Prime reciprocal sum

Prime reciprocal sum is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Generate the sequence of primes where each term is the smallest prime whose reciprocal can be added to the cumulative sum and remain smaller than 1.

E.G.

The cumulative reciprocal sum with zero terms is 0. The smallest prime whose reciprocal can be added to the sum without reaching or exceeding 1 is 2, so the first term is 2 and the new cumulative reciprocal sum is 1/2.

The smallest prime whose reciprocal can be added to the sum without reaching or exceeding 1 is 3. (2 would cause the cumulative reciprocal sum to reach 1.) So the next term is 3, and the new cumulative sum is 5/6.

and so on...

Task

• Find and display the first 10 terms of the sequence. (Or as many as reasonably supported by your language if it is less.) For any values with more than 40 digits, show the first and last 20 digits and the overall digit count.

Stretch

• Find and display the next 5 terms of the sequence. (Or as many as you have the patience for.) Show only the first and last 20 digits and the overall digit count.

See also

• OEIS:A075442 - Slowest-growing sequence of primes whose reciprocals sum to 1

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

Uses Algol 68G's LONG LONG INT which has programmer defined precision.
Re-uses code from the Arithmetic/Rational task, modified to use LONG LONG INT.

Library: ALGOL 68-primes

BEGIN # find a sequence of primes whose members are the smallest prime whose #
      # reciprocal can be added to the sum or the reciprocals of the         #
      # previous primes and the sum remain below 1                           #

    PR precision 5000 PR                # set the precision of LONG LONG INT #
    PR read "primes.incl.a68" PR                   # include prime utilities #

    # iterative Greatest Common Divisor routine, returns the gcd of m and n  #
    PROC gcd = ( LONG LONG INT m, n )LONG LONG INT:
         BEGIN
            LONG LONG INT a := ABS m, b := ABS n;
            WHILE b /= 0 DO
                LONG LONG INT new a = b;
                b        := a MOD b;
                a        := new a
            OD;
            a
        END # gcd # ;

# code from the Arithmetic/Rational task modified to use LONG LONG INT       #
 MODE FRAC = STRUCT( LONG LONG INT num #erator#,  den #ominator# );

  PROC lcm = ( LONG LONG INT a, b )LONG LONG INT:    # least common multiple #
   a OVER gcd(a, b) * b;
 
  PRIO // = 9;                                 # higher then the ** operator #
  OP // = ( LONG LONG INT num, den )FRAC: (       # initialise and normalise #
   LONG LONG INT common = gcd( num, den );
   IF den < 0 THEN
     ( -num OVER common, -den OVER common )
   ELSE
     ( num OVER common, den OVER common )
   FI
 );
 
 OP + = (FRAC a, b)FRAC: (
   LONG LONG INT common = lcm( den OF a, den OF b );
   FRAC result := ( common OVER den OF a * num OF a + common OVER den OF b * num OF b, common );
   num OF result//den OF result
 );

 OP +:= = (REF FRAC a, FRAC b)REF FRAC: ( a := a + b );
 
# end code from the Arithmetic/Rational task modified to use LONG LONG INT   #

    # the sequence starts with the reciprocal of the first prime > 0, i.e. 2 #
    LONG LONG INT one = 1;
    FRAC sum := one // LONG LONG 2;
    print( ( " 1: 2", newline ) );

    # each succeeding prime is the next prime > the denominator of the sum   #
    # divided by the difference of the denominator and the numerator         #
    FOR i FROM 2 TO 15 DO
        LONG LONG INT next := IF num OF sum + 1 = den OF sum
                              THEN den OF sum + 1
                              ELSE ( den OF sum OVER ( den OF sum - num OF sum ) ) + 1
                              FI;
        IF NOT ODD next THEN next +:= 1 FI;
        WHILE NOT is probably prime( next ) DO next +:= 2 OD;
        print( ( whole( i, -2 ), ": " ) );
        STRING prime text = whole( next, 0 );
        IF INT digits = ( UPB prime text - LWB prime text ) + 1;
            digits <= 40
        THEN
            print( ( prime text ) )
        ELSE
            print( ( prime text[ : LWB prime text + 19 ], "..."
                   , prime text[ UPB prime text - 19 : ], "   "
                   , whole( digits, -6 ), " digits"
                   )
                 )
        FI;
        print( ( newline ) );
        sum +:= one // next
    OD

END

Output:

 1: 2
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191...13803869133407474043       76 digits
11: 20300753813848234767...91313959045797597991      150 digits
12: 20323705381471272842...21649394434192763213      297 digits
13: 12748246592672078196...20708715953110886963      592 digits
14: 46749025165138838243...65355869250350888941     1180 digits
15: 11390125639471674628...31060548964273180103     2358 digits

C

Library: GMP

Translation of: Wren

#include <stdio.h>
#include <stdbool.h>
#include <string.h>
#include <gmp.h>

void abbreviate(char a[], const char *s) {
    size_t len = strlen(s);
    if (len < 40) {
        strcpy(a, s);
        return;
    }
    strncpy(a, s, 20);
    strcpy(a + 20, "...");
    strncpy(a + 23, s + len - 20, 21);
}

int main() {
    mpq_t q, r, s, u;
    mpz_t p, t;
    int count = 0, limit = 16;
    size_t len;
    bool isInteger;
    char *ps, a[44];
    mpq_inits(q, r, s, u, NULL);
    mpq_set_ui(u, 1, 1);
    mpz_inits(p, t, NULL);
    printf("First %d elements of the sequence:\n", limit);
    while (count < limit) {
        mpq_sub(q, u, s);
        mpq_inv(q, q);
        mpq_get_den(t, q);
        isInteger = !mpz_cmp_ui(t, 1);
        mpz_set_q(p, q);
        if (!isInteger) mpz_add_ui(p, p, 1);
        mpz_nextprime(p, p);
        ++count;
        ps = mpz_get_str(NULL, 10, p);
        len = strlen(ps);
        if (len <= 40) {
            printf("%2d: %s\n", count, ps);
        } else {
            abbreviate(a, ps);
            printf("%2d: %s (digits: %ld)\n", count, a, len);
        }
        mpq_set_z(r, p);
        mpq_inv(r, r);
        mpq_add(s, s, r);
    }
    mpq_clears(q, r, s, u, NULL);
    mpz_clears(p, t, NULL);
    return 0;
}

Output:

First 16 elements of the sequence:
 1: 2
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191...13803869133407474043 (digits: 76)
11: 20300753813848234767...91313959045797597991 (digits: 150)
12: 20323705381471272842...21649394434192763213 (digits: 297)
13: 12748246592672078196...20708715953110886963 (digits: 592)
14: 46749025165138838243...65355869250350888941 (digits: 1180)
15: 11390125639471674628...31060548964273180103 (digits: 2358)
16: 36961763505630520555...02467094377885929191 (digits: 4711)

J

Given:

taskfmt=: {{
  if. 40<#t=. ":y do.
    d=. ":#t
    (20{.t),'..',(_20{.t),' (',d,' digits)'
  else.
    t
  end.
}}@>

The task and stretch could be:

   taskfmt (, 4 p:1%1-+/@:%)^:(15)i.0
2                                                       
3                                                       
7                                                       
43                                                      
1811                                                    
654149                                                  
27082315109                                             
153694141992520880899                                   
337110658273917297268061074384231117039                 
84241975970641143191..13803869133407474043 (76 digits)  
20300753813848234767..91313959045797597991 (150 digits) 
20323705381471272842..21649394434192763213 (297 digits) 
12748246592672078196..20708715953110886963 (592 digits) 
46749025165138838243..65355869250350888941 (1180 digits)
11390125639471674628..31060548964273180103 (2358 digits)

Here, +/@:% is the sum of reciprocals, so 1%1-+/@:% is the reciprocal of the amount remaining, and 4 p:1%1-+/@:% is the smallest prime which is larger than that value.

Tested in J9.4

Julia

""" rosettacode.org/wiki/Prime_reciprocal_sum """

using Primes
using ResumableFunctions

""" Abbreviate a large string by showing beginning / end and number of chars """
function abbreviate(s; ds = "digits", t = 40, x = (t - 1) ÷ 2)
    wid = length(s)
    return wid < t ? s : s[begin:begin+x] * ".." * s[end-x:end] * " ($wid $ds)"
end

@resumable function generate_oeis75442()
    psum = big"0" // big"1"
    while true
        n = BigInt(ceil(big"1" // (1 - psum)))
        while true 
            n = nextprime(n + 1)
            if psum + 1 // n < 1
                psum += 1 // n
                @yield n
                break
            end
        end
    end
end

for (i, n) in enumerate(Iterators.take(generate_oeis75442(), 17))
    println(lpad(i, 2), ": ", abbreviate(string(n)))
end

Output:

 1: 2   
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191..13803869133407474043 (76 digits)
11: 20300753813848234767..91313959045797597991 (150 digits)
12: 20323705381471272842..21649394434192763213 (297 digits)
13: 12748246592672078196..20708715953110886963 (592 digits)
14: 46749025165138838243..65355869250350888941 (1180 digits)
15: 11390125639471674628..31060548964273180103 (2358 digits)
16: 36961763505630520555..02467094377885929191 (4711 digits)
17: 21043364724798439508..14594683820359204509 (9418 digits)

Nim

Library: bignum

import std/strformat
import bignum

iterator a075442(): Int =
  let One = newRat(1)
  var sum = newRat(0)
  var p = newInt(0)
  while true:
    let q = reciprocal(One - sum)
    p = nextPrime(if q.isInt: q.num else: q.toInt + 1)
    yield p
    sum += newRat(1, p)

func compressed(str: string; size: int): string =
  ## Return a compressed value for long strings of digits.
  if str.len <= 2 * size: str
  else: &"{str[0..<size]}...{str[^size..^1]} ({str.len} digits)"

var count = 0
for p in a075442():
  inc count
  echo &"{count:2}: {compressed($p, 20)}"
  if count == 15: break

Output:

 1: 2
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191...13803869133407474043 (76 digits)
11: 20300753813848234767...91313959045797597991 (150 digits)
12: 20323705381471272842...21649394434192763213 (297 digits)
13: 12748246592672078196...20708715953110886963 (592 digits)
14: 46749025165138838243...65355869250350888941 (1180 digits)
15: 11390125639471674628...31060548964273180103 (2358 digits)

Pascal

Free Pascal

Most time consuming is finding the next prime.
Now pre-sieving with the primes til 65535, to reduce tests.
That is the same as checking all numbers in sieve as divisible by the small primes.Since the prime are in big distance in that region, that's an improvement.

program primeRezSum;
{$IFDEF FPC}  {$MODE DELPHI}{$Optimization ON,ALL} {$ENDIF}
{$IFDEF WINDOWS}{$APPTYPE CONSOLE}{$ENDIF}
uses
  sysutils,
  gmp;
const
  PrimeCount =   6542;
  PRIMEMAXVAL = 65535;
  SIEVESIZE = 65536;
type
  Tsieveprime = record
                  prime,
                  offset : Uint16;
                end;
  tSievePrimes = array[0..PrimeCount-1] of Tsieveprime;
  tSieve       = array[0..SIEVESIZE-1] of byte;
var
  s : AnsiString;
  MyPrimes : tSievePrimes;
  sieve : tSieve;
procedure OutStr(const s: AnsiString);
var
  myString : AnsiString;
  l : Integer;
begin
  myString := copy(s,1,length(s));
  l :=length(pChar(@s[1]));
  setlength(myString,l);
  IF l< 41 then
    writeln(myString)
  else
  begin
    myString[20]:= #0;
    myString[21]:= #0;
    writeln(pChar(@myString[1]),'...',pChar(@myString[l-19]),' (',l:6,' digits)');
  end;
end;

function InitPrimes:Uint32;
var
  f : extended;
  idx,p,pr_cnt : Uint32;
Begin
  fillchar(sieve,Sizeof(sieve),#0);
  pr_cnt := 0;
  p := 2;
  f := 1.0;
  repeat
     while Sieve[p]<> 0 do
       inc(p);
     MyPrimes[pr_cnt].prime := p;
     f := f*(p-1)/p;
     inc(pr_cnt);
     idx := p*p;
     if idx > PRIMEMAXVAL then
       Break;
     repeat
       sieve[idx] := 1;
       inc(idx,p);
     until idx > high(sieve);
     inc(p);
  until sqr(p)>PRIMEMAXVAL;

  while (pr_cnt<= High(MyPrimes)) AND (p<PRIMEMAXVAL)  do
  begin
    while Sieve[p]<> 0 do
      inc(p);
    MyPrimes[pr_cnt].prime := p;
    f := f*(p-1)/p;
    inc(p);
    inc(pr_cnt);
  end;
  Writeln ('reducing factor ',f:10:8);
  result := pr_cnt-1;
end;

procedure DoSieveOffsetInit(var tmp:mpz_t);
var
  dummy :mpz_t;
  i,j,p : Uint32;
Begin
  mpz_init(dummy);
  for i := 0 to High(MyPrimes) do
    with MyPrimes[i] do
    Begin
      if prime = 0 then  Begin  writeln(i);halt;end;
      offset := prime-mpz_tdiv_q_ui(dummy,tmp,prime);
    end;
  mpz_set(dummy,tmp);
  repeat
    //one sieve
    fillchar(sieve,Sizeof(sieve),#0);
    //sieve
    For i := 0 to High(MyPrimes) do
    begin
      with MyPrimes[i] do
      begin
        p := prime;
        j := offset;
      end;
      repeat
        sieve[j] := 1;
        j += p;
      until j >= SIEVESIZE;
      MyPrimes[i].offset := j-SIEVESIZE;
    end;

    j := 0;
    For i := 0 to High(sieve) do
    begin
//  function mpz_probab_prime_p(var n: mpz_t; reps: longint): longint;1 = prime
      if (sieve[i]= 0) then
      begin
        mpz_add_ui(dummy,dummy,i-j);
        j := i;
        if (mpz_probab_prime_p(dummy,1) >0)  then
        begin
          mpz_set(tmp,dummy);
          mpz_clear(dummy);
          EXIT;
        end;
      end;
    end;
  until false;
end;

var
  nominator,denominator,tmp,tmpDemP,p : mpz_t;
  T1,T0:Int64;
  cnt : NativeUint;
begin
  InitPrimes;
  setlength(s,100000);
  cnt := 1;
  mpz_init(nominator);
  mpz_init(tmp);
  mpz_init(tmpDemP);
  mpz_init_set_ui(denominator,1);
  mpz_init_set_ui(p,1);

  repeat
    mpz_set(tmpDemP,p);
    T0 := GetTickCount64;
    if cnt > 9 then
      DoSieveOffsetInit(p)
    else
      mpz_nextprime(p,p);
    T1 := GetTickCount64;
    write(cnt:3,'  ',T1-T0,' ms ,delta ');
    mpz_sub(tmpDemP,p,tmpDemP);
    mpz_get_str(pChar(@s[1]),10, tmpDemP); OutStr(s);
    mpz_get_str(pChar(@s[1]),10, p); OutStr(s);
    if cnt >=15 then
      Break;
    repeat
      mpz_mul(tmp,nominator,p);
      mpz_add(tmp,tmp,denominator);
      mpz_mul(tmpDemP,denominator,p);
      if mpz_cmp(tmp,tmpDemP)< 0 then
        BREAK;
      mpz_add_ui(p,p,1);
    until false;
    mpz_set(nominator,tmp);
    mpz_mul(denominator,denominator,p);

    //next smallest possible number denominator/delta
    mpz_sub(tmp,denominator,nominator);
    mpz_fdiv_q(p,denominator,tmp);

    inc(cnt);
  until cnt> 17;
end.

@TIO.RUN:

reducing factor 0.05041709
  1  0 ms ,delta 1
2
  2  0 ms ,delta 1
3
  3  0 ms ,delta 1
7
  4  0 ms ,delta 1
43
  5  0 ms ,delta 5
1811
  6  0 ms ,delta 16
654149
  7  0 ms ,delta 5
27082315109
  8  0 ms ,delta 71
153694141992520880899
  9  1 ms ,delta 14
337110658273917297268061074384231117039
 10  1 ms ,delta 350
8424197597064114319...13803869133407474043 (    76 digits)
 11  1 ms ,delta 203
2030075381384823476...91313959045797597991 (   150 digits)
 12  3 ms ,delta 33
2032370538147127284...21649394434192763213 (   297 digits)
 13  69 ms ,delta 348
1274824659267207819...20708715953110886963 (   592 digits)
 14  234 ms ,delta 192
4674902516513883824...65355869250350888941 (  1180 digits)
 15  20391 ms ,delta 3510
1139012563947167462...31060548964273180103 (  2358 digits)

Real time: 21.024 s User time: 20.768 s Sys. time: 0.067 s CPU share: 99.10 %

@home (4.4Ghz 5600G  ) modified with primes up to 1E6 ( Uint16-> Uint32 )
     78498    999979 reducing factor 0.04059798
...
 15  10015 ms ,delta 3510  // first guess than searching for next prime 
1139012563947167462...31060548964273180103 (  2358 digits)
   1731172    999979
 16  110030 ms ,delta 6493
3696176350563052055...02467094377885929191 (  4711 digits)
 17 5098268 ms ,delta 55552 // first guess than searching for next prime 
2104336472479843950...14594683820359204509 (  9418 digits)

Perl

use strict; use warnings; use feature 'state';
use Math::AnyNum <next_prime ceil>;

sub abbr { my $d=shift; my $l = length $d; $l < 41 ? $d : substr($d,0,20) . '..' . substr($d,-20) . " ($l digits)" }

sub succ_prime {
    state $sum = 0;
    my $next = next_prime ceil( 1 / (1-$sum) );
    $sum += 1/$next;
    $next
}

printf "%2d: %s\n", $_, abbr succ_prime for 1..14;

Output:

 1: 2
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191..13803869133407474043 (76 digits)
11: 20300753813848234767..91313959045797597991 (150 digits)
12: 20323705381471272842..21649394434192763213 (297 digits)
13: 12748246592672078196..20708715953110886963 (592 digits)
14: 46749025165138838243..65355869250350888941 (1180 digits)

Phix

The Julia entry took over 4 hours to get the 17th on this box, so I just went with "keep it all under 10s"...
(As others have alluded, it is all about getting the next prime. Even should you land on one straightaway, it still takes quite some time to prove an n-thousand digit number is prime.)

with javascript_semantics
constant limit = iff(platform()=JS?12:14)
include mpfr.e
mpq {q, r, s, u} = mpq_inits(4,{0,0,0,1})
mpz {p, t} = mpz_inits(2)
printf(1,"First %d elements of the sequence:\n", limit);
for count=1 to limit do
    mpq_sub(q, u, s)
    mpq_inv(q, q)
    mpq_get_den(t, q)
    mpz_set_q(p, q)
    if mpz_cmp_si(t, 1) then mpz_add_si(p, p, 1) end if
    mpz_nextprime(p, p)
    printf(1,"%2d: %s\n", {count, mpz_get_short_str(p)})
    mpq_set_z(r, p)
    mpq_inv(r, r)
    mpq_add(s, s, r)
end for

Output:

First 14 elements of the sequence:
 1: 2
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191...13803869133407474043 (76 digits)
11: 20300753813848234767...91313959045797597991 (150 digits)
12: 20323705381471272842...21649394434192763213 (297 digits)
13: 12748246592672078196...20708715953110886963 (592 digits)
14: 46749025165138838243...65355869250350888941 (1,180 digits)

If you've really got nothing better to do, you can also get, in 10 mins on 64-bit, thrice that on 32-bit:

15: 11390125639471674628...31060548964273180103 (digits: 2358)
16: 36961763505630520555...02467094377885929191 (digits: 4711)

Python

''' rosettacode.org/wiki/Prime_reciprocal_sum '''
from fractions import Fraction
from sympy import nextprime


def abbreviated(bigstr, label="digits", maxlen=40, j=20):
    ''' Abbreviate string by showing beginning / end and number of chars '''
    wid = len(bigstr)
    return bigstr if wid <= maxlen else bigstr[:j] + '..' + bigstr[-j:] + f' ({wid} {label})'


def ceil(rat):
    ''' ceil function for Fractions '''
    return rat.numerator if rat.denominator == 1 else rat.numerator // rat.denominator + 1


psum = Fraction(0, 1)
for i in range(1, 15):  # get first 14 in sequence
    next_in_seq = nextprime(ceil(Fraction(1, 1 - psum)))
    psum += Fraction(1, next_in_seq)
    print(f'{i:2}:', abbreviated(str(next_in_seq)))

Output:

 1: 2
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191..13803869133407474043 (76 digits)
11: 20300753813848234767..91313959045797597991 (150 digits)
12: 20323705381471272842..21649394434192763213 (297 digits)
13: 12748246592672078196..20708715953110886963 (592 digits)
14: 46749025165138838243..65355869250350888941 (1180 digits)

Raku

The sixteenth is very slow to emerge. Didn't bother trying for the seventeenth. OEIS only lists the first fourteen values.

sub abbr ($_) { .chars < 41 ?? $_ !! .substr(0,20) ~ '..' ~ .substr(*-20) ~ " ({.chars} digits)" }

sub next-prime {
    state $sum = 0;
    my $next = ((1 / (1 - $sum)).ceiling+1..*).hyper(:2batch).grep(&is-prime)[0];
    $sum += FatRat.new(1,$next);
    $next;
}

printf "%2d: %s\n", $_, abbr next-prime for 1..15;

Output:

 1: 2
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191..13803869133407474043 (76 digits)
11: 20300753813848234767..91313959045797597991 (150 digits)
12: 20323705381471272842..21649394434192763213 (297 digits)
13: 12748246592672078196..20708715953110886963 (592 digits)
14: 46749025165138838243..65355869250350888941 (1180 digits)
15: 11390125639471674628..31060548964273180103 (2358 digits)
16: 36961763505630520555..02467094377885929191 (4711 digits)

RPL

Limited floating-point precision prevents from finding the correct 7th term. Program starts with a non-empty sequence to avoid the ∑LIST calculation bug that occurs when the input list has less than two items.

Works with: HP version 49g

≪ {2 3} 
   1 4 START 1 OVER INV ∑LIST - INV →NUM IP NEXTPRIME + NEXT
≫ '∑INVPR' STO

Output:

1: {2 3 7 43 1811 654149} 

Sidef

var A075442 = Enumerator({|callback|
    var sum = 0
    loop {
        var p = next_prime(ceil(1/(1-sum)))
        sum += 1/p
        callback(p)
    }
})

A075442.first(15).each_kv {|k,n|
    var s = Str(n)
    s = "#{s.first(20)}..#{s.last(20)} (#{s.len} digits)" if (s.len > 50)
    say "#{'%2d' % k+1}: #{s}"
}

Output:

 1: 2
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191..13803869133407474043 (76 digits)
11: 20300753813848234767..91313959045797597991 (150 digits)
12: 20323705381471272842..21649394434192763213 (297 digits)
13: 12748246592672078196..20708715953110886963 (592 digits)
14: 46749025165138838243..65355869250350888941 (1180 digits)
15: 11390125639471674628..31060548964273180103 (2358 digits)

Wren

Library: Wren-gmp

Library: Wren-fmt

Even with GMP takes about 4½ minutes to find the first 16.

import "./gmp" for Mpz, Mpq
import "./fmt" for Fmt

var q = Mpq.new()
var p = Mpz.new()
var r = Mpq.new()
var s = Mpq.new()
var count = 0
var limit = 16
Fmt.print("First $d elements of the sequence:", limit)
while (count < limit) {
     q.set(Mpq.one.sub(s)).inv
     var isInteger = (q.den == Mpz.one)
     if (isInteger) p.set(q.toMpz) else p.set(q.toMpz.inc)
     p.nextPrime(p)
     count = count + 1
     var ps = p.toString
     Fmt.write("$2d: $20a", count, p)
     if (ps.count > 40) {
        Fmt.print(" (digits: $d)", ps.count)
     } else {
        System.print()
     }
     r.set(p).inv
     s.add(r)
}

Output:

First 16 elements of the sequence:
 1: 2
 2: 3
 3: 7
 4: 43
 5: 1811
 6: 654149
 7: 27082315109
 8: 153694141992520880899
 9: 337110658273917297268061074384231117039
10: 84241975970641143191...13803869133407474043 (digits: 76)
11: 20300753813848234767...91313959045797597991 (digits: 150)
12: 20323705381471272842...21649394434192763213 (digits: 297)
13: 12748246592672078196...20708715953110886963 (digits: 592)
14: 46749025165138838243...65355869250350888941 (digits: 1180)
15: 11390125639471674628...31060548964273180103 (digits: 2358)
16: 36961763505630520555...02467094377885929191 (digits: 4711)