How to resolve the algorithm Sequence: smallest number greater than previous term with exactly n divisors step by step in the PL/M programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Sequence: smallest number greater than previous term with exactly n divisors step by step in the PL/M programming language
Table of Contents
Problem Statement
Calculate the sequence where each term an is the smallest natural number greater than the previous term, that has exactly n divisors.
Show here, on this page, at least the first 15 terms of the sequence.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Sequence: smallest number greater than previous term with exactly n divisors step by step in the PL/M programming language
Source code in the pl/m programming language
100H: /* FIND THE SMALLEST NUMBER > THE PREVIOUS ONE WITH EXACTLY N DIVISORS */
/* CP/M BDOS SYSTEM CALL */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
/* CONSOLE OUTPUT ROUTINES */
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$STRING( .( 0DH, 0AH, '$' ) ); END;
PR$NUMBER: PROCEDURE( N );
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR( 6 ) BYTE INITIAL( '.....$' ), W BYTE;
N$STR( W := LAST( N$STR ) - 1 ) = '0' + ( ( V := N ) MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
/* TASK */
/* RETURNS THE DIVISOR COUNT OF N */
COUNT$DIVISORS: PROCEDURE( N )ADDRESS;
DECLARE N ADDRESS;
DECLARE ( I, I2, COUNT ) ADDRESS;
COUNT = 0;
I = 1;
DO WHILE( ( I2 := I * I ) < N );
IF N MOD I = 0 THEN COUNT = COUNT + 2;
I = I + 1;
END;
IF I2 = N THEN RETURN ( COUNT + 1 ); ELSE RETURN ( COUNT );
END COUNT$DIVISORS ;
DECLARE MAX LITERALLY '15';
DECLARE ( I, NEXT ) ADDRESS;
CALL PR$STRING( .'THE FIRST $' );
CALL PR$NUMBER( MAX );
CALL PR$STRING( .' TERMS OF THE SEQUENCE ARE:$' );
NEXT = 1;
I = 1;
DO WHILE( NEXT <= MAX );
IF NEXT = COUNT$DIVISORS( I ) THEN DO;
CALL PR$CHAR( ' ' );
CALL PR$NUMBER( I );
NEXT = NEXT + 1;
END;
I = I + 1;
END;
EOF
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