Random Sequences

Hi,

Picking 4 numbers from 1-100 each more than the previous, what is the probability that you have chosen 4 which fit no pattern (using any function you choose).

0.

ObPuzzle - what is the cardinality of the set of patterns?

Hi,

You say zero - what about 19 89 90 191 can anyone find me a function whose successive terms are these? (or have I mis-defined the puzzle?)

Dan

Let f(x) = (169*x^3 - 1221*x^2 + 2900*x - 1734)/6, then f(1) = 19, f(2) = 89, f(3) = 90 and f(4) = 191.

And f(5) = 561, f(6) = 1369, and so on.

Given any finite sequence of values, there is a polynomial which takes those values for successive integer arguments.

I agree with Virgil's reasoning, but I also notice that you originally referred to numbers from 1 to 100. The number 191 that you now mention is greater than 100

Here's a more general solution: Let the four numbers, in order, be A, B, C, D; these are the solutions for X = 0, 1, 2, 3 of:

F(X) = (1/6 D - 1/2 C + 1/2 B - 1/6 A) * X^3 - (1/2 D + 2 C - 5/2 B + A) * X^2 + (1/3 D - 3/2 C + 3 B - 11/6 A) * X + A

(Note that's 0, 1, 2, 3, and not 1, 2, 3, 4)

For A = 19, B = 89, C = 90, D = 191:

F(X) = 28 1/6 * X^3 - 119 * X^2 + 160 5/6 * X + 19 F(0) = 19 F(1) = 28 1/6 - 119 + 160 5/6 + 19 = 89 F(2) = 28 1/6 * 8 - 119 * 4 + 160 5/6 * 2 + 19 = 90 F(3) = 28 1/6 * 27 - 119 * 9 + 160 5/6 * 3 + 19 = 191

p=1, choosing a, b, c & d, and the reqd function is the roots of (x-a)(x-b)(x-c)(x-d)(...)

I say zero, too, and you have not mis-defined the puzzle.

Then Don Del Grande wrote:

From the FAQ:

'...there are infinitely many formulas that will fit any finite sequence'

Here's a template for an infinite number of functions that will do the job (and requires no root-taking). Elements numbered starting at 1.

f_j(n) = (n-2) * (n-3) * (n-4) * 19 / ((1-2) * (1-3) * (1-4)) + (n-1) * (n-3) * (n-4) * 89 / ((2-1) * (2-3) * (2-4)) + (n-1) * (n-2) * (n-4) * 90 / ((3-1) * (3-2) * (3-4)) + (n-1) * (n-2) * (n-3) * 191 / ((4-1) * (4-2) * (4-3)) + (n-1) * (n-2) * (n-3) * (n-4) * j

It's not hard to generalize it so it will produce any 'next integer in the sequence' that you desire. That means that if you give me *any* finite series and I can claim *any* integer is next, and prove it.

ObPuzzle: given the series {19,89,90,191} and the next integer in the sequence, m, write a function f(i) that will produce the series for i=1 to 4 and m for i=5.

ObMetaPuzzle: given *any* series of *any* finite length and the next integer m, write a template for a function f_m(i) that will produce the series i=1 to n and m for i=n+1.

ObMetaMetaPuzzle: given *any* *two* series of *any* *two* finite lengths, write a template for a function g(i) that will produce the first series followed by the second series.

Bob Morris Three-time winner of the 8th and 9th grade district puzzle competition.

A suitable LaGrange polynomial is the appropriate response to all of these questions.

erm, got meself confused... p=0, in this case, or alternatively p=1 in the case of f(n)=-14, and f(n+1)=f(n) for n>0