Explicit coefficients of Lucas Polynomials

Yasuaki dot Honda at gmail dot com

Lucas Polynomials are defined using a recurrence relation and initial values. Then, the explicit coefficients of the Lucas Polynomials are given a pri ori and we prove polynomials with such coefficients satisfy the original recurrence relation.

In [1]:
LP[0](x):=2;
Out[1]:
\[\tag{${\it \%o}_{1}$}{\it LP}_{0}(x):=2\]
In [2]:
LP[1](x):=x;
Out[2]:
\[\tag{${\it \%o}_{2}$}{\it LP}_{1}(x):=x\]
In [3]:
LP[n](x):=x*LP[n-1](x)+LP[n-2](x);
Out[3]:
\[\tag{${\it \%o}_{3}$}{\it LP}_{n}(x):=x\,{\it LP}_{n-1}(x)+{\it LP}_{n-2}(x)\]
In [4]:
for n:1 thru 12 do print(expand('LP[n](x)=LP[n](x)));
\({\it LP}_{1}(x)=x\)
\({\it LP}_{2}(x)=x^2+2\)
\({\it LP}_{3}(x)=x^3+3\,x\)
\({\it LP}_{4}(x)=x^4+4\,x^2+2\)
\({\it LP}_{5}(x)=x^5+5\,x^3+5\,x\)
\({\it LP}_{6}(x)=x^6+6\,x^4+9\,x^2+2\)
\({\it LP}_{7}(x)=x^7+7\,x^5+14\,x^3+7\,x\)
\({\it LP}_{8}(x)=x^8+8\,x^6+20\,x^4+16\,x^2+2\)
\({\it LP}_{9}(x)=x^9+9\,x^7+27\,x^5+30\,x^3+9\,x\)
\({\it LP}_{10}(x)=x^{10}+10\,x^8+35\,x^6+50\,x^4+25\,x^2+2\)
\({\it LP}_{11}(x)=x^{11}+11\,x^9+44\,x^7+77\,x^5+55\,x^3+11\,x\)
\({\it LP}_{12}(x)=x^{12}+12\,x^{10}+54\,x^8+112\,x^6+105\,x^4+36\,x^2+2\)
Out[4]:
\[\tag{${\it \%o}_{4}$}\mathbf{done}\]
In [145]:
GLP[n](x):=intosum(sum(n/(n-k)*binomial(n-k,k)*x^(n-2*k),k,0,floor(n/2)));
Out[145]:
\[\tag{${\it \%o}_{142}$}{\it GLP}_{n}(x):={\it intosum}\left({\it sum}\left(\frac{n}{n-k}\,{{n-k}\choose{k}}\,x^{n-2\,k} , k , 0 , \left \lfloor \frac{n}{2} \right \rfloor\right)\right)\]
In [146]:
'GLP[n](x)=GLP[n](x);
Out[146]:
\[\tag{${\it \%o}_{143}$}{\it GLP}_{n}(x)=n\,\sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor}{\frac{{{n-k}\choose{k}}\,x^{n-2\,k}}{n-k}}\]
In [148]:
for i:1 thru 12 do print(ev(GLP[i](x),nouns,expand));
\(x\)
\(x^2+2\)
\(x^3+3\,x\)
\(x^4+4\,x^2+2\)
\(x^5+5\,x^3+5\,x\)
\(x^6+6\,x^4+9\,x^2+2\)
\(x^7+7\,x^5+14\,x^3+7\,x\)
\(x^8+8\,x^6+20\,x^4+16\,x^2+2\)
\(x^9+9\,x^7+27\,x^5+30\,x^3+9\,x\)
\(x^{10}+10\,x^8+35\,x^6+50\,x^4+25\,x^2+2\)
\(x^{11}+11\,x^9+44\,x^7+77\,x^5+55\,x^3+11\,x\)
\(x^{12}+12\,x^{10}+54\,x^8+112\,x^6+105\,x^4+36\,x^2+2\)
Out[148]:
\[\tag{${\it \%o}_{145}$}\mathbf{done}\]
In [72]:
C:n/(n-k)*binomial(n-k,k);
Out[72]:
\[\tag{${\it \%o}_{71}$}\frac{n\,{{n-k}\choose{k}}}{n-k}\]
In [75]:
subst(n-1,n,C)+subst(k-1,k,subst(n-2,n,C));
Out[75]:
\[\tag{${\it \%o}_{74}$}\frac{\left(n-1\right)\,{{n-k-1}\choose{k}}}{n-k-1}+\frac{\left(n-2\right)\,{{n-k-1}\choose{k-1}}}{n-k-1}\]
In [79]:
rec:C-(subst(n-1,n,C)+subst(k-1,k,subst(n-2,n,C)));
Out[79]:
\[\tag{${\it \%o}_{78}$}\frac{n\,{{n-k}\choose{k}}}{n-k}-\frac{\left(n-1\right)\,{{n-k-1}\choose{k}}}{n-k-1}-\frac{\left(n-2\right)\,{{n-k-1}\choose{k-1}}}{n-k-1}\]
In [80]:
makefact(rec),factorial_expand:true,factor;
Out[80]:
\[\tag{${\it \%o}_{79}$}0\]