Primes whose first and last number is 3: Difference between revisions

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=={{header|ALGOL 68}}==

{{Trans|ALGOL W}} With added stretch goal. As with the Go and other samples, generates the candidate sequence.

<lang algol68>BEGIN # find some primes whose first and last digits are 3 #

INT max prime = 1 000 000; # largest number to consider #

# sieve the primes to max prime #

PR read "primes.incl.a68" PR

[ 1 : max prime ]BOOL prime;

~~prime~~[ 1 ] ~~:= FALSE;~~ prime[ 2 ] := ~~TRUE~~;

[]BOOL prime = PRIMESIEVE max prime;

FOR i FROM 3 BY 2 TO UPB prime DO prime[ i ] := TRUE OD;

FOR i FROM 4 BY 2 TO UPB prime DO prime[ i ] := FALSE OD;

FOR i FROM 3 BY 2 TO ENTIER sqrt( max prime ) DO

IF prime[ i ] THEN FOR s FROM i * i BY i + i TO UPB prime DO prime[ s ] := FALSE OD FI

OD;

INT p3count := 0;

# prints n, if it is prime, handles newlines #

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IF ( p3count +:= 1 ) MOD 12 = 0 THEN print( ( newline ) ) FI

FI # p #;

# find 1, 2, 3 ~~and~~ 4 ~~digit~~ ~~3x3~~ ~~primes~~ #

# find 3x3 primes #

p( 3 ); p( 33 ); ~~FOR~~ i ~~FROM~~ 0 BY 10 TO 90 DO p( ~~303~~ + i ) ~~OD;~~ ~~FOR~~ i ~~FROM~~ 0 BY 10 TO ~~990~~ DO p( ~~3003~~ + i ) ~~OD;~~

p( 3 ); p( 33 ); # a & 2 digit 3x3 primes #

FOR i FROM 0 BY 10 TO 90 DO p( 303 + i ) OD; # 3 digit 3x3 primes #

FOR i FROM 0 BY 10 TO 990 DO p( 3003 + i ) OD; # 4 digit 3x3 primes #

# 5 and 6 digit 3x3 primes #

FOR i FROM 0 BY 10 TO 9 990 DO IF prime[ 30 003 + i ] THEN p3count +:= 1 FI OD;