[seqfan] Re: numbers whose cube is a palindrome

[Matevž Markovič]

> On Fri, May 20, 2011 at 5:41 PM, Vladimir Shevelev <shevelev at bgu.ac.il> wrote:
> > Let a(n) be (k+1)-digit number. Let us show that for sufficiently large k, a(n) has (0,1)-digits only.
> > Let a(n)=A002780(n)=a_0*10^k+...+a_k.  Then a(n)^3=Sum{i_0+...+i_k=3}6/(i_0!*...*i_k!)*a_0^i_0*...*a_k^i_k*10^(k*i_0+(k-1)*i_1+...+1*i_(k-1)).
> > Coefficient of 10^r equals to Sum{i_0+...+i_k=3; k*i_0+(k-1)*i_1+...+1*i_(k-1)=r}6/(i_0!*...*i_k!)*a_0^i_0*...*a_k^i_k.
> > Coefficient of 10^(3*k-r) equals to
> > Sum{i_0+...+i_k=3; k*i_0+ (k-1)*i_1+...+1*i_(k-1)=3*k-r}6/(i_0!*...*i_k!)*a_0^i_0*...*a_k^i_k.
> > If i_j=i_m=i_s=1 and other i_h=0, then, by the condition of summing of the first sum, we have
> > (k-j)*i_j+(k-m)*i_m+(k-s)*i_s=r, or (k-j)+(k-m)+(k-s)=r, while, by the condition of summing of the second sum, we have (k-j)+(k-m)+(k-s)=3*k-r, or j+m+s=r. Analogously, if i_j=1, i_m=2,
> > then (k-j)+2*(k-m)=r in the first sum, while j+2*m=r in the second sum. Thus to every i_j of the second sum corresponds k- i_j  in the first one. Note that the condition
> > k*i_0+ (k-1)*i_1+...+1*i_(k-1)=3*k-r is equivalent to i_1+2*i_2+...+(k-1)*i_(k-1)=r and the condition k*i_0+ (k-1)*i_1+...+1*i_(k-1)=r is equivalent to i_1+2*i_2+...+(k-1)*i_(k-1)=3*k-r.
> >  Therefore, since a(n)^3 is a palindrome, then we should have
> > Sum{i_0+...+i_k=3; 1*i_1+...+(k-1)*i_(k-1)=r, i_j>0}1/(i_0!*...*i_k!)*a_0^i_0*...*a_k^i_k=Sum{i_0+...+i_k=3; 1*i_1+...+(k-1)*i_(k-1)=3*k-r, i_j>0}1/(k-i_0)!*...*(k-i_k!)!*a_0^(k-i_0)*...*a_k^(k-i_k).
>
> You did not take into account possible carries.
> The coefficients of 10^r and 10^(3*k-r) in the multinomial expansion
> are *not* necessasry the decimal digits of the resulting number.
>
> Regards,
> Max

Dear Mr. Vladimir Shevelev, would you be so kind to try to prove (or
disprove) also the A186080 conjecture, please? It is very close to
this one.

Thank you

--
Matevž Markovič