# [seqfan] Re: numbers whose cube is a palindrome

Wed May 25 08:46:53 CEST 2011

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Dear  Matevž,

Below I give a proof that A069748 (Numbers n such that n and n^3 are both palindromes)
contains no terms with digit sum 5  The proof for this case is shorter than for A002780 but based on the same idea.

Consider compositions of 3 with the positive parts: 3, 2+1, 1+2, 1+1+1.  Let delta (i,j) be Kronecker's symbol. From the multinomial formula it follows that the coefficient of 10^r in (a_0+a_1*10+...+a_k*10^k)^3 is Sum{ t=0,...,k, a_t=1}delta(r,3*t)+3*Sum{j=0,...,k}Sum{t=0,...,k, t differs from j, a_j=a_t=1}delta(r,2*t+j)+6*Sum{1<=j<t<s<=k, a_j=a_t=a_s=1}delta(r,j+t+s). These sums correspond to the compositions of 3. Put (a_0+a_1*10+...+a_k*10^k)^3 = b_0+b_1*10+...+b_(3*k)*10^(3*k).
Palindrome rootnumber with (0,1)-digits
Let N be palindrome of the form 1+10^(m+1)+10^(m+r+2)+10^(m+2*r+3)+10^(2*m+2*r+4),
such that between ajacent 1's we have m,r,r,m 0's.
According to the first sum of the formula for coefficient of 10^r, we notice that a_0=1 (t=0) contributes 1 to b_0, analogously, a_(m+1)=1, a_(m+r+2)=1, a_(m+2*r+3)=1 and a_(2*m+2*r+4)=1 cotribute 1 to b_(3*m+3), b_(3*m+3*r+6)=1, b_(3*m+6*r+9)=1 and b_(6*m+6*r+12) correspondingly. So I write coefficient b as I receive the first contribution to it.  For example, since in the second sum we have the indices 2*(m+1)+0 and 2*0+(m+1), then I write b_(2*m+2)=3 and   b_(m+1)=3. So, at the end I find
b_0=1, b_(3*m+3)=1, b_(3*m+3*r+6)=13, b_(3*m+6*r+9)=1, b_(6*m+6*r+12)=1, b_(2*m+2)=3,  b_(m+1)=3, b_(m+r+2)=3, b_(2*m+2*r+4)=12, b_(2*m+2*r+3)=3, b_(4*m+4*r+6)=3,
b_(3*m+r+4)=3, b_(3*m+2*r+5)=12, b_(3*m+4*r+7)=12, b_(4*m+4*r+8)=12,
b_(4*m+2*r+6)=3, b_(5*m+4*r+9)=3, b_(3*m+5*r+8)=3, b_(5*m+5*r+10)=3,
b_(6*m+6*r+10)=3, b_(2*m+r+3)=6, b_(2*m+3*r+5)=6, b_(4*m+3*r+7)=6, b_(4*m+5*r+9)=6.
Till now I wrire these values without any ordering over increasing of indices, since such an order is possible only after accepting a ratio between m and r.
Note that a necessary condition for the palindromness of b_0+b_1*10+...+b_(3*k)*10^(3*k) is the following : ignoring zeros, we should have the same equidistant from the ends values of the positive b.
Note that from the left the first 4 positive values of b should be 1,3,3,3. Indeed , in any ratio between m and r, we have 0<m+1<2*m+2?m+r+2. But b_0=1, b_(m+1)=3 and b_(2*m+2)=b_(m+r+2)=3. Since all other indices of positive b are absolutely more than max (2*m+2,m+r+2), then our statement is true.
From the right we have the following absolutely maximal values of indices of positive b: 6*m+6*r+12>6*m+6*r+10>5*m+5*r+10 such that, since b_(6*m+6*r+12)=1, b_(6*m+6*r+10)=3,  b_(5*m+5*r+10)=3, then the first 3 digits of positive  values of b are 1,3,3.  Now we should
obtain 3 as the following digit. The maximal following index u when b_u=3 is u=5*m+4*r+9. Since b_(3*m+6*r+9)=1, then u could be the following after the first 3 most large digits from the right, if the condition 5*m+4*r+9>3*m+6*r+9 holds, or m>r.
Now, only for a better orientation, let us order positive b_j over the increasing of the coefficient of m in its indices, and, if such coefficients are the same, then over the increasing of the coefficient of r.
b_0=1, b_(m+1)=3, b_(m+r+2)=3, b_(2*m+2)=3, b_(2*m+r+3)=6, b_(2*m+2*r+3)=3, b_(2*m+2*r+4)=12, b_(2*m+3*r+5)=6, b_(3*m+3)=1,  b_(3*m+r+4)=3,
b_(3*m+2*r+5)=12,  b_(3*m+3*r+6)=13, b_(3*m+4*r+7)=12, b_(3*m+5*r+8)=3, b_(3*m+6*r+9)=1, b_(4*m+2*r+6)=3,   b_(4*m+3*r+7)=6, b_(4*m+4*r+6)=3, b_(4*m+4*r+8)=12, b_(4*m+5*r+9)=6,  b_(5*m+4*r+9)=3,  b_(5*m+5*r+10)=3, b_(6*m+6*r+10)=3, b_(6*m+6*r+12)=1.

Now from the left we obtain the following digit 6. Indeed, evidently, 2*m+r+3<2*m+2*r+3<
2*m+2*r+4<2*m+3*r+5 and, although not necessary 2*m+3*r+5<3*m+3, however, 2*m+r+3<3*m+3. Thus b_(2*m+r+3)=6 is the following digit from the left. From the right we have digit 6 as well. Further, the following maximal index is 4*m+4*r+8. Since b_(4*m+4*r+8)=12, then we should do a carry. Thus we have new values b_(4*m+4*r+8)=2 and b_(4*m+4*r+9)=1. Therefore, from the right we have digits 1,3,3,3,6,1,...The nearest 1 from the left appears after carry
in b_(2*m+2*r+4)=12, such that we have new values b_(2*m+2*r+4)=2 and b_(2*m+2*r+5)=1.
Since 2*m+2*r+3<2*m+2*r+5, then the following digit from the left is 3, i.e. , from the left we have digits 1,3,3,3,6,3,... This means that N is not a palindrome.
In cases, when a palindrome rootnumber has non-zero digit 2,1,2 or 1,3,1, a search  is almost trivial.
Palindrome rootnumber with (0,1,2) or (0,1,3)-digits
Let, for m>=2, rootnumber is N=2*10^(2*m)+10^m+2. Using formula
(a+b+c)^3=a^3+b^3+c^3+3*(a*b^2+a*c^2+a^2*b+a^2*c+b*c^2+b^2*c)+6*a*b*c,
we have
N^3=8*10^(6*m)+10^(3*m)+8+6*10^(4*m)+24*10^(2*m)+12*10^(5*m)+24*10^(4*m)+12*10^m+6*10^(2*m)+24*10^(3*m)=8+2*10^m+10^(m+1)+3*10^(2*m+1)+5*10^(3*m)+2*10^(3*m+1)+3*10^(4*m+1)+2*10^(5*m)+10^(5*m+1)+8*10^(6*m). Since positive digits from the left 8,2,..., while from the right 8,1,..., then it is not a palindrome. Case m=1 is trivial.
Finally, let, for m>=2, N= 10^(2*m)+3*10^m+1. Then quite analogously we see that  positive digits from the left 1,1,..., while from the right 1,9,..., then it is not a palindrome. Case m=1 is trivial.

Regards,

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Monday, May 23, 2011 1:17
Subject: [seqfan] Re: numbers whose cube is a palindrome
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Using table by Noe-De Geest, I noticed, that all numbers
> {a(n)=A002780(n);  11<=a(n)<=10^17+10^16+11}, except
> of 2201, allow a partition into 3 disjoint classes of terms of
> the following forms: 1) 10^k+1; 2) 10^(2*k)+10^k+1; 3)
> (10^u+1)*(10^v+1).
> The question arises, whether exists a term a(n)>10^17+10^16+11
> which is in none of these classes?
> If such terms do not exist, then we conclude that the sum of
> digits of a(n) not exceeds 4 (more exactly, it is i+1,where
> i  is the number of class).
> This problem has not an absurd character. E.g., I have a long
> proof that there does not exist a(n) (other than 2201) with the
> sum of digits 5.
>
> Regards,
>
>
----- Original Message -----
From: Matevž Markovič <matevz.markovic.v at gmail.com>
Date: Monday, May 23, 2011 18:01
Subject: about the A002780 sequence on oeis.org and discussion on the seqfan mailing list
To: shevelev at bgu.ac.il

> Dear Mr. Shevelev,
>
> I saw your replies on the seqfan mailing list regarding the
> palindromic cubes (A002780 sequence on the oeis). Would you mind
> if I
> asked you to send me the proof that there does not exist a(n) (other
> than 2201) with the sum of its digits being 5?
> I am just starting to learn how one can proove such conjectures and
> there is still a long way to go for me.
>
> Thank you!
>
> Matevž Markovič