## Re: puzzle for a counterexample

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• ... Here are 5 such counterexamples, in the format [n, a, t]: [921244609, 52198420, 515951943] [2544329101, 263357820, 100205789] [11273067541, 703848073,
Message 1 of 66 , Nov 1, 2012
"bhelmes_1" <bhelmes@...> wrote:

> If you find one counterexample of the form n=1 mod 4
> i would be glad to know it.

Here are 5 such counterexamples, in the format [n, a, t]:

[921244609, 52198420, 515951943]
[2544329101, 263357820, 100205789]
[11273067541, 703848073, 6861362057]
[27536216921, 12158783674, 18501786352]
[30048858209, 9925760819, 22855838720]

David
• ... Here are some scores out of 5: {A(k,x)=sum(j=0,k/2,(-1)^j*binomial(k-j,j)*x^(k-2*j));} {B(k,x)=sum(j=0,(k-1)/2,(-1)^j*binomial(k-j-1,j)*x^(k-2*j-1));}
Message 66 of 66 , Nov 22, 2012
paulunderwooduk" <paulunderwood@...> wrote:

> At least one of the evaluations of x at -1,1,0,-2 or 2
> should be -1,1,0,-2, or 2

Here are some scores out of 5:

{A(k,x)=sum(j=0,k/2,(-1)^j*binomial(k-j,j)*x^(k-2*j));}
{B(k,x)=sum(j=0,(k-1)/2,(-1)^j*binomial(k-j-1,j)*x^(k-2*j-1));}
{L=[-1,1,0,-2,2];S=Set(L);for(k=2,40,f=factor(A(k,x)-B(k,x))[,1];
g=f[#f];c=0;for(j=1,#L,if(setsearch(S,subst(g,x,L[j])),c++));
print1(c","));}

4,4,3,4,4,4,4,4,4,4,4,3,4,4,4,5,4,4,4,4,5,4,4,4,4,5,4,4,4,5,5,4,4,4,4,4,5,4,3,

David
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