Hi Mapleprimes,

I have made this little procedure with Maple. 

check_g_conjecture_10.pdf

similar to this next one check_g_conjecture_10.mw

This may be worth a look.

Regards,

Matthew

P.S.   see  https://en.wikipedia.org/wiki/Goldbach%27s_conjecture

In My Humble Opinion, Wikipedia is a good crowd sourced resource.

> restart; for j to nops(n) do sys := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))+1-(diff(f(eta), eta))^2 = 0, (diff(diff(theta(eta), eta), eta))/pr+f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta) = 0; bcs := f(0) = 0, (D(f))(0) = l+b*((D@@2)(f))(0), (D(f))(-.5) = 1, theta(0) = 1+s*(D(theta))(-.5), theta(2) = 0; n := [1, 2, 3, 4, 5, 6]; pr := .71; p := 0; q := 0; b := 0; l := 0; s := 0; L := [red, blue, orange]; R1 := 2*n[j]/(1+n[j]); R2 := 2*p/(1+n); p := proc (f1, th1, { output::name := 'number' }) local res1, fvals, thvals, res2; option remember; res1 := dsolve({sys, f(1) = 0, theta(0) = 1+th1, (D(f))(0) = f1, (D(theta))(0) = th1, ((D@@2)(f))(0) = f1-1}, numeric, :-output = listprocedure); fvals := (subs(res1, [seq(diff(f(eta), [`$`(eta, i)]), i = 0 .. 2)]))(0); thvals := (subs(res1, [seq(diff(theta(eta), [`$`(eta, i)]), i = 0 .. 1)]))(0); res2 := dsolve({sys, f(0) = fvals[1], theta(0) = thvals[1], theta(5) = 0, (D(f))(0) = fvals[2], (D(f))(5) = 1}, numeric, :-output = listprocedure); if output = 'number' then [fvals[3]-(subs(res2, diff(f(eta), `$`(eta, 2))))(0), thvals[2]-(subs(res2, diff(theta(eta), eta)))(0)] else res1, res2 end if end proc; p1 := proc (f1, th1) p(args)[1] end proc; p2 := proc (f1, th1) p(args)[2] end proc; p(.3, -.2); par := fsolve([p1, p2], [.3, -.2]); res1, res2 := p(op(par), output = xxx); plots:-display(plots:-odeplot(res1, [[eta, f(eta)], [eta, theta(eta)]]), plots:-odeplot(res2, [[eta, f(eta)], [eta, theta(eta)]])); plots:-display(plots:-odeplot(res1, [[eta, diff(f(eta), eta)], [eta, diff(theta(eta), eta)]]), plots:-odeplot(res2, [[eta, diff(f(eta), eta)], [eta, diff(theta(eta), eta)]])); plots:-display(plots:-odeplot(res1, [[eta, diff(f(eta), eta, eta)]]), plots:-odeplot(res2, [[eta, diff(f(eta), eta, eta)]])); plots:-display(plots:-odeplot(res1, [[eta, diff(f(eta), eta)]])); fplt[j] := plots[odeplot](res1, [eta, diff(diff(f(eta), eta), eta)], color = L[j], axes = boxed); tplt[j] := plots[odeplot](res1, [[eta, theta(eta)]], color = L[j], axes = boxed) end do;
> plots:-display([seq(fplt[j], j = 1 .. nops(n))]);

Dear Sir

In my above problem i trying to plot for set of values of n but in plot command it not executing , can you do this why it is not executing ??

> restart;
> n := [1, 2, 3, 4, 5]; pr := .71; p := 0; q := 0; b := 0; l := 0; s := 0;
> for j to nops(n) do R1 := 2*n[j]/(1+n[j]); R2 := 2*p/(1+n); sys := diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))+1-(diff(f(eta), eta))^2 = 0, (diff(diff(theta(eta), eta), eta))/pr+f(eta)*(diff(theta(eta), eta))-R2*(diff(f(eta), eta))*theta(eta) = 0; bcs := f(0) = 0, (D(f))(0) = l+b*((D@@2)(f))(0), (D(f))(-.5) = 1, theta(0) = 1+s*(D(theta))(0), theta(-.5) = 0; proc (f1, th1, { output::name := 'number' }) local res1, fvals, thvals, res2; option remember; res1 := dsolve({sys, f(0) = 0, theta(0) = 1+th1, (D(f))(-2) = f1, (D(theta))(-2) = th1, ((D@@2)(f))(0) = f1-1}, numeric, :-output = listprocedure); fvals := (subs(res1, [seq(diff(f(eta), [`$`(eta, i)]), i = 0 .. 2)]))(0); thvals := (subs(res1, [seq(diff(theta(eta), [`$`(eta, i)]), i = 0 .. 1)]))(0); res2 := dsolve({sys, f(0) = fvals[1], theta(0) = thvals[1], theta(1) = 0, (D(f))(0) = fvals[2], (D(f))(1) = 0}, numeric, :-output = listprocedure); if output = 'number' then [fvals[3]-(subs(res2, diff(f(eta), `$`(eta, 2))))(0), thvals[2]-(subs(res2, diff(theta(eta), eta)))(0)] else res1, res2 end if end proc; p1 := proc (f1, th1) p(args)[1] end proc; p2 := proc (f1, th1) p(args)[2] end proc; p(.3, -.2); par := fsolve([p1, p2], [.3, -.2]); res1, res2 := p(op(par), output = xxx); plots:-display(plots:-odeplot(res1, [[eta, f(eta)], [eta, theta(eta)]]), plots:-odeplot(res2, [[eta, f(eta)], [eta, theta(eta)]])); plots:-display(plots:-odeplot(res1, [[eta, diff(f(eta), eta)], [eta, diff(theta(eta), eta)]]), plots:-odeplot(res2, [[eta, diff(f(eta), eta)], [eta, diff(theta(eta), eta)]])); plots:-display(plots:-odeplot(res1, [[eta, diff(f(eta), eta, eta)]]), plots:-odeplot(res2, [[eta, diff(f(eta), eta, eta)]])); fplt[j] := plots[odeplot](sol1, [eta, diff(diff(f(eta), eta), eta)], color = L[j], axes = boxed); tplt[j] := plots[odeplot](sol1, [[eta, theta(eta)]], color = L[j], axes = boxed) end do;

Dear Sir

In this above problem it showing that error as  Error, cannot split rhs for multiple assignment please can you tell why it is showing like this  ?? and where i did multiple assignments ??

Dear sir

In my ode problem i do not know that how to set range (eta) from -2 to 2 please can  you help me.

> restart;
> with(plots);
> n := [0, .5, 1, 5]; pr := .71; p := 0; l := [1, 2, 3]; b := 0; s := 0; L := [green, blue, black, gold];
                         [green, blue, black, gold]
> R1 := 2*n/(n+1);
                              2 [0, 0.5, 1, 5]
                             ------------------
                             [0, 0.5, 1, 5] + 1
for j from 1 to nops(l) do; for j from 1 to nops(n) do        R1 := 2*n[j]/(1+n[j]);        R2 := 2*p/(1+n[j]); sol1 := dsolve([diff(diff(diff(f(eta),eta),eta),eta)+f(eta)*diff(diff(f(eta),eta),eta)+R1*(1-diff(f(eta),eta)^2) = 0, (1/pr)*diff(diff(theta(eta),eta),eta)+f(eta)*diff(theta(eta),eta)-R2*diff(f(eta),eta)*theta(eta) = 0, f(0) = 0, (D(f))(0) = l+b*((D@@2)(f))(0), (D(f))(-2) =1, theta(0) = 1+s*(D(theta))(0), theta(-2) = 0], numeric, method = bvp); fplt[j]:= plots[odeplot](sol1,[eta,diff(diff(f(eta),eta),eta)],color=["blue","black","orange"]);         tplt[j]:= plots[odeplot](sol1, [eta,theta(eta)],color=L[j]); fplt[j]:= plots[odeplot](sol1,[eta,diff(f(eta),eta)],color=L[j]);      od:od:

Error, (in dsolve/numeric/bvp) unable to store 'Limit([eta, 2*eta, 3*eta]+eta^2*[.250000000000000, .500000000000000, .750000000000000]-.250000000000000*eta^2, eta = -2., left)' when datatype=float[8]
> plots:-display([seq(fplt[j], j = 1 .. nops(n))], color = [green, red], [seq(fplt[j], j = 1 .. nops(l))]);

> sol1(0);

Dear sir

In the  above problem i tried to write a nested program but its not executing and showing the error as Error, (in dsolve/numeric/bvp) unable to store 'Limit([eta, 2*eta, 3*eta]+eta^2*, i want the plot range from -2 to 2 but taking only 0 to -2 ,and -2.5 to 3 but taking only 0 to 1

> restart;
> with(plots);
> pr := .72; p := 0; n := [2, 3, 4, 5]; s := 1; a := .2; b := 1;
> R1 := 2*n/(n+1);
                               2 [2, 3, 4, 5]
                              ----------------
                              [2, 3, 4, 5] + 1
> R2 := 2*p/(n+1);
                                      0
>
>
> for j to nops(n) do R1 := 2*n[j]/(1+n[j]); R2 := 2*p/(1+n[j]); sol1 := dsolve([diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))+R1*(1-(diff(f(eta), eta))^2) = 0, diff(diff(theta(eta), eta), eta)+pr*s^f(eta)*(diff(theta(eta), eta))+R2*pr*s*(diff(f(eta), eta))*theta(eta)+2*(a*(diff(f(eta), eta))+b*theta(eta))/(n[j]-1) = 0, f(0) = 0, (D(f))(0) = 1+b*((D@@2)(f))(0), (D(f))(5) = 0, theta(0) = 1+s*(D(theta))(0), theta(5) = 0], numeric, method = bvp); fplt[j] := plots[odeplot](sol1, [eta, diff(diff(f(eta), eta), eta)], axes = boxed); tplt[j] := plots[odeplot](sol1, [[eta, theta(eta)]], axes = boxed) end do;
>
> plots:-display([seq(fplt[j], j = 1 .. nops(n))]);

> plots:-display([seq(tplt[j], j = 1 .. nops(n))]);

Dear sir

In the above problem graph, i am getting all the lines are in same color then how to identify the lines of different values like n=2,3,4,5,6(or can we set different color for different values of n for each line)

> restart;
> with(plots);
> pr := .72; p := 0; n := 1; s := 1; a := [-0.5,0.0,0.5]; b := 1;
> R1 := 2*n/(n+1);
                                      1
> R2 := 2*p/(n+1);
                                      0
>
>
> for j to nops(a) do R1 := 2*n/(n+1); R2 := 2*p/(n+1); sol1 := dsolve([diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))+R1*(1-(diff(f(eta), eta))^2) = 0, diff(diff(theta(eta), eta), eta)+pr*s^f(eta)*(diff(theta(eta), eta))+R2*pr*s*(diff(f(eta), eta))*theta(eta)+2*(a[j]*(diff(f(eta), eta))+b*theta(eta))/(n+1) = 0, f(-.5) = 0, (D(f))(0) = 1+b*((D@@2)(f))(0), (D(f))(5) = 0, theta(-.5) = 1+s*(D(theta))(0), theta(5) = 0], numeric, method = bvp); fplt[j] := plots[odeplot](sol1, [eta, diff(diff(f(eta), eta), eta)], color = ["blue", "black", "orange"]); tplt[j] := plots[odeplot](sol1, [eta, theta(eta)], color = setcolors(["red", "Coral"])) end do;
Error, (in dsolve/numeric/process_input) boundary conditions specified at too many points: {0, 5, -1/2}, can only solve two-point boundary value problems
>
> plots:-display([seq(fplt[j], j = 1 .. nops(a))], color = [green, red]);

> plots:-display([seq(tplt[j], j = 1 .. nops(a))], color = [green, red]);

Dear sir,

In this program i m not getting the solution for decimal values and i do not have idea about the how to set different color for multiple lines(i tried for different set of colors but it shows that only for first color )

hy

i have to develop a code i which i have system of nonlinear equation 

i have to generate the matrix of that nonlinear equation then i want to do or apply any method say newton method and make a loop which help us to find a solution using some tolerance 

at the end i get a result in form of a table which give nth matrix then value of function matrix at nth value then error i-e xn-x(n-1) 

thanx in advance

The 196 algorithm goes like this.  Start with an integer.  Reverse the digits.  Add the reversed number to the integer.  For most numbers, this eventually leads to a palendrome.  That is to say the number is equal to the reversed number.  I wrote a little Maple procedure to explore 196, the smallest number that will probrably never become a palendrome when put into the algorithm.

Let me know if you like my code.

Regards,
Matt

proc4.pdf

proc4.mw

http://mathworld.wolfram.com/196-Algorithm.html

if i m working in maple 13 i have to solve a non linear integral equation then what will be the steps to use the do loop.

> restart;
> with(plots);
> setoptions(title = `Family Plot`, axes = boxed);
> pr := .71; n := 1; p := 0; q := 0; b := 0; l := 0; s := 0; m := 0;
>
>
> R1 := 2.*n/(1+n);
                                 1.000000000
> R2 := 2.*p/(1+n);
                                     0.
> sol1 := dsolve([diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))+R1*(1-(diff(f(eta), eta))^2) = 0, diff(diff(theta(eta), eta), eta)+.71*f(eta)*(diff(theta(eta), eta))-.71*(diff(f(eta), eta))*theta(eta)*R2 = 0, f(0) = 0, (D(f))(0) = 1.8+b*((D@@2)(f))(0), (D(f))(18) = 0, theta(0) = 1+s*(D(theta))(0), theta(18) = 0], numeric, method = bvp[midrich]); plots[odeplot](sol1, [eta, theta(eta)], color = red, axes = boxed);

Dear sir/madam

In my program i want to plot multiple lines for diffrent values of n, but i do not have such type of idea please can you help me.

Dear sir

I try the below program and it is not executing because of error and that error is " unable to match delimiters" 

Please can you varify it.

sol1 := dsolve([diff(diff(diff(f(eta),eta),eta),eta)+f(eta)*diff(diff(f(eta),eta),eta)+((2*n)/((n+1)))*(1-diff(f(eta),eta)^2) = 0, 1/(Pr) *diff(diff(theta(eta),eta),eta)+f(eta)*diff(theta(eta),eta)-((2*p)/((n+1)))diff(f(eta),eta)*theta(eta) = 0, f(0) = 0, D(f)(0) = 1+lambda*'@@'(D,2)(f)(0), D(f)(10) = 1, theta(0)=0, theta(0)=1+sigma*D(theta(0)),theta(10)=1], numeric, method = bvp);  plots[odeplot](sol1, [eta, `@@`(D,2)(f)(eta)], color = red,axes=boxed);

> restart;
> libname = [shootlib, libname];
> with(shoot);
Error, invalid input: with expects its 1st argument, pname, to be of type {`module`, package}, but received shoot
> with(plots);
Pr := 10; s = -.1; lambda := 0; Gr := 1.0; Gm := 1.0; beta := -1.20;
10
s = -0.1
0
1.0
1.0
-1.20
> M := 0.; z := .1; Xi := .5; Nt := .5; Nb := .2; l := 5; Nr := .5; epsilon1 := .2; epsilon2 := .2;
0.
0.1
0.5
0.5
0.2
5
0.5
0.2
0.2
> Prff := Pr/(1+4.*N*(1/3));
10
-----------------
1 + 1.333333333 N
> FNS := {f(eta), h(eta), r(eta), u(eta), v(eta), theta(eta), `ϕ`(eta)};
{f(eta), h(eta), r(eta), u(eta), v(eta), theta(eta), ϕ(eta)}
> ODE := {diff(h(eta), eta)+.75*l*f(eta)*h(eta)-(1/4)*l*u(eta)*epsilon2-Nt*(.75*f(eta)*r(eta)-(1/4)*u(eta)*epsilon1+Nb*r(eta)*h(eta)+Nt*r(eta)*r(eta))/Nb = 0, .75*f(eta)*r(eta)+diff(r(eta), eta)-(1/4)*u(eta)*epsilon1+Nb*r(eta)*h(eta)+Nt*r(eta)*r(eta) = 0, diff(v(eta), eta)+3*(f(eta)*v(eta)-u(eta)*u(eta))/(4*Pr)-(M+lambda)*u(eta)+theta(eta)-Nr*`ϕ`(eta) = 0, diff(f(eta), eta) = u(eta), diff(u(eta), eta) = v(eta), diff(theta(eta), eta) = r(eta), diff(`ϕ`(eta), eta) = h(eta)};
/ / d \
{ 0.75 f(eta) r(eta) + |----- r(eta)| - 0.05000000000 u(eta)
\ \ deta /

2 / d \ 3
+ 0.2 r(eta) h(eta) + 0.5 r(eta) = 0, |----- v(eta)| + -- f(eta) v(eta)
\ deta / 40

3 2 / d \
- -- u(eta) + theta(eta) - 0.5 ϕ(eta) = 0, |----- h(eta)|
40 \ deta /

+ 3.75 f(eta) h(eta) - 0.1250000000 u(eta) - 1.875000000 f(eta) r(eta)

2
- 0.5000000000 r(eta) h(eta) - 1.250000000 r(eta) = 0,

d d d
----- f(eta) = u(eta), ----- u(eta) = v(eta), ----- theta(eta) = r(eta),
deta deta deta

d \
----- ϕ(eta) = h(eta) }
deta /
> IC := {f(0) = s, h(0) = xi, r(0) = tau, u(0) = 0, v(0) = alpha(0), theta(0) = 1-(1/4)*epsilon1, `ϕ`(0) = (1/4)*epsilon2};
{f(0) = s, h(0) = xi, r(0) = tau, u(0) = 0, v(0) = alpha(0),

theta(0) = 0.9500000000, ϕ(0) = 0.05000000000}
> L := 2;
2
> BC = {u(L) = 0, theta(L) = 0, `ϕ`(L) = 0};
BC = {u(2) = 0, theta(2) = 0, ϕ(2) = 0}
> S := Shoot(ODE, IC, BC, FNS, [alpha = .42453091564332, tau = -.21166705749821127, xi = -.4944583739651814]);
/ / / d \
Shoot|{ 0.75 f(eta) r(eta) + |----- r(eta)| - 0.05000000000 u(eta)
\ \ \ deta /

2 / d \ 3
+ 0.2 r(eta) h(eta) + 0.5 r(eta) = 0, |----- v(eta)| + -- f(eta) v(eta)
\ deta / 40

3 2 / d \
- -- u(eta) + theta(eta) - 0.5 ϕ(eta) = 0, |----- h(eta)|
40 \ deta /

+ 3.75 f(eta) h(eta) - 0.1250000000 u(eta) - 1.875000000 f(eta) r(eta)

2
- 0.5000000000 r(eta) h(eta) - 1.250000000 r(eta) = 0,

d d d
----- f(eta) = u(eta), ----- u(eta) = v(eta), ----- theta(eta) = r(eta),
deta deta deta

d \
----- ϕ(eta) = h(eta) }, {f(0) = s, h(0) = xi, r(0) = tau, u(0) = 0,
deta /

v(0) = alpha(0), theta(0) = 0.9500000000, ϕ(0) = 0.05000000000}, BC,

{f(eta), h(eta), r(eta), u(eta), v(eta), theta(eta), ϕ(eta)}, [

alpha = 0.42453091564332, tau = -0.21166705749821127,

\
xi = -0.4944583739651814]|
/
RungeKutta(ODE, BC, alpha = .42453091564332, tau = -.21166705749821127, xi = -.4944583739651814, output=plot);
/ / / d \
RungeKutta|{ 0.75 f(eta) r(eta) + |----- r(eta)| - 0.05000000000 u(eta)
\ \ \ deta /

2 / d \ 3
+ 0.2 r(eta) h(eta) + 0.5 r(eta) = 0, |----- v(eta)| + -- f(eta) v(eta)
\ deta / 40

3 2 / d \
- -- u(eta) + theta(eta) - 0.5 ϕ(eta) = 0, |----- h(eta)|
40 \ deta /

+ 3.75 f(eta) h(eta) - 0.1250000000 u(eta) - 1.875000000 f(eta) r(eta)

2
- 0.5000000000 r(eta) h(eta) - 1.250000000 r(eta) = 0,

d d d
----- f(eta) = u(eta), ----- u(eta) = v(eta), ----- theta(eta) = r(eta),
deta deta deta

d \
----- ϕ(eta) = h(eta) }, BC, alpha = 0.42453091564332,
deta /

\
tau = -0.21166705749821127, xi = -0.4944583739651814, output = plot|
/
>

Dear sir 

in the above problem im geiitng the problem with , with(shoot) command and even it is not executing at

S := Shoot(ODE, IC, BC, FNS, [alpha = .42453091564332, tau = -.21166705749821127, xi = -.4944583739651814]) this command, here alpha,tau and zi variable should change.

> restart;
> with(plots);
> Eql := diff(f(eta), eta, eta, eta)+.5*f(eta)*(diff(f(eta), eta, eta)) = 0;
/ d / d / d \\\ / d / d \\
|----- |----- |----- f(eta)||| + 0.5 f(eta) |----- |----- f(eta)|| = 0
\ deta \ deta \ deta /// \ deta \ deta //
> blt := 10;
10
> bcs1 := f(0) = f0, (D(f))(0) = 0, (D(f))(blt) = 1;
f(0) = f0, D(f)(0) = 0, D(f)(10) = 1
> L := [0];
[0]
> for k to 1 do R := dsolve(eval({Eql, bcs1}, f0 = L[k]), f(eta), numeric, output = listprocedure); X1 || k := rhs(R[3]); X2 || k := rhs(R[4]) end do;
[
[eta = proc(eta) ... end;, f(eta) = proc(eta) ... end;,
[

d
----- f(eta) = proc(eta) ... end;,
deta

d / d \ ]
----- |----- f(eta)| = proc(eta) ... end;]
deta \ deta / ]
proc(eta) ... end;
proc(eta) ... end;
> print([X2], [1 .. 1, 0]);

dear sir/madam

in the above problem i should get the asnser (at print line) but its not getting so please can you tell me why it is not getting.

A prime producing polynomial.

Observations on the trinomial n² + n + 41.

by Matt C. Anderson

September 3, 2016

The story so far

We assume that n is an integer.  We focus our attention on the polynomial n^2 + n + 41.

Furthur, we analyze the behavior of the factorization of integers of the form

h(n) = n2 + n + 41                                          (expression 1)

where n is a non-negative integer.  It was shown by Legendre, in 1798 that if 0 ≤ n < 40 then h(n) is a prime number.

Certain patterns become evident when considering points (a,n) where

h(n) ≡ 0 mod a.                                             (expression 2)

The collection of all such point produces what we are calling a "graph of discrete divisors" due to certain self-similar features.  From experimental data we find that the integer points in this bifurcation graph lie on a collection of parabolic curves indexed by pairs of relatively prime integers.  The expression for the middle parabolas is –

p(r,c) = (c*x – r*y)² – r*(c*x – r*y) – x + 41*r².           (expression 3)

The restrictions are that 0<r<c and gcd(r,c) = 1 and all four of r,c,x, and y are integers.

Each such pair (r,c) yields (again determined experimentally and by observation of calculations) an integer polynomial a*z2 + b*z + c, and the quartic h(a*z2 + b*z + c) then factors non-trivially over the integers into two quadratic expressions.  We call this our "parabola conjecture".  Certain symmetries in the bifurcation graph are due to elementary relationships between pairs of co-prime integers.  For instance if m<n are co-prime integers, then there is an observable relationship between the parabola it determines that that formed from (n-m, n).

We conjecture that all composite values of h(n) arise by substituting integer values of z into h(a*z² + b*z + c), where this quartic factors algebraically over Z for a*z² + b*z + c a quadratic polynomial determined by a pair of relatively prime integers.  We name this our "no stray points conjecture" because all the points in the bifurcation graph appear to lie on a parabola.

We further conjecture that the minimum x-values for parabolas corresponding to (r, c) with gcd(r, c) = 1 are equal for fixed n.  Further, these minimum x-values line up at 163*c^2/4 where c = 2, 3, 4, ...  The numerical evidence seems to support this.  This is called our "parabolas line up" conjecture.

The notation gcd(r, c) used above is defined here.  The greatest common devisor of two integers is the smallest whole number that divides both of those integers.

Theorem 1 - Consider h(n) with n a non negative integer. 

h(n) never has a factor less than 41.

We prove Theorem 1 with a modular construction.  We make a residue table with all the prime factors less than 41.  The fundamental theorem of arithmetic states that any integer greater than one is either a prime number, or can be written as a unique product of prime numbers (ignoring the order).  So if h(n) never has a prime factor less than 41, then by extension it never has an integer factor less than 41.

For example, to determine that h(n) is never divisible by 2, note the first column of the residue table.  If n is even, then h(n) is odd.  Similarly, if n is odd then h(n) is also odd.  In either case, h(n) does not have factorization by 2.

Also, for divisibility by 3, there are 3 cases to check.  They are n = 0, 1, and 2 mod 3. h(0) mod 3 is 2.  h(1) mod 3 is 1. and h(2) mod 3 is 2.  Due to these three cases, h(n) is never divisible by 3.  This is the second column of the residue table.

The number 0 is first found in the residue table for the cases h(0) mod 41 and h(40) mod 41.  This means that if n is congruent to 0 mod 41 then h(n) will be divisible by 41.  Similarly, if n is congruent to 40 mod 41 then h(n) is also divisible by 41.

After the residue table, we observe a bifurcation graph which has points when h(y) mod x is divisible by x.  The points (x,y) can be seen on the bifurcation graph.

< insert residue table here >

Thus we have shown that h(n) never has a factor less than 41.

Theorem 2

Since h(a) = a^2 + a + 41, we want to show that h(a) = h( -a -1).

Proof of Theorem 2

Because h(a) = a*(a+1) + 41,

Now h(-a -1) = (-a -1)*(-a -1 +1) + 41.

So h(-a -1) = (-a -1)*(-a) +41,

And h(-a -1) = h(a).

Which was what we wanted.

End of proof of theorem 2.

Corrolary 1

Further, if h(b) mod c ≡ = then h(c –b -1) mod c ≡ 0.

We can observe interesting patterns in the “graph of discrete divisors” on a following page.