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A118594
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Palindromes in base 3 (written in base 3).
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13
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0, 1, 2, 11, 22, 101, 111, 121, 202, 212, 222, 1001, 1111, 1221, 2002, 2112, 2222, 10001, 10101, 10201, 11011, 11111, 11211, 12021, 12121, 12221, 20002, 20102, 20202, 21012, 21112, 21212, 22022, 22122, 22222, 100001, 101101, 102201, 110011, 111111, 112211, 120021
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,3
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COMMENTS
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The number of n-digit terms is given by A225367. - M. F. Hasler, May 05 2013 [Moved here on May 08 2013]
Digit-wise application of A000578 (and also superposition of a(n) with its horizontal OR vertical reflection) yields A006072. - M. F. Hasler, May 08 2013
Equivalently, palindromes k (written in base 10) such that 4*k is a palindrome. - Bruno Berselli, Sep 12 2018
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LINKS
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Eric Weisstein's World of Mathematics, Ternary
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MATHEMATICA
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(* get NextPalindrome from A029965 *) Select[NestList[NextPalindrome, 0, 1110], Max@IntegerDigits@# < 3 &] (* Robert G. Wilson v, May 09 2006 *)
Select[FromDigits/@Tuples[{0, 1, 2}, 8], IntegerDigits[#]==Reverse[ IntegerDigits[ #]]&] (* Harvey P. Dale, Apr 20 2015 *)
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PROG
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(PARI) {for(l=1, 5, u=vector((l+1)\2, i, 10^(i-1)+(2*i-1<l)*10^(l-i))~; forvec(v=vector((l+1)\2, i, [l>1&&i==1, 2]), print1(v*u", ")))} \\ The n-th term could be produced by using (partial sums of) A225367 to skip all shorter terms, and then skipping the adequate number of vectors v until n is reached. - M. F. Hasler, May 08 2013
(Sage)
[int(n.str(base=3)) for n in (0..757) if Word(n.digits(3)).is_palindrome()] # Peter Luschny, Sep 13 2018
(Python)
from itertools import count, islice, product
def agen(): # generator of terms
yield from [0, 1, 2]
for d in count(2):
for start in "12":
for rest in product("012", repeat=d//2-1):
left = start + "".join(rest)
for mid in [[""], ["0", "1", "2"]][d%2]:
yield int(left + mid + left[::-1])
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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