What is the next number in this sequence? 95, 36, 15,?

Question

My question really is, what is the proof of the following:?

The answer to the above sequence question can be any number N and rationalized by the following equation:

f(x) = 192 - 97C(x,1) + 38C(x,2) - [32-N]C(x,4)

where C(a,b) = a! / [(a-b)! b!]

and which implies that f(1) = 95, f(2) = 36, f(3) = 15 f(4) = N

The question was originally asked a little while ago at:
http://answers.yahoo.com/question/index;_ylt=Al4bT9wa8eL0PiWC14B6WUPsy6IX?qid=20060618090802AAxjV1V

Eulercrosser
where do you need to evaluate N! ?

The function f only needs to have values at 1,2,3,4,....

Eulercrosser-
Gotcha...
C'mon tho, you're a math guy. C is the combinitorial function and C(a,b)=0 when a

The real point here is that C(x,k) is a polynomial in x of degree k, since
C(x,k) = x(x-1)(x-2)...(x-k+1)/k!

Actually, you can get away with
f(x-1) = 95 - 59C(x,1) + 38C(x,2) -[32-N]C(x,3)
A polynomial of degree 3.

Or,
F(y) = 95 - 59C(y+1,1) + 38C(y+1,2) -[32-N]C(y+1,3)

Eulercrosser
Die Antwort ist richtig.

Yea Lagrange no good unless you change variable and define for larger x; Newton Interpolation is probably the best as here.

Answer 1

how do you have n! defined when n<0 integer?

f(1)=192-97C(1,1)+38C(1,2)- [32-N](C(x,4)

C(1,1)=1
C(1,2)=?
C(1,4)=?

Unless I'm missing something really stupid, f is only defined for x≥4
Sorry, knew I was missing something really obvious . . .

f(1)= 192-97C(1,1)+38C(1,2)- [32-N](C(1,4) =192-97=95;
f(2)= 192-97C(2,1)+38C(2,2)- [32-N](C(2,4) =192-97(2) + 38(1)= 36;
f(3)= 192-97C(3,1)+38C(3,2)- [32-N](C(3,4) =192-97(3) + 38(3)= 15;
f(4)= 192-97C(4,1)+38C(4,2)- [32-N](C(4,4) =192-97(4) + 38(6)- [32-N](1)= 192-388+ 228 - (32-N)= N.


Just plug and chug, oder?
And of course you have to mention that the function is defined for all integers.

I was spending my time to see if I could use legrange interpolation to get another formula, but you can't get integer values from a 4th degree Legrange interpolation (unless 6 divides x).

Answer 2

16