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A056810
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Numbers whose fourth power is a palindrome.
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3
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0, 1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001, 100000000001
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OFFSET
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1,3
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COMMENTS
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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 last two 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.
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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LINKS
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G. J. Simmons, Palindromic powers, J. Rec. Math., 3 (No. 2, 1970), 93-98. [Annotated scanned copy]
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MATHEMATICA
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palQ[n_] := Block[{}, Reverse[idn = IntegerDigits@ n] == idn]; k = 0; lst = {}; While[k < 1000000002, If[ palQ[k^4], AppendTo[lst, k]]; k++]; lst (* Robert G. Wilson v, Oct 23 2015 *)
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PROG
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(Python)
def ispal(n): s = str(n); return s == s[::-1]
def afind(limit):
for k in range(limit+1):
if ispal(k**4): print(k, end=", ")
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CROSSREFS
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KEYWORD
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nonn,base,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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