[seqfan] Re: numbers whose cube is a palindrome

Vladimir Shevelev shevelev at bgu.ac.il
Fri May 20 15:41:59 CEST 2011

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).  Let at least one digit >1.  Let us take k_0 such that, for k>k_0, the following conditions satisfy
g^(k-1)/(k-1)!<g, g^(k-2)/(k-2)!<g^2/2, g^(k-3)<g^3/6, g=2,3,...,9.
Such k_0 exists since g^n/n! tends to 0, when n increases. For such k we have a_0^(k-i_0)*...*a_k^(k-i_k)/((k-i_0)!*...*(k-i_k)!)<a_0^i_0*...*a_k^i_k/(i_0!*...*i_k!) ,  i.e., every summand of the second sum is less than the correspondent summand of the first one. Besides, since to i_j corresponds k-i_j,  then the numbers of summands in both sums are the same. Thus, a(n)^3 is not a palindrome. 

----- Original Message -----
From: Hans Havermann <pxp at rogers.com>
Date: Wednesday, May 18, 2011 15:33
Subject: [seqfan]  Re: numbers whose cube is a palindrome
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Matevž Markovič:
> > ... did you two search only among those numbers, which are 
> constructed> with digits {0,1,2}?
> For me, yes. Restricting my search to those three digits reduced 
> my  
> search space from 10^20 to 3^20, which is about all that my 
> eight-year- 
> old computer can handle in a relatively short period of time.
> _______________________________________________
> Seqfan Mailing list - http://list.seqfan.eu/

 Shevelev Vladimir‎

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