Index Nr
L base 10
L base 2
L base 3
L base 4
L base 5
L base 6
L base 7
L base 8
L base 9
L base 11
L base 12
L base 13
L base 14
L base 15
L base 16
L base 17
L base 18
L base 19
L base 20
L base 21
L base 22
L base 23
L base 24
L base 25
L base 26
L base 27
L base 28
L base 29
L base 30
L base 31
L base 32
L base 33
L base 34
L base 35
L base 36
L base 60
Palindromes in bases 2 and 10. The main source for this table comes from the following weblink. Binary/Decimal Palindromes by Charlton Harrison (email) from Austin, Texas. See also Sloane's sequences A007632 and A046472.
" I just finished writing a distributed client/server program for finding these numbers, and I currentlyhave it running on 4 different machines at the same time and I'm finding them A LOT faster. That's howI was able to come up with those new ones. I'd say there are more to come in the near future, too."
[ April 11, 2003 ] Dw (email) wrote me the following :
"Using a backtracking solver, I have found larger numbers. The first of these, which is the next after the one mentioned above, is 50824513851188115831542805 (86 bit, 3*5*11*11*17*17*83*11974799*97488286319). There are no other 86 bit double palindromes. I also have found two 89 bit double palindromes, but I am not sure if they are the next ones. There may be some lower ones in 89 bit (haven't searched that space completely yet) or in 87 bit that I haven't found. These numbers are 532079161251434152161970235 (5*29*85839676103*42748430836381) and 552963956270141072659369255 (5*7*7*71*441607*71984077228507867) As can be seen by their factor representation, they are all composite."
I also have found two 89 bit double palindromes, but I am not sure if they are the next ones. There may be some lower ones in 89 bit (haven't searched that space completely yet) or in 87 bit that I haven't found. These numbers are 532079161251434152161970235 (5*29*85839676103*42748430836381) and 552963956270141072659369255 (5*7*7*71*441607*71984077228507867) As can be seen by their factor representation, they are all composite."
50824513851188115831542805 {26} 10101000001010100000100110101010101001100000000110010101010101100100000101010000010101 {86} 532079161251434152161970235 {27} 11011100000100000001001010010000011001110101010101110011000001001010010000000100000111011 {89} 552963956270141072659369255 {27} 11100100101100110101011000010001001111100100100100111110010001000011010101100110100100111 {89}
532079161251434152161970235 {27} 11011100000100000001001010010000011001110101010101110011000001001010010000000100000111011 {89}
552963956270141072659369255 {27} 11100100101100110101011000010001001111100100100100111110010001000011010101100110100100111 {89}
[ April 13, 2003 ] Dw (email) found four new ones including the 6th double palindromic prime :
"They are, in opposite sorted order: 138758321383797383123857831 (87 bit, composite, the only 87 bit double palindrome) 390714505091666190505417093 (89 bit, prime) 351095331428353824133590153 (89 bit, composite) 795280629691202196926082597 (90 bit, composite, not sure if this is the next one)."
When asking Dw for an explanation or description in a few words what he meant by 'Using a backtracking solver' he replied :
"A backtracking solver is one that solves a problem made up of smaller problems by trying every one except if it knows its earlier guesses make all later ones depending on them impossible. For instance, if you're in a maze and know that all corridors with green walls eventually lead to a dead end, you can turn around as soon as you find such a wall (backtrack) if you're trying to find the exit. The smaller problem is avoiding a dead end, and the larger one is finding the exit. If you want the details: My general strategy goes that to find a double palindromic number, the solution to its palindromic decimal representation minus its binary representation must be zero (since they are equal). Furthermore, you can write a decimal palindrome like 101 * a + 10 * b, where a and b are digits. The same can be done for binary, and you end up with a giant (linear diophantine) equation of positive decimal and negative binary factors. The factors can then be solved, one at a time, using the extended euclidean algorithm. These are the smaller problems. I also have a table of maximum and minimum values for each step. That way, if it's impossible for the binary factors left to subtract enough from the decimal factors to get zero (or the other way around), the solver backtracks."
For instance, if you're in a maze and know that all corridors with green walls eventually lead to a dead end, you can turn around as soon as you find such a wall (backtrack) if you're trying to find the exit.
The smaller problem is avoiding a dead end, and the larger one is finding the exit.
If you want the details:
My general strategy goes that to find a double palindromic number, the solution to its palindromic decimal representation minus its binary representation must be zero (since they are equal).
Furthermore, you can write a decimal palindrome like 101 * a + 10 * b, where a and b are digits. The same can be done for binary, and you end up with a giant (linear diophantine) equation of positive decimal and negative binary factors.
The factors can then be solved, one at a time, using the extended euclidean algorithm. These are the smaller problems.
I also have a table of maximum and minimum values for each step. That way, if it's impossible for the binary factors left to subtract enough from the decimal factors to get zero (or the other way around), the solver backtracks."
138758321383797383123857831 {27} 111001011000111001101111010110010111011100111001110111010011010111101100111000110100111 {87} 351095331428353824133590153 {27} 10010001001101011010101000000110111100000100000100000111101100000010101011010110010001001 {89} 390714505091666190505417093 {27} 10100001100110001000000111100001100011000111011100011000110000111100000010001100110000101 {89} 795280629691202196926082597 {27} 101001000111010111001110010101001010000101000000101000010100101010011100111010111000100101 {90}
351095331428353824133590153 {27} 10010001001101011010101000000110111100000100000100000111101100000010101011010110010001001 {89}
390714505091666190505417093 {27} 10100001100110001000000111100001100011000111011100011000110000111100000010001100110000101 {89}
795280629691202196926082597 {27} 101001000111010111001110010101001010000101000000101000010100101010011100111010111000100101 {90}
[ May 21, 2003 ] Dw (email) found a new Binary/Decimal Palindrome :
"I have been trying to find a polynomial time (growth of time needed is a polynomial of number of bits) algorithm for finding double palindromes of base 2 and 10. I haven't succeeded yet (my backtracking solver being exponential), but I have found some interesting things. For one, to find double palindromes in base 2 and 8 is very simple. Each base 8 digit maps to three bits. Therefore, every double palindrome must consist of digits who themselves are double palindromes. For instance, 757 as well as 575 is double palindromic. (These values are 495 and 381 in decimal respectively). 5 maps to 101 in binary, and 7 to 111 in binary. I have found 1609061098335005338901609061 (91 bits composite), and this is the only 91 bit one. No 92 bit double palindrome exists, and 93 bits seems to require several days of searching; therefore I'm trying to find a polynomial time algorithm as mentioned above. Another approach could be to create a networked version (to do the search on multiple computers), but I haven't done that yet."
For one, to find double palindromes in base 2 and 8 is very simple. Each base 8 digit maps to three bits. Therefore, every double palindrome must consist of digits who themselves are double palindromes.
For instance, 757 as well as 575 is double palindromic. (These values are 495 and 381 in decimal respectively). 5 maps to 101 in binary, and 7 to 111 in binary.
I have found 1609061098335005338901609061 (91 bits composite), and this is the only 91 bit one. No 92 bit double palindrome exists, and 93 bits seems to require several days of searching; therefore I'm trying to find a polynomial time algorithm as mentioned above.
Another approach could be to create a networked version (to do the search on multiple computers), but I haven't done that yet."
1609061098335005338901609061 {28} 1010011001011111011111100001110110010000000100010000000100110111000011111101111101001100101 {91}
[ June 12, 2003 ] Dw (email) found new Binary/Decimal Palindromes :
"I rewrote my program to use another strategy at finding the numbers, and this let me search somewhat faster. As a result, I have found binary/decimal palindromes up to 102 bits -- broke the 100 bit barrier so to speak. These are: None at 92 or 93 bits. 17869806142184248124160896871 (94 bits) 19756291244127372144219265791 (94 bits) 30000258151173237115185200003 (95 bits) 30658464822225352222846485603 (95 bits) 56532345659072227095654323565 (96 bits) None at 97 or 98 bits. 378059787464677776464787950873 (99 bits) 1115792035060833380605302975111 (100 bits) None at 101 bits. 3390741646331381831336461470933 (102 bits) There may be higher ones at 102 bits; I haven't completed the search there."
These are:
None at 92 or 93 bits. 17869806142184248124160896871 (94 bits) 19756291244127372144219265791 (94 bits) 30000258151173237115185200003 (95 bits) 30658464822225352222846485603 (95 bits) 56532345659072227095654323565 (96 bits) None at 97 or 98 bits. 378059787464677776464787950873 (99 bits) 1115792035060833380605302975111 (100 bits) None at 101 bits. 3390741646331381831336461470933 (102 bits)
There may be higher ones at 102 bits; I haven't completed the search there."
17869806142184248124160896871 {29} 1110011011110110001110101001000001000000000011001100000000001000001001010111000110111101100111 {94}
19756291244127372144219265791 {29} 1111111101011000000101011001100101101101100001111000011011011010011001101010000001101011111111 {94}
30000258151173237115185200003 {29} 11000001110111110100001110001010010000110100011011000101100001001010001110000101111101110000011 {95}
30658464822225352222846485603 {29} 11000110001000000010110011101000100010000101111111110100001000100010111001101000000010001100011 {95}
56532345659072227095654323565 {29} 101101101010101001110101110101101111011010100000000001010110111101101011101011100101010101101101 {96}
378059787464677776464787950873 {30} 100110001011001001110111010000000001110111000111010111000111011100000000010111011100100110100011001 {99}
1115792035060833380605302975111 {31} 1110000101010101000110001001111111101011111110110110110111111101011111111001000110001010101010000111 {100}
3390741646331381831336461470933 {31} 101010110011000001001111000000100001000111101001011110100101111000100001000000111100100000110011010101 {102}
[ June 17, 2003 ] Dw (email) adds :
" There are no numbers for 103 bits."
Binary/Decimal Palindromes by Charlton Harrison (email) : the longest list in existence ?
[ September 30, 2015 ] Three more stunning numbers could be retrieved from Sloane's OEIS database. See index numbers 124, 125 & 126.
[ March 8, 2020 ] Many more numbers could be retrieved from Sloane's OEIS database. See index numbers up to 147. The record dates from the end of 2015.
I sampled the following base X palindromic numbers sequences from the table :
Click here to view some of the author's [P. De Geest] entries to the table. Click here to view some entries to the table about palindromes.
A043001 n-th base 3 palindrome that starts with 1. - Clark Kimberling A043002 n-th base 3 palindrome that starts with 2. - Clark Kimberling A043003 n-th base 4 palindrome that starts with 1. - Clark Kimberling A043004 n-th base 4 palindrome that starts with 2. - Clark Kimberling A043005 n-th base 4 palindrome that starts with 3. - Clark Kimberling A043006 n-th base 5 palindrome that starts with 1. - Clark Kimberling A043007 n-th base 5 palindrome that starts with 2. - Clark Kimberling A043008 n-th base 5 palindrome that starts with 3. - Clark Kimberling A043009 n-th base 5 palindrome that starts with 4. - Clark Kimberling A043010 n-th base 6 palindrome that starts with 1. - Clark Kimberling A043011 n-th base 6 palindrome that starts with 2. - Clark Kimberling A043012 n-th base 6 palindrome that starts with 3. - Clark Kimberling A043013 n-th base 6 palindrome that starts with 4. - Clark Kimberling A043014 n-th base 6 palindrome that starts with 5. - Clark Kimberling A043015 n-th base 7 palindrome that starts with 1. - Clark Kimberling A043016 n-th base 7 palindrome that starts with 2. - Clark Kimberling A043017 n-th base 7 palindrome that starts with 3. - Clark Kimberling A043018 n-th base 7 palindrome that starts with 4. - Clark Kimberling A043019 n-th base 7 palindrome that starts with 5. - Clark Kimberling A043020 n-th base 7 palindrome that starts with 6. - Clark Kimberling A043021 n-th base 8 palindrome that starts with 1. - Clark Kimberling A043022 n-th base 8 palindrome that starts with 2. - Clark Kimberling A043023 n-th base 8 palindrome that starts with 3. - Clark Kimberling A043024 n-th base 8 palindrome that starts with 4. - Clark Kimberling A043025 n-th base 8 palindrome that starts with 5. - Clark Kimberling A043026 n-th base 8 palindrome that starts with 6. - Clark Kimberling A043027 n-th base 8 palindrome that starts with 7. - Clark Kimberling A043028 n-th base 9 palindrome that starts with 1. - Clark Kimberling A043029 n-th base 9 palindrome that starts with 2. - Clark Kimberling A043030 n-th base 9 palindrome that starts with 3. - Clark Kimberling A043031 n-th base 9 palindrome that starts with 4. - Clark Kimberling A043032 n-th base 9 palindrome that starts with 5. - Clark Kimberling A043033 n-th base 9 palindrome that starts with 6. - Clark Kimberling A043034 n-th base 9 palindrome that starts with 7. - Clark Kimberling A043035 n-th base 9 palindrome that starts with 8. - Clark Kimberling A043036 n-th base 10 palindrome that starts with 1. - Clark Kimberling A043037 n-th base 10 palindrome that starts with 2. - Clark Kimberling A043038 n-th base 10 palindrome that starts with 3. - Clark Kimberling A043039 n-th base 10 palindrome that starts with 4. - Clark Kimberling A043040 n-th base 10 palindrome that starts with 5. - Clark Kimberling A043041 n-th base 10 palindrome that starts with 6. - Clark Kimberling A043042 n-th base 10 palindrome that starts with 7. - Clark Kimberling A043043 n-th base 10 palindrome that starts with 8. - Clark Kimberling A043044 n-th base 10 palindrome that starts with 9. - Clark Kimberling
A043045 a(n)=(s(n)+2)/3, where s(n)=n-th base 3 palindrome that starts with 1. - Clark Kimberling A043046 a(n)=(s(n)+1)/3, where s(n)=n-th base 3 palindrome that starts with 2. - Clark Kimberling A043047 a(n)=(s(n)+3)/4, where s(n)=n-th base 4 palindrome that starts with 1. - Clark Kimberling A043048 a(n)=(s(n)+2)/4, where s(n)=n-th base 4 palindrome that starts with 2. - Clark Kimberling A043049 a(n)=(s(n)+1)/4, where s(n)=n-th base 4 palindrome that starts with 3. - Clark Kimberling A043050 a(n)=(s(n)+4)/5, where s(n)=n-th base 5 palindrome that starts with 1. - Clark Kimberling A043051 a(n)=(s(n)+3)/5, where s(n)=n-th base 5 palindrome that starts with 2. - Clark Kimberling A043052 a(n)=(s(n)+2)/5, where s(n)=n-th base 5 palindrome that starts with 3. - Clark Kimberling A043053 a(n)=(s(n)+1)/5, where s(n)=n-th base 5 palindrome that starts with 4. - Clark Kimberling A043054 a(n)=(s(n)+5)/6, where s(n)=n-th base 6 palindrome that starts with 1. - Clark Kimberling A043055 a(n)=(s(n)+4)/6, where s(n)=n-th base 6 palindrome that starts with 2. - Clark Kimberling A043056 a(n)=(s(n)+3)/6, where s(n)=n-th base 6 palindrome that starts with 3. - Clark Kimberling A043057 a(n)=(s(n)+2)/6, where s(n)=n-th base 6 palindrome that starts with 4. - Clark Kimberling A043058 a(n)=(s(n)+1)/6, where s(n)=n-th base 6 palindrome that starts with 5. - Clark Kimberling A043059 a(n)=(s(n)+6)/7, where s(n)=n-th base 7 palindrome that starts with 1. - Clark Kimberling A043060 a(n)=(s(n)+5)/7, where s(n)=n-th base 7 palindrome that starts with 2. - Clark Kimberling A043061 a(n)=(s(n)+4)/7, where s(n)=n-th base 7 palindrome that starts with 3. - Clark Kimberling A043062 a(n)=(s(n)+3)/7, where s(n)=n-th base 7 palindrome that starts with 4. - Clark Kimberling A043063 a(n)=(s(n)+2)/7, where s(n)=n-th base 7 palindrome that starts with 5. - Clark Kimberling A043064 a(n)=(s(n)+1)/7, where s(n)=n-th base 7 palindrome that starts with 6. - Clark Kimberling A043065 a(n)=(s(n)+7)/8, where s(n)=n-th base 8 palindrome that starts with 1. - Clark Kimberling A043066 a(n)=(s(n)+6)/8, where s(n)=n-th base 8 palindrome that starts with 2. - Clark Kimberling A043067 a(n)=(s(n)+5)/8, where s(n)=n-th base 8 palindrome that starts with 3. - Clark Kimberling A043068 a(n)=(s(n)+4)/8, where s(n)=n-th base 8 palindrome that starts with 4. - Clark Kimberling A043069 a(n)=(s(n)+3)/8, where s(n)=n-th base 8 palindrome that starts with 5. - Clark Kimberling A043070 a(n)=(s(n)+2)/8, where s(n)=n-th base 8 palindrome that starts with 6. - Clark Kimberling A043071 a(n)=(s(n)+1)/8, where s(n)=n-th base 8 palindrome that starts with 7. - Clark Kimberling A043072 a(n)=(s(n)+8)/9, where s(n)=n-th base 9 palindrome that starts with 1. - Clark Kimberling A043073 a(n)=(s(n)+7)/9, where s(n)=n-th base 9 palindrome that starts with 2. - Clark Kimberling A043074 a(n)=(s(n)+6)/9, where s(n)=n-th base 9 palindrome that starts with 3. - Clark Kimberling A043075 a(n)=(s(n)+5)/9, where s(n)=n-th base 9 palindrome that starts with 4. - Clark Kimberling A043076 a(n)=(s(n)+4)/9, where s(n)=n-th base 9 palindrome that starts with 5. - Clark Kimberling A043077 a(n)=(s(n)+3)/9, where s(n)=n-th base 9 palindrome that starts with 6. - Clark Kimberling A043078 a(n)=(s(n)+2)/9, where s(n)=n-th base 9 palindrome that starts with 7. - Clark Kimberling A043079 a(n)=(s(n)+1)/9, where s(n)=n-th base 9 palindrome that starts with 8. - Clark Kimberling A043080 a(n)=(s(n)+9)/10, where s(n)=n-th base 10 palindrome that starts with 1. - Clark Kimberling A043081 a(n)=(s(n)+8)/10, where s(n)=n-th base 10 palindrome that starts with 2. - Clark Kimberling A043082 a(n)=(s(n)+7)/10, where s(n)=n-th base 10 palindrome that starts with 3. - Clark Kimberling A043083 a(n)=(s(n)+6)/10, where s(n)=n-th base 10 palindrome that starts with 4. - Clark Kimberling A043084 a(n)=(s(n)+5)/10, where s(n)=n-th base 10 palindrome that starts with 5. - Clark Kimberling A043085 a(n)=(s(n)+4)/10, where s(n)=n-th base 10 palindrome that starts with 6. - Clark Kimberling A043086 a(n)=(s(n)+3)/10, where s(n)=n-th base 10 palindrome that starts with 7. - Clark Kimberling A043087 a(n)=(s(n)+2)/10, where s(n)=n-th base 10 palindrome that starts with 8. - Clark Kimberling A043088 a(n)=(s(n)+1)/10, where s(n)=n-th base 10 palindrome that starts with 9. - Clark Kimberling
A016038 Strictly non-palindromic numbers: n is not palindromic in any base b with 2 <= b <= n-2. - N. J. A. Sloane. A100563 Number of bases less than sqrt(n) in which n is a palindrome. - Gordon Robert Hamilton
Kevin Brown informed me that he has more info about tetrahedral palindromes in other base representations. Link to his article : On General Palindromic Numbers
Alain Bex (email) sent me the first palindromic squares in base 12 - go to topic.
Dw (email) found several binary/decimal palindromes of record lengths - go to topic.
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