Firstly, we need to establish what the exponential generating function is for the Fibonacci numbers. It turns out to be: ∞∑n=0Fntnn!
where Fn=αn−βnα−β and α=1+√52 and β=1−√52
Now the convolution of the generating function with itself yields new terms that can be found from the original Fibonacci numbers. Let's define two exponential generating functions: A(t)=∞∑n=0antnn! and B(t)=∞∑n=0bntnn!
then A(t).B(t)=∞∑n=0(n∑k=0(nk).ak.bn−k)tnn!
Let's remember that an=bn=Fn and so all the coefficients of this generating function can be found by setting n=0,1,2,3,.... For example, when n=4 the coefficient is (40).F0.F4+(41).F1.F3+(42).F2.F2+(43).F3.F1+(44).F4.F0
Remember that F0=0,F1=1,F2=1,F3=2,F4=3 and so substituting these values, the coefficient becomes 1∗0∗3+4∗1∗2+6∗1∗1+4∗2∗1+1∗3∗0=22. The magnitude of these coefficients rises quickly as sequence OEIS A014334 shows. So this is how the number 25190 can be calculated (by setting n=10).
REFERENCE: https://www.mathstat.dal.ca/FQ/Scanned/11-3/church.pdf
No comments:
Post a Comment