reverse numerical patterns

Question

what is a reverse numerical pattern for the numbers 3,4,7,13,23

Answer 1

Here are some successive differences

3 4 7 13 23 38

1 3 6 10 15

2 3 4 5

1 1 1

Since the third set of difference is constant, y = ax³ + bx² + cx + d

Some background:

x ax³ + bx² + cx + d 1st difference 2nd difference 3rd difference

0 d

1 a + b + c + d a + b + c

2 8a + 4b + 2c + d 7a + 3b + c 6a + 2b

3 27a + 9b + 3c + d 19a + 5b + c 12a + 2b 6a

The third difference in the given pattern is 1. Therefore 6a = 1 and a = (1/6)

From here you should be able to determine values of b, c, and d.

Answer 2

A good starting strategy for such series problems is to assume that they represent some kind of polynomial function (i.e. the y values for integer x in the function).

To determine polynomiality, take successive differences between terms. Representing the nth difference as delta(n), we have

delta(1) [differences between successive terms in the original series]: 1, 3, 6, 10

delta(2) [differences between successive terms in delta(1) series : 2, 3, 4

delta (3) .... 1, 1, 1

So you have determined that the series COULD represent a third degree polynomial. All that remains is to determine the coefficients for the various terms of that polynomial.

By comparison with x^3 : 0, 1, 8, 27, 64 .... for which delta(1) = 1, 7, 19, 37 ; delta(2) = 6, 12, 18 ; delta (3) = 6, 6 we find that we have a term (1/6)x^3 (do you see the reasoning why?). Subtract that series from the originally given series; you will now have a reduced series with a constant delta(2). Compare the value of that delta(2) with that of the series from x^2, to similarly find the coefficient of the x^2 term. Repeat to get a further reduced series, with a constant delta(1). Compare that delta(1) with that for the series for x (==1), to find the coefficient for the x^1 term. Subtract that series from the further reduced series, to get an even further reduced series, which should be all constant; that is the x^0 term. Add up all your coefficiented x^n terms; you now have a polynomial expression which yields your original series, for integer values of x. From this high-level representation, you should be able to formulate a "reverse numerical pattern", as it were!

Hope this helps you get a grasp of how to "iron out" polynomial-function series. If you DO NOT obtain a constant delta(n) for some value n, then perhaps you have a series with non-integer powers of x, or negative powers of x, or perhaps a Fibonacci-type relationship, among other possibilities! And also, do not forget that there are an infinite number of functions which CAN generate an arbitrary finite set of successive values; the simplest polynomial representation is just one of these (albeit frequently the desired solution).