Perl

Translation of: Raku
Library: ntheory
`use strict;use warnings;use feature 'say';use ntheory 'primes';use List::AllUtils <indexes max>; my \$limit  = 1000000;my @primes = @{primes( \$limit )}; sub runs {    my(\$op) = @_;    my @diff = my \$diff = my \$run = 1;    push @diff, map {        my \$next = \$primes[\$_] - \$primes[\$_ - 1];        if (\$op eq '>') { if (\$next > \$diff) { ++\$run } else { \$run = 1 } }        else            { if (\$next < \$diff) { ++\$run } else { \$run = 1 } }        \$diff = \$next;        \$run    } 1 .. \$#primes;     my @prime_run;    my \$max = max @diff;    for my \$r ( indexes { \$_ == \$max } @diff ) {        push @prime_run, join ' ', map { \$primes[\$r - \$_] } reverse 0..\$max    }    @prime_run} say   "Longest run(s) of ascending prime gaps up to \$limit:\n"  . join "\n", runs('>');say "\nLongest run(s) of descending prime gaps up to \$limit:\n" . join "\n", runs('<');`
Output:
```Longest run(s) of ascending prime gaps up to 1000000:
128981 128983 128987 128993 129001 129011 129023 129037
402581 402583 402587 402593 402601 402613 402631 402691
665111 665113 665117 665123 665131 665141 665153 665177

Longest run(s) of descending prime gaps up to 1000000:
322171 322193 322213 322229 322237 322243 322247 322249
752207 752251 752263 752273 752281 752287 752291 752293```

## Phix

```integer pn = 1, -- prime numb
lp = 2, -- last prime
lg = 0, -- last gap
pd = 0  -- prev d
sequence cr = {0,0},    -- curr run [a,d]
mr = {{0},{0}} -- max runs  ""
while true do
pn += 1
integer p = get_prime(pn), gap = p-lp,
d = compare(gap,lg)
if p>1e6 then exit end if
if d then
integer i = (3-d)/2
cr[i] = iff(d=pd?cr[i]:lp!=2)+1
if cr[i]>mr[i][1] then mr[i] = {cr[i],pn} end if
end if
{pd,lp,lg} = {d,p,gap}
end while

for run=1 to 2 do
integer {l,e} = mr[run]
sequence p = apply(tagset(e,e-l),get_prime),
g = sq_sub(p[2..\$],p[1..\$-1])
printf(1,"longest %s run length %d: %v gaps: %v\n",
{{"ascending","descending"}[run],length(p),p,g})
end for
```
Output:
```longest ascending run length 8: {128981,128983,128987,128993,129001,129011,129023,129037} gaps: {2,4,6,8,10,12,14}
longest descending run length 8: {322171,322193,322213,322229,322237,322243,322247,322249} gaps: {22,20,16,8,6,4,2}
```

## Python

` from sympy import sieve primelist = list(sieve.primerange(2,1000000)) listlen = len(primelist) # ascending pindex = 1old_diff = -1curr_list=[primelist[0]]longest_list=[] while pindex < listlen:     diff = primelist[pindex] - primelist[pindex-1]    if diff > old_diff:        curr_list.append(primelist[pindex])        if len(curr_list) > len(longest_list):            longest_list = curr_list    else:        curr_list = [primelist[pindex-1],primelist[pindex]]     old_diff = diff    pindex += 1 print(longest_list) # descending pindex = 1old_diff = -1curr_list=[primelist[0]]longest_list=[] while pindex < listlen:     diff = primelist[pindex] - primelist[pindex-1]    if diff < old_diff:        curr_list.append(primelist[pindex])        if len(curr_list) > len(longest_list):            longest_list = curr_list    else:        curr_list = [primelist[pindex-1],primelist[pindex]]     old_diff = diff    pindex += 1 print(longest_list) `
Output:
```[128981, 128983, 128987, 128993, 129001, 129011, 129023, 129037]
[322171, 322193, 322213, 322229, 322237, 322243, 322247, 322249]
```

## Raku

`use Math::Primesieve;use Lingua::EN::Numbers; my \$sieve = Math::Primesieve.new; my \$limit = 1000000; my @primes = \$sieve.primes(\$limit); sub runs (&op) {    my \$diff = 1;    my \$run = 1;     my @diff = flat 1, (1..^@primes).map: {        my \$next = @primes[\$_] - @primes[\$_ - 1];        if &op(\$next, \$diff) { ++\$run } else { \$run = 1 }        \$diff = \$next;        \$run;    }     my \$max = max @diff;    my @runs = @diff.grep: * == \$max, :k;     @runs.map( {        my @run = (0..\$max).reverse.map: -> \$r { @primes[\$_ - \$r] }        flat roundrobin(@run».&comma, @run.rotor(2 => -1).map({[R-] \$_})».fmt('(%d)'));    } ).join: "\n"} say "Longest run(s) of ascending prime gaps up to {comma \$limit}:\n" ~ runs(&infix:«>»); say "\nLongest run(s) of descending prime gaps up to {comma \$limit}:\n" ~ runs(&infix:«<»);`
Output:
```Longest run(s) of ascending prime gaps up to 1,000,000:
128,981 (2) 128,983 (4) 128,987 (6) 128,993 (8) 129,001 (10) 129,011 (12) 129,023 (14) 129,037
402,581 (2) 402,583 (4) 402,587 (6) 402,593 (8) 402,601 (12) 402,613 (18) 402,631 (60) 402,691
665,111 (2) 665,113 (4) 665,117 (6) 665,123 (8) 665,131 (10) 665,141 (12) 665,153 (24) 665,177

Longest run(s) of descending prime gaps up to 1,000,000:
322,171 (22) 322,193 (20) 322,213 (16) 322,229 (8) 322,237 (6) 322,243 (4) 322,247 (2) 322,249
752,207 (44) 752,251 (12) 752,263 (10) 752,273 (8) 752,281 (6) 752,287 (4) 752,291 (2) 752,293```

## REXX

`/*REXX program finds the longest sequence of consecutive primes where the differences   *//*──────────── between the primes are strictly ascending;  also for strictly descending.*/parse arg hi cols .                              /*obtain optional argument from the CL.*/if   hi=='' |   hi==","  then   hi= 1000000      /* "      "         "   "   "     "    */if cols=='' | cols==","  then cols=      10      /* "      "         "   "   "     "    */call genP                                        /*build array of semaphores for primes.*/w= 10                                            /*width of a number in any column.     */call fRun 1;  call show 1                        /*find runs with ascending prime diffs.*/call fRun 0;  call show 0                        /*  "    "    " descending   "     "   */exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/commas: parse arg ?;  do jc=length(?)-3  to 1  by -3; ?=insert(',', ?, jc); end;  return ?/*──────────────────────────────────────────────────────────────────────────────────────*/fRun: parse arg ?;    mxrun=0;     seq.=         /*max run length;  lists of prime runs.*/                                                 /*search for consecutive primes  <  HI.*/        do j=2  for #-2;   cp= @.j;   jn= j+1    /*CP: current prime;  JN:  next j      */        diff= @.jn - cp                          /*get difference between last 2 primes.*/        cnt= 1;                       run=       /*initialize the   CNT   and   RUN.    */               do k= jn+1  to #-2;    km= k-1    /*look for more primes in this run.    */               if ?  then if @.[email protected].km<=diff  then leave  /*Diff. too small? Stop looking*/                                             else nop                     else if @.[email protected].km>=diff  then leave  /*  "    "  large?   "     "   */               run= run  @.k;         cnt= cnt+1 /*append a prime to the run; bump count*/               diff= @.k - @.km                  /*calculate difference for next prime. */               end   /*k*/        if cnt<=mxrun  then iterate              /*This run too short? Then keep looking*/        mxrun= max(mxrun, cnt)                   /*define a new maximum run (seq) length*/        seq.mxrun= cp  @.jn  run                 /*full populate the sequence (RUN).    */        end   /*j*/;                   return/*──────────────────────────────────────────────────────────────────────────────────────*/genP: @.1=2; @.2=3; @.3=5; @.4=7; @.5=11; @.6=13; @.7=17; @.8=19    /*define low primes.*/                           #=8;   sq.#= @.# ** 2 /*number of primes so far; prime sqiare*/                                                 /* [↓]  generate more  primes  ≤  high.*/        do [email protected].#+2  by 2  to hi;  parse var j '' -1 _    /*find odd primes from here on.*/        if     _==5  then iterate;  if j// 3==0  then iterate   /*J ÷  5?    J ÷ by  3? */        if j// 7==0  then iterate;  if j//11==0  then iterate   /*" "  7?    " "  " 11? */        if j//13==0  then iterate;  if j//17==0  then iterate   /*" " 13?    " " "  17? */               do k=8  while sq.k<=j             /* [↓]  divide by the known odd primes.*/               if j // @.k == 0  then iterate j  /*Is  J ÷ X?  Then not prime.     ___  */               end   /*k*/                       /* [↑]  only process numbers  ≤  √ J   */        #= #+1;             @.#= j;    sq.#= j*j /*bump # of Ps; assign next P; P square*/        end          /*j*/;            return/*──────────────────────────────────────────────────────────────────────────────────────*/show: parse arg ?;  if ?  then AorD= 'ascending'     /*choose which literal for display.*/                          else AorD= 'descending'    /*   "     "      "     "     "    */      title= ' longest run of consecutive primes whose differences between primes are'  ,                              'strictly'         AorD         "and  < "         commas(hi)      say;    say;    say      if cols>0  then say ' index │'center(title,   1 + cols*(w+1)     )      if cols>0  then say '───────┼'center(""   ,   1 + cols*(w+1), '─')      found= 0;                idx= 1            /*initialize # of consecutive primes.  */      \$=                                         /*a list of consecutive primes (so far)*/         do o=1  for words(seq.mxrun)            /*show all consecutive primes in seq.  */         c= commas( word(seq.mxrun, o) )         /*obtain the next prime in the sequence*/         found= found + 1                        /*bump the number of consecutive primes*/         if cols<=0            then iterate      /*build the list  (to be shown later)? */         \$= \$  right(c, max(w, length(c) ) )     /*add a nice prime ──► list, allow big#*/         if found//cols\==0    then iterate      /*have we populated a line of output?  */         say center(idx, 7)'│'  substr(\$, 2)     /*display what we have so far  (cols). */         idx= idx + cols;              \$=        /*bump the  index  count for the output*/         end   /*o*/      if \$\==''  then say center(idx, 7)"│"  substr(\$, 2)   /*maybe show residual output*/      if cols>0  then say '───────┴'center(""   ,   1 + cols*(w+1), '─')      say;            say commas(Cprimes)  ' was the'title;        return`
output   when using the default inputs:
``` index │  longest run of consecutive primes whose differences between primes are strictly ascending and  <  1,000,000
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
1   │    128,981    128,983    128,987    128,993    129,001    129,011    129,023    129,037
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────

8  was the longest run of consecutive primes whose differences between primes are strictly ascending and  <  1,000,000

index │  longest run of consecutive primes whose differences between primes are strictly descending and  <  1,000,000
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
1   │    322,171    322,193    322,213    322,229    322,237    322,243    322,247    322,249
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────

8  was the longest run of consecutive primes whose differences between primes are strictly descending and  <  1,000,000
```

## Rust

`// [dependencies]// primal = "0.3" fn print_diffs(vec: &[usize]) {    for i in 0..vec.len() {        if i > 0 {            print!(" ({}) ", vec[i] - vec[i - 1]);        }        print!("{}", vec[i]);    }    println!();} fn main() {    let limit = 1000000;    let mut asc = Vec::new();    let mut desc = Vec::new();    let mut max_asc = Vec::new();    let mut max_desc = Vec::new();    let mut max_asc_len = 0;    let mut max_desc_len = 0;    for p in primal::Sieve::new(limit)        .primes_from(2)        .take_while(|x| *x < limit)    {        let alen = asc.len();        if alen > 1 && p - asc[alen - 1] <= asc[alen - 1] - asc[alen - 2] {            asc = asc.split_off(alen - 1);        }        asc.push(p);        if asc.len() >= max_asc_len {            if asc.len() > max_asc_len {                max_asc_len = asc.len();                max_asc.clear();            }            max_asc.push(asc.clone());        }        let dlen = desc.len();        if dlen > 1 && p - desc[dlen - 1] >= desc[dlen - 1] - desc[dlen - 2] {            desc = desc.split_off(dlen - 1);        }        desc.push(p);        if desc.len() >= max_desc_len {            if desc.len() > max_desc_len {                max_desc_len = desc.len();                max_desc.clear();            }            max_desc.push(desc.clone());        }    }    println!("Longest run(s) of ascending prime gaps up to {}:", limit);    for v in max_asc {        print_diffs(&v);    }    println!("\nLongest run(s) of descending prime gaps up to {}:", limit);    for v in max_desc {        print_diffs(&v);    }}`
Output:
```Longest run(s) of ascending prime gaps up to 1000000:
128981 (2) 128983 (4) 128987 (6) 128993 (8) 129001 (10) 129011 (12) 129023 (14) 129037
402581 (2) 402583 (4) 402587 (6) 402593 (8) 402601 (12) 402613 (18) 402631 (60) 402691
665111 (2) 665113 (4) 665117 (6) 665123 (8) 665131 (10) 665141 (12) 665153 (24) 665177

Longest run(s) of descending prime gaps up to 1000000:
322171 (22) 322193 (20) 322213 (16) 322229 (8) 322237 (6) 322243 (4) 322247 (2) 322249
752207 (44) 752251 (12) 752263 (10) 752273 (8) 752281 (6) 752287 (4) 752291 (2) 752293
```

## Sidef

Translation of: Raku
`func runs(f, arr) {     var run = 0    var diff = 0    var diffs = []     arr.each_cons(2, {|p1,p2|        var curr_diff = (p2 - p1)        f(curr_diff, diff) ? ++run : (run = 1)        diff = curr_diff        diffs << run    })     var max  = diffs.max    var runs = []     diffs.indices_by { _ == max }.each {|i|        runs << arr.slice(i - max + 1, i + 1)    }     return runs} var limit = 1e6var primes = limit.primes say "Longest run(s) of ascending prime gaps up to #{limit.commify}:"say runs({|a,b| a > b }, primes).join("\n") say "\nLongest run(s) of descending prime gaps up to #{limit.commify}:"say runs({|a,b| a < b }, primes).join("\n")`
Output:
```Longest run(s) of ascending prime gaps up to 1,000,000:
[128981, 128983, 128987, 128993, 129001, 129011, 129023, 129037]
[402581, 402583, 402587, 402593, 402601, 402613, 402631, 402691]
[665111, 665113, 665117, 665123, 665131, 665141, 665153, 665177]

Longest run(s) of descending prime gaps up to 1,000,000:
[322171, 322193, 322213, 322229, 322237, 322243, 322247, 322249]
[752207, 752251, 752263, 752273, 752281, 752287, 752291, 752293]
```

## Wren

Library: Wren-math
`import "/math" for Int var LIMIT = 999999var primes = Int.primeSieve(LIMIT) var longestSeq = Fn.new { |dir|    var pd = 0    var longSeqs = [[2]]    var currSeq = [2]    for (i in 1...primes.count) {        var d = primes[i] - primes[i-1]        if ((dir == "ascending" && d <= pd) || (dir == "descending" && d >= pd)) {            if (currSeq.count > longSeqs[0].count) {                longSeqs = [currSeq]            } else if (currSeq.count == longSeqs[0].count) longSeqs.add(currSeq)            currSeq = [primes[i-1], primes[i]]        } else {            currSeq.add(primes[i])        }        pd = d    }    if (currSeq.count > longSeqs[0].count) {        longSeqs = [currSeq]    } else if (currSeq.count == longSeqs[0].count) longSeqs.add(currSeq)    System.print("Longest run(s) of primes with %(dir) differences is %(longSeqs[0].count):")    for (ls in longSeqs) {        var diffs = []        for (i in 1...ls.count) diffs.add(ls[i] - ls[i-1])        for (i in 0...ls.count-1) System.write("%(ls[i]) (%(diffs[i])) ")        System.print(ls[-1])    }    System.print()} System.print("For primes < 1 million:\n")for (dir in ["ascending", "descending"]) longestSeq.call(dir)`
Output:
```For primes < 1 million:

Longest run(s) of primes with ascending differences is 8:
128981 (2) 128983 (4) 128987 (6) 128993 (8) 129001 (10) 129011 (12) 129023 (14) 129037
402581 (2) 402583 (4) 402587 (6) 402593 (8) 402601 (12) 402613 (18) 402631 (60) 402691
665111 (2) 665113 (4) 665117 (6) 665123 (8) 665131 (10) 665141 (12) 665153 (24) 665177

Longest run(s) of primes with descending differences is 8:
322171 (22) 322193 (20) 322213 (16) 322229 (8) 322237 (6) 322243 (4) 322247 (2) 322249
752207 (44) 752251 (12) 752263 (10) 752273 (8) 752281 (6) 752287 (4) 752291 (2) 752293
```

## XPL0

`func IsPrime(N);        \Return 'true' if N > 2 is a prime numberint  N, I;[if (N&1) = 0 \even number\ then return false;for I:= 3 to sqrt(N) do    [if rem(N/I) = 0 then return false;    I:= I+1;    ];return true;]; proc ShowSeq(Dir, Str); \Show longest sequence of distances between primesint  Dir, Str;int  Count, MaxCount, N, P, P0, D, D0, I, AP(1000), MaxAP(1000);[Count:= 0;  MaxCount:= 0;P0:= 2;  D0:= 0;                \preceding prime and distanceAP(Count):= P0;  Count:= Count+1;for N:= 3 to 1_000_000-1 do    if IsPrime(N) then        [P:= N;                 \got a prime number        D:= P - P0;             \distance from preceding prime        if D*Dir > D0*Dir then            [AP(Count):= P;  Count:= Count+1;            if Count > MaxCount then        \save best sequence                [MaxCount:= Count;                for I:= 0 to MaxCount-1 do                    MaxAP(I):= AP(I);                ];            ]        else            [Count:= 0;     \restart sequence            AP(Count):= P0;  Count:= Count+1; \possible beginning            AP(Count):= P;   Count:= Count+1;            ];        P0:= P;  D0:= D;        ];Text(0, "Longest sequence of ");  Text(0, Str);Text(0, " distances between primes: ");  IntOut(0, MaxCount);  CrLf(0);for I:= 0 to MaxCount-2 do    [IntOut(0, MaxAP(I));    Text(0, " (");    IntOut(0, MaxAP(I+1) - MaxAP(I));    Text(0, ") ");    ];IntOut(0, MaxAP(I));  CrLf(0);]; [ShowSeq(+1,  "ascending");     \main ShowSeq(-1, "descending");]`
Output:
```Longest sequence of ascending distances between primes: 8
128981 (2) 128983 (4) 128987 (6) 128993 (8) 129001 (10) 129011 (12) 129023 (14) 129037
Longest sequence of descending distances between primes: 8
322171 (22) 322193 (20) 322213 (16) 322229 (8) 322237 (6) 322243 (4) 322247 (2) 322249
```