Consider the generating function f(n) that produces the following values:

f(1) = 1

f(2) = 2

f(3) = 4

Obviously these values can be generated by f(n)= 2^(n-1).

These values can equally well be generated by f(n) = (n^2-n+2)/2, a second order polynomial.

Many (all?) integer series f(k), where k = 1,2,3,..K-1,K can be generated by a polynomial of order K-1.

The integer series 2^(n-1), where n = 1,2,3...K can also be generated by a polynomial of order K-1.

The following interesting thing happens.

If we describe the series 1,2,4 by f(n) = (n^2-n+2)/2 then f(4) = 7

For 1,2,4,8, f(n) = (n^3-3n^2+8n)/6 and f(5) = 15

For 1,2,4,8,16, f(n) = (n^4-6n^3+23n^2-18n) /24 and f(6)=31

For 1,2,4,8,16,32, f(n) = (n^5-10n^4+55n^3-110n^2+184n)/120 and f(7)=63

For 1,2,4,8,16,32,64, f(n) = (n^6-15n^5+115n^4-405n^3+964n^2-660n+720)/720) and f(8)=127

I have verified this till order 14.

Lets add the series "1" and "1,2" for completeness:

For 1, f(n) = 1 and f(2) =1. f(2) = 2* f(1)-1

For 1,2 f(n) = n and f(3) = 3. f(3) = 2 * f(2)-1

This suggests that f(k+1) = 2 * f(k) -1 when f(n) is the k-1 th order polynomial function that generates the values 1,2,4,...2^(k-1).

This raises the question if this is true for all integer series of this type and if so, how to prove it. I hope that I am not overlooking the obvious.

Another observation is that if you write the polynomials that describe the series 1,2,4,8... in a fractional form where all coefficients of n^k in the numerator are integers, then the denominator always seems to be

(K-1)! (1,1,2,6,24,120 etc.)

Can anybody shine some light on these observations please? Thanks a lot in advance

Bob Andriesse

Consider the generating function $f(n)$ that produces the following values:

$$f(1) = 1$$

$$f(2) = 2$$

$$f(3) = 4$$

Obviously these values can be generated by $f(n)= 2^{n-1}$.

These values can equally well be generated by $f(n) = (n^2-n+2)/2$, a second order polynomial.

Many (all?) integer series $f(k)$, where $k = 1,2,3,...,K-1,K$ can be generated by a polynomial of order $K-1$.

The integer series $2^{n-1}$, where $n = 1,2,3,...,K$ can also be generated by a polynomial of order $K-1$.

The following interesting thing happens.

If we describe the series $1,2,4$ by $f(n) = \frac{n^2-n+2}{2}$ then $f(4) = 7$

For $1,2,4,8$, $f(n) = \frac{n^3-3n^2+8n}{6}$ and $f(5) = 15$

For $1,2,4,8,16$, $f(n) = \frac{n^4-6n^3+23n^2-18n}{24}$ and $f(6)=31$

For $1,2,4,8,16,32$, $f(n) = \frac{n^5-10n^4+55n^3-110n^2+184n}{120}$ and $f(7)=63$

For $1,2,4,8,16,32,64$, $f(n) = \frac{n^6-15n^5+115n^4-405n^3+964n^2-660n+720}{720}$ and $f(8)=127$

I have verified this till order 14.

Lets add the series "1" and "1,2" for completeness:

For $1$, $f(n) = 1$ and $f(2) =1$. $f(2) = 2 \cdot f(1)-1$

For $1,2$, $f(n) = n$ and $f(3) = 3$. $f(3) = 2 \cdot f(2)-1$

This suggests that $f(k+1) = 2 \cdot f(k) -1$ when $f(n)$ is the $k-1$ th order polynomial function that generates the values $1,2,4,...2^{k-1}$.

This raises the question if this is true for all integer series of this type and if so, how to prove it. I hope that I am not overlooking the obvious.

Another observation is that if you write the polynomials that describe the series $1,2,4,8,...$ in a fractional form where all coefficients of $n^k$ in the numerator are integers, then the denominator always seems to be $(K-1)!$ ($1,1,2,6,24,120,$ etc.)

Can anybody shine some light on these observations please?