Prime progressions...

[mfgoode]

Originally Posted by cheesehead

I'm pretty sure that's what koal meant.

Nope. Only 40 distinct prime values for the range n = 0 through 79. (Though who knows how many for n outside that range!) :)

The function is symmetrical around the line n = 39.5

f(0) = 0 - 0 + 1601 = 1601
f(79) = 79^2 - 79^2 + 1601 = 1601 = f(0)

f(39) = 39*39 - 79*39 +1601 = -40*39 +1601 = 41
f(40) = 40*40 -79*40 + 1601 = -39*40 + 1601 = 41 = f(39)

f(79-m) = (79-m)*(79-m) - 79*(79-m) + 1601
= 79^2 - 79m - 79m + m^2 - 79^2 + 79m + 1601
= m^2 - 79m + 1601


= f(m)

:surprised
My thanks to you koal for rekindling my interest in the polynomial x^2-79x+1601 and to take a closer look at it. Im sorry I did not follow your correct presentation on the duplicate primes though after that I considered the possibility of the existence of such a function.

Hats off Cheesehead for your very clear and concise presentation and the general proof after which there is no argument or doubt.
You have to spoon feed an oldtimer like me who finished my formal education
some 50 years ago, and Im still learning!

To show my appreciation of your help I describe the function in question as a matter of interest.
Yes the function represents a parabola one form of which is
(x - 39.5)^2 = (y -40.75 )
Put X =(x -39.5 ) and Y = (y - 40.75 ) we get
X^2 = Y therefore X^2 = 4 (1/4 ) Y
Particulars of the parabola:
vertex is at ( 39.5 , 40.75 )
Latus Rectum = 1
focus at (39.5 , 41 )
Symmetrical about x = 39.5.
eqn of directrix y =40.5

I am working on the peculiar properties of polynomials with regard to primes.
I will appreciate it if anyone could add to this store of prime generating polynomials. Thanks.
Mally