I have found the Fibonacci series as a function. The function is as follows :- $$F(x) = 1 - 0×f_1(x) + 1×f_2(x) - 1×f_3(x) + 2×f_4(x) - 3×f_5(x) + 5×f_6(x) - 8×f_7(x) + 13×f_8(x) - 21×f_9(x) + 34×f_{10}(x) - 55×f_{11}(x) + .................$$

where $$f_n(x) = ((x-1)×(x-2)×(x-3)×.....×(x-n)) / n!$$ and $F(x)=$ the $x^{th}$ term of the Fibonacci series.

And to my amusement, this formula contains the terms of the fibonacci series itself starting from the third term onwards(1,1,2,3,5,8,13,21,34,55, .......).

I want to know why is the series appearing again in the function.

I can assure you that the above formula is correct. Actually, I have made a formula(which I call meta formula) which can generate the formula of any sequence and this formula is a consequence of that meta formula. Please don't ask me how I made it. I am new to this forum, so please excuse if the post does not seem to be research oriented.


closed as off-topic by David Loeffler, Franz Lemmermeyer, Daniel Loughran, Ben Linowitz, Peter Humphries Jul 11 '16 at 18:06

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  • 14
    $\begingroup$ The function f_n(x) is the binomial $\binom{x-1}{n}$. Your identity can be written as $F_k=\sum_{i=0}^k \binom{k}{i} (-1)^i F_{i-1}$. It is equivalent to a special case ($n=-2$, $c=k$) of formula (50) in mathworld.wolfram.com/FibonacciNumber.html $\endgroup$ – user35593 Jul 11 '16 at 12:06