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Suppose a number is of the form a=10...01 then a^2=10..020..01, so a^2 is always a palindrome. a^3=10..030..030..01, so a^3 is always a palindrome. Similarly we also have a^4=10..040..060..040..01, so a^4 is always a palindrome. However, a^5 is in general not a palindrome, for example 101^5=10510100501. - Dmitry Kamenetsky, Apr 17 2009
The sequence contains no term with digit sum 3. - Vladimir Shevelev, May 23 2011. Proof: There are four possibilities for n:
1) 1+10^k+10^m, 0<k<m, 2) 1+2*10^r, r>0, 3) 2+10^s, s>0, 4) 3*10^t, t>=0.
In the two last cases n^4 is trivially not a palindrome.
For r>=2, in the second case we have n^4 = (1 + 2*10^r)^4 = 1 + 8*10^r + 4*10^(2*r) + 2*10^(2*r + 1) + 2*10^(3*r) + 3*10^(3*r + 1) + 6*10^(4*r) + 10^(4*r + 1)
which cannot be a palindrome.
If r=1, we have 1+8*10+...9*10^4+10^5 which also is not a palindrome.
The proof for the first case is similar. QED Vladimir Shevelev, Oct 24 2015
Does every term have the structure 100...0001? Referring to the Simmons (1972) paper, we can also ask, if n is a number whose cube is a palindrome in base 4, must the base-4 expansion of n have the form 100...0001? - N. J. A. Sloane, Oct 22 2015
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