A262527 Conjectured minimum number of ways to represent a number >= A262526(i) by sums of three base-10 palindromes.

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

Offset

1,2

Comments

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).

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.

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

Links

Hugo Pfoertner, Table of n, a(n) for n = 1..1072

Hugo Pfoertner, Least number of representations by sums of three palindromes.

Example

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

Crossrefs

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

See A261422, A262544, A262545 for another approach.

Sequence in context: A350097 A046889 A026267 * A253773 A124067 A308806

Adjacent sequences: A262524 A262525 A262526 * A262528 A262529 A262530

Keyword

nonn,base

Author

Hugo Pfoertner, Sep 25 2015

Status

approved