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101 gives two palindromes : 101 and 011 = 11 hence a(2) = 101.
a(6) = 112233, The the digit permutation gives six palindromes : 123321, 132231, 213312, 231132, 312213, 321123.
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Max Alekseyev, <a href="/A082276/b082276.txt">Table of n, a(n) for n = 1..130</a>
1, 101, 1001, 10001, 100001, 112233, 10000001, 100122, 10000111, 1111112222, 100000000001, 1000122, 10000000000001, 1000011111, 10011122, 1000000111, 100000000000000001, 10000122, 10000000000000000001, 1111112233, 11111111112222, 10000000000000000000001, 100000000000000000000001, 11223344
a(n) <= 10^n + 1.
Note that Any number C(i+j,j) is the number of palindromes from 2i 1's and 2j 2's, so in particular a(10^n + ) <= 1111112222 and a(15) <= 111111112222. If a number in this sequence has an odd number of digits, the odd digit must be 0 or 1 , with all other digits in pairs; if the number of digits is always an upper boundeven, all must be in pairs. The counts of the nonzero digits must be monotonically decreasing (i.e., at least as many 1's as 2's, etc.) - _Franklin T. Adams-Watters_, Oct 26 2006
a(12) = 1000122, a(18) = 10000122, a(30) = 10012233; probably a(24) = 11223344. Any number C(i+j,j) is the number of palindromes from 2i 1's and 2j 2's, so in particular a(10) <= 1111112222 and a(15) <= 111111112222. If a number in this sequence has an odd number of digits, the odd digit must be 0 or 1, with all other digits in pairs; if the number of digits is even, all must be in pairs. The counts of the nonzero digits must be monotonically decreasing (i.e., at least as many 1's as 2's, etc.) - Franklin T. Adams-Watters, Oct 26 2006
base,more,nonn
More terms Terms to a(9), and a(12), a(18), a(24), a(30) from Franklin T. Adams-Watters, Oct 26 2006
Terms from a(10) onward from Max Alekseyev, Feb 17 2023
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editing
_Amarnath Murthy (amarnath_murthy(AT)yahoo.com), _, Apr 13 2003
a(12) = 1000122, a(18) = 10000122, a(30) = 10012233; probably a(24) = 11223344. Any number C(i+j,j) is the number of palindromes from 2i 1's and 2j 2's, so in particular a(10) <= 1111112222 and a(15) <= 111111112222. If a number in this sequence has an odd number of digits, the odd digit must be 0 or 1, with all other digits in pairs; if the number of digits is even, all must be in pairs. The counts of the nonzero digits must be monotonically decreasing (i.e., at least as many 1's as 2's, etc.) - _Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), _, Oct 26 2006
More terms from _Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), _, Oct 26 2006
1, 101, 1001, 10001, 100001, 112233, 10000001, 100122, 10000111
Note that 10^n + 1 is always an upper bound.
Needs to be checked. Though base dependent, it is a combinatorial problem. Note: 10^n + 1 qualifies to be (12) = 1000122, a(18) = 10000122, a term but (30) = 10012233; probably a(624) = 11223344. Any number C(i+j,j) is not equal to 1000001. Perhaps the number of palindromes from 2i 1's and 2j 2's, so in particular a(7) = 10^7 + 1, ) <= 1111112222 and a(15) <= 111111112222. If a number in this sequence has an odd number of digits, the odd digit must be 0 or a(p) = 10^p + 1 , with all other digits in pairs; if p the number of digits is a primeeven, all must be in pairs. The counts of the nonzero digits must be monotonically decreasing (i.e., at least as many 1's as 2's, etc.) - Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Oct 26 2006
base,more,nonn,new
More terms from Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Oct 26 2006