[seqfan] Re: numbers whose cube is a palindrome

[Vladimir Shevelev]

Dear  Matevž,
 
Unfortunately, my proof from Fri, May 20, 2011 at 5:41 PM is not true but the reason is another than indicated Max. The reason is a wrong change of variables. Therefore, I ask you and others to ignore this proof.  On the other hand, one can  prove  that sequence A056810 does not contain any term with sum of digits 3 (it is much more easy than to prove that sequence A002780 does not contain any term with sum of digits 5 (cf. my to-day message).

Best regards, 
Vladimir


----- Original Message -----
From: Matevž Markovič <matevz.markovic.v at gmail.com>
Date: Monday, May 23, 2011 1:16
Subject: [seqfan] Re: numbers whose cube is a palindrome
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> > 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
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/
> 

 Shevelev Vladimir‎