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Conjectured minimum number of ways to represent a number >= A262526(i) by sums of three base-10 palindromes.
5

%I #16 Oct 17 2015 12:07:54

%S 1,2,3,4,5,6,8,11,13,14,15,16,17,18,21,27,29,35,41,45,48,51,54,55,56,

%T 61,63,64,65,74,75,79,80,82,83,85,86,87,91,111,112,113,114,115,116,

%U 118,120,121,124,127,133,134,138,140,141,142,145,147,150,153,165,169,171,174,175,177,179,180,183,184,185

%N Conjectured minimum number of ways to represent a number >= A262526(i) by sums of three base-10 palindromes.

%C The sequence is obtained by sorting the counts A261132 into increasing order together with their positions of occurrence. If a new record in the sorted A261132 is found, the index of its latest occurrence in A261132 becomes the next term in A262526 and the corresponding value of A261132 becomes a(n).

%C 7 is not in the sequence, because the latest occurrence of 7 is at A261132(64), whereas the latest occurrence of 6 had already set the record to A262526(6) = 120.

%C a(7) = 8 corresponds to the latest occurrence of 8 at A261132(142), thus A262526(7) = 142.

%H Hugo Pfoertner, <a href="/A262527/b262527.txt">Table of n, a(n) for n = 1..1072</a>

%H Hugo Pfoertner, <a href="/A262527/a262527.pdf">Least number of representations by sums of three palindromes.</a>

%e a(5) = 5 because A261132(k) > 5 for all k > A262526(5) = 98.

%Y Cf. A002113, A261132, A262524, A262525, A262526.

%Y See A261422, A262544, A262545 for another approach.

%K nonn,base

%O 1,2

%A _Hugo Pfoertner_, Sep 25 2015