[seqfan] Re: numbers whose cube is a palindrome

Matevž Markovič matevz.markovic.v at gmail.com
Fri May 20 23:17:19 CEST 2011

> 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

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