Hey Grok, Can you write the simplest recursive fibonacci-generator in COBOL?<p><pre><code> IDENTIFICATION DIVISION.
PROGRAM-ID. FIBONACCI.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 NUM PIC 9(10).
01 RESULT PIC 9(10).
PROCEDURE DIVISION.
MAIN-PARA.
MOVE 10 TO NUM
CALL 'FIB' USING NUM RESULT
DISPLAY 'FIBONACCI OF ' NUM ' IS ' RESULT
STOP RUN.
IDENTIFICATION DIVISION.
PROGRAM-ID. FIB.
DATA DIVISION.
LINKAGE SECTION.
01 N PIC 9(10).
01 FIB-RESULT PIC 9(10).
PROCEDURE DIVISION USING N FIB-RESULT.
IF N <= 1
MOVE N TO FIB-RESULT
ELSE
COMPUTE FIB-RESULT = FUNCTION FIB(N - 1) + FUNCTION FIB(N - 2)
END-IF
GOBACK.
END PROGRAM FIB.
END PROGRAM FIBONACCI.</code></pre>