|
|
A351491
|
|
Irregular triangle read by rows: T(n,k) is the minimum number of alphabetic symbols in a regular expression for the k lexicographically first palindromes of length 2*n over a ternary alphabet, n >= 0, 1 <= k <= 3^n.
|
|
0
|
|
|
0, 2, 4, 6, 4, 6, 8, 12, 14, 16, 20, 22, 24, 6, 8, 10, 14, 16, 18, 22, 24, 26, 32, 34, 36, 40, 42, 44, 48, 50, 52, 58, 60, 62, 66, 68, 70, 74, 76, 78, 8, 10, 12, 16, 18, 20, 24, 26, 28, 34, 36, 38, 42, 44, 46, 50, 52, 54, 60, 62, 64, 68, 70, 72, 76, 78, 80, 88
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
Analogous to A351489 (which is the corresponding sequence for palindromes over binary alphabet).
|
|
REFERENCES
|
Hermann Gruber and Markus Holzer, Optimal Regular Expressions for Palindromes of Given Length. Extended journal version, in preparation, 2022.
|
|
LINKS
|
|
|
FORMULA
|
Let SumOfDigitsInBase(m,b) denote the digit sum of nonnegative integer m in base b. Then the general formula for alphabet size q reads as
T(n,k) = 2*n + (2*q*(k-1))/(q-1) - (2*SumOfDigitsInBase(k-1,q))/(q-1). [Gruber and Holzer 2022 theorem 27]
|
|
EXAMPLE
|
Triangle T(n,k) begins:
k=1 2 3 4 5 6 ...
n=0: 0,
n=1: 2, 4, 6;
n=2: 4, 6, 8, 12, 14, 16, 20, 22, 24;
n=3: 6, 8, 10, 14, 16, 20, 22, 24, 26, 32, 34, 36, 40, 42, 44, 48, 50, 52, 58, 60, 62, 66, 68, 70, 74, 76, 78;
...
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy,tabf
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|