I Note how the successor function σ0(n) = n + 1 is built directly into the definition of primitive recursion. I The class PR is built up from σ0 the constant functions and the projections πi (a1; :::; ai ; :::; an) = ai by substitution and primitive recursion. Primitive recursion Definition Given a binary function β(m; n) and any a 2 N define the function '(n) as follows; '(0) := a and '(n + 1) := β(n;'(n)): DMACC John M. Gillespie 1 I The class PR is built up from σ0 the constant functions and the projections πi (a1; :::; ai ; :::; an) = ai by substitution and primitive recursion. Primitive recursion Definition Given a binary function β(m; n) and any a 2 N define the function '(n) as follows; '(0) := a and '(n + 1) := β(n;'(n)): I Note how the successor function σ0(n) = n + 1 is built directly into the definition of primitive recursion. DMACC John M. Gillespie 2 substitution and primitive recursion. Primitive recursion Definition Given a binary function β(m; n) and any a 2 N define the function '(n) as follows; '(0) := a and '(n + 1) := β(n;'(n)): I Note how the successor function σ0(n) = n + 1 is built directly into the definition of primitive recursion. I The class PR is built up from σ0 the constant functions and the projections πi (a1; :::; ai ; :::; an) = ai by DMACC John M. Gillespie 3 and primitive recursion. Primitive recursion Definition Given a binary function β(m; n) and any a 2 N define the function '(n) as follows; '(0) := a and '(n + 1) := β(n;'(n)): I Note how the successor function σ0(n) = n + 1 is built directly into the definition of primitive recursion. I The class PR is built up from σ0 the constant functions and the projections πi (a1; :::; ai ; :::; an) = ai by substitution DMACC John M. Gillespie 4 Primitive recursion Definition Given a binary function β(m; n) and any a 2 N define the function '(n) as follows; '(0) := a and '(n + 1) := β(n;'(n)): I Note how the successor function σ0(n) = n + 1 is built directly into the definition of primitive recursion. I The class PR is built up from σ0 the constant functions and the projections πi (a1; :::; ai ; :::; an) = ai by substitution and primitive recursion. DMACC John M. Gillespie 5 sg sg¯ I Exponent notation is used e.g. 11 = 1 and 2 = 0. Two surprisingly useful functions We define by primitive recursion sg andsg ¯ . I sg(0) = 0 with sg(n + 1) = 1. I sg¯ (0) = 1 withsg ¯ (n + 1) = 0. DMACC John M. Gillespie 6 Two surprisingly useful functions We define by primitive recursion sg andsg ¯ . I sg(0) = 0 with sg(n + 1) = 1. I sg¯ (0) = 1 withsg ¯ (n + 1) = 0. sg sg¯ I Exponent notation is used e.g. 11 = 1 and 2 = 0. DMACC John M. Gillespie 7 2. The exponent expl (n) of the l-th prime in the prime factorization of n 2 N. 3. The residue res(a; n) of a modulo n. 4. Restricted subtraction a −· b. Examples Well known primitive recursive functions. Takes some work! 1. The n-th prime pn DMACC John M. Gillespie 8 3. The residue res(a; n) of a modulo n. 4. Restricted subtraction a −· b. Examples Well known primitive recursive functions. Takes some work! 1. The n-th prime pn 2. The exponent expl (n) of the l-th prime in the prime factorization of n 2 N. DMACC John M. Gillespie 9 4. Restricted subtraction a −· b. Examples Well known primitive recursive functions. Takes some work! 1. The n-th prime pn 2. The exponent expl (n) of the l-th prime in the prime factorization of n 2 N. 3. The residue res(a; n) of a modulo n. DMACC John M. Gillespie 10 Examples Well known primitive recursive functions. Takes some work! 1. The n-th prime pn 2. The exponent expl (n) of the l-th prime in the prime factorization of n 2 N. 3. The residue res(a; n) of a modulo n. 4. Restricted subtraction a −· b. DMACC John M. Gillespie 11 then S(n) yields the number of divisors of n. I If we let n X π(n) = sg¯ (S(n) −· 2) i=2 then π(n) yields the number of primes among 2; :::; n. Then the nth prime is the least j such that π(j) = n + 1. Examples I If we let n X S(n) = sg¯ (res(n; i)) i=1 DMACC John M. Gillespie 12 I If we let n X π(n) = sg¯ (S(n) −· 2) i=2 then π(n) yields the number of primes among 2; :::; n. Then the nth prime is the least j such that π(j) = n + 1. Examples I If we let n X S(n) = sg¯ (res(n; i)) i=1 then S(n) yields the number of divisors of n. DMACC John M. Gillespie 13 then π(n) yields the number of primes among 2; :::; n. Then the nth prime is the least j such that π(j) = n + 1. Examples I If we let n X S(n) = sg¯ (res(n; i)) i=1 then S(n) yields the number of divisors of n. I If we let n X π(n) = sg¯ (S(n) −· 2) i=2 DMACC John M. Gillespie 14 Then the nth prime is the least j such that π(j) = n + 1. Examples I If we let n X S(n) = sg¯ (res(n; i)) i=1 then S(n) yields the number of divisors of n. I If we let n X π(n) = sg¯ (S(n) −· 2) i=2 then π(n) yields the number of primes among 2; :::; n. DMACC John M. Gillespie 15 Examples I If we let n X S(n) = sg¯ (res(n; i)) i=1 then S(n) yields the number of divisors of n. I If we let n X π(n) = sg¯ (S(n) −· 2) i=2 then π(n) yields the number of primes among 2; :::; n. Then the nth prime is the least j such that π(j) = n + 1. DMACC John M. Gillespie 16 Then the smallest n for which τ(a1; :::; ar ; n) = 0 is given by the expression below. B i X Y τ(~a; j)sg i=0 j=0 = τ(~a; 0)sg + τ(~a; 0)sgτ(~a; 1)sg + τ(~a; 0)sgτ(~a; 1)sgτ(~a; 2)sg + ::: + τ(~a; 0)sgτ(~a; 1)sg : : : τ(~a; B)sg Bounded search Suppose a function τ(a1; :::; ar ; n) is given which satisfies the following formula. (8a1 ::: 8ar 9n)(τ(~a; n) = 0) DMACC John M. Gillespie 17 = τ(~a; 0)sg + τ(~a; 0)sgτ(~a; 1)sg + τ(~a; 0)sgτ(~a; 1)sgτ(~a; 2)sg + ::: + τ(~a; 0)sgτ(~a; 1)sg : : : τ(~a; B)sg Bounded search Suppose a function τ(a1; :::; ar ; n) is given which satisfies the following formula. (8a1 ::: 8ar 9n)(τ(~a; n) = 0) Then the smallest n for which τ(a1; :::; ar ; n) = 0 is given by the expression below. B i X Y τ(~a; j)sg i=0 j=0 DMACC John M. Gillespie 18 Bounded search Suppose a function τ(a1; :::; ar ; n) is given which satisfies the following formula. (8a1 ::: 8ar 9n)(τ(~a; n) = 0) Then the smallest n for which τ(a1; :::; ar ; n) = 0 is given by the expression below. B i X Y τ(~a; j)sg i=0 j=0 = τ(~a; 0)sg + τ(~a; 0)sgτ(~a; 1)sg + τ(~a; 0)sgτ(~a; 1)sgτ(~a; 2)sg + ::: + τ(~a; 0)sgτ(~a; 1)sg : : : τ(~a; B)sg DMACC John M. Gillespie 19 I The bound B is specific to the function τ. 2n I For example the nth prime pn satisfies pn < 2 . 2n I This provides the bound B = 2 . Bounded search The following notation is fairly standard. B i X Y sg µj [τ(~a; j) = 0] = τ(~a; j) i=0 j=0 DMACC John M. Gillespie 20 2n I For example the nth prime pn satisfies pn < 2 . 2n I This provides the bound B = 2 . Bounded search The following notation is fairly standard. B i X Y sg µj [τ(~a; j) = 0] = τ(~a; j) i=0 j=0 I The bound B is specific to the function τ. DMACC John M. Gillespie 21 2n I This provides the bound B = 2 . Bounded search The following notation is fairly standard. B i X Y sg µj [τ(~a; j) = 0] = τ(~a; j) i=0 j=0 I The bound B is specific to the function τ. 2n I For example the nth prime pn satisfies pn < 2 . DMACC John M. Gillespie 22 Bounded search The following notation is fairly standard. B i X Y sg µj [τ(~a; j) = 0] = τ(~a; j) i=0 j=0 I The bound B is specific to the function τ. 2n I For example the nth prime pn satisfies pn < 2 . 2n I This provides the bound B = 2 . DMACC John M. Gillespie 23 I Take the sign of the sum sg(expl(n) + exp2(n) + ::: + expn(n)): I What we seek is the smallest j such that sg(expj+1(n) + expj+2(n) + ::: + expn(n)) = 0: The length of n 2 N Consider the problem of finding the index l of the largest prime dividing n 2 N and suppose n > 1.