[seqfan] Re: A158415

[Vladimir Reshetnikov]

On Sun, Mar 22, 2009 at 1:00 PM, Robert Gerbicz <robert.gerbicz at gmail.com>wrote:

> 2009/3/22 Vladimir Reshetnikov <v.reshetnikov at gmail.com>
>
> > Dear all,
> >
> > Can anybody suggest an algorithm for calculating the terms of A158415?
> >
> > Thanks
> > Vladimir
> >
> > _______________________________________________
> >
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
> Hi!
>
> There are two possibilities for N symbols: u=sqrt(x) where x is using N-1
> symbols, or u=x+y where x is using 1<=j<N and y is using 1<=N-1-j<N
> symbols.
> Extension of the sequence (I've already submitted):
> 1,1,2,3,5,8,13,20,33,54,91,154,264,455,791,1379,2424,4277,7588,13513,
> 24162,43336,77978,140683,254487,461409,838433,1526536
>
> A quick PARI-Gp code for it:
> allocatemem(2*10^8);\
> a=L=vector(28);eps=10^(-20);a[1]=[1];L[1]=1;print1(1",");\
> for(i=2,28,b=vector(L[i-1]+sum(j=1,(i-1)\2,L[j]*L[i-j-1]));\
> up=0;for(j=1,L[i-1],up++;b[up]=sqrt(a[i-1][j]));\
> for(j=1,(i-1)\2,for(k=1,L[j],for(l=1,L[i-1-j],\
> up++;b[up]=a[j][k]+a[i-1-j][l])));\
> c=vector(up,j,b[j]);c=vecsort(c);s=0;\
> for(j=1,up,if((j==1)||(c[j]>c[j-1]+eps),s++));\
> a[i]=vector(s);s=0;\
> for(j=1,up,if((j==1)||(c[j]>c[j-1]+eps),s++;a[i][s]=c[j]));\
> L[i]=s;print1(L[i]","))
>
> _______________________________________________
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>

How did you prove that eps=10^(-20) is sufficient? In my view, the main
problem here is how to rigorously determine equality of two expressions.

Thanks
Vladimir